Bài tập phương pháp tọa độ trong không gian – Diệp Tuân Toán 12

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PHƯƠNG PHÁP TỌA Đ TRONG KHÔNG GIAN
A. THUYT
I. Hệ trục tọa độ trong không gian
Oxyz
1. Định nghĩa: Hệ gồm ba trục
đôi một
vuông góc được gọi là hệ trục tọa độ vuông góc trong
không gian.
Điểm
O
gọi là gốc của hệ tọa độ,
Trục
Ox
là trục hoành,
Oy
là trục tung và
Oz
là trục cao.
2. Véctơ đơn vị trên các trục
lần lượt là
, , ,i j k
Ta có:
. . . 0.i j j k k i
1i j k
3. Tọa độ củac tơ:
Với véctơ
u
trong hệ tọa độ
Oxyz
luôn tồn tại duy nhất bộ
( ; ; )x y z
thỏa:
. . . .u x i y j z k
Khi đó: Véc tơ
u
có tọa độ
( ; ; )x y z
4. Tọa độ của điểm:
Xét điểm
M
thỏa mãn
. . .OM x i y j z k
thì điểm
( ; ; ).M x y z
Ngược lại, điểm
( ; ; )M x y z
thì
. . .OM xi y j z k
.
5. Tính chất – quy tắc – phép toán trên c tơ.
Cho
1 1 1 2 2 2
( ; ; ), ( ; ; )a x y z b x y z
và số thực
k
. Khi đó
Tổng và hiệu của hai vec tơ:
1 2 1 2 1 2
( ; ; )a b x x y y z z
Tích của véc tơ với một số thực
k
1 1 1
( ; ; )ka kx ky kz
Sự cùng phương của hai vec tơ
12
1 1 1
12
2 2 2
12
//
xx
x y z
a b a kb k a b y y
x y z
zz
.
Chú ý: Nếu
2 2 2
0 0, 0x y z
thì
1 1 1
0 0, 0x y z
Độ dài một vec tơ
2 2 2
1 1 1
||a x y z
Tích vô hướng của hai vec tơ
1 2 1 2 1 2
.ab x x y y z z
Nhậnt:
1 2 1 2 1 2
0a b x x y y z z
Góc của hai véctơ
.
cos( , )
| || |
ab
ab
ab
6. Tính chất Phép toán của điểm.
Cho
( ; ; ), ( ; ; ), ( ; ; ), ( ; ; )
A A A B B B C C C D D D
A x y z B x y z C x y z D x y z
. Khi đó:
Véc tơ của hai điểm
( ; ; )
B A B A B A
AB x x y y z z
Khoảng cách hay độ dài của hai điểm
2 2 2
( ) ( ) ( )
B A B A B A
AB AB x x y y z z
z
y
x
k
j
i
O
§BI 1. H TA ĐỘ TRONG KHÔNG GIAN OXYZ
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Trung điểm
I
của đoạn
AB
:
;;
2 2 2
A B A B A B
x x y y z z
I



Trọng tâm
G
của
ABC
:
;;
3 3 3
A B C A B C A B C
x x x y y y z z z
G



Trọng tâm
G
của tứ diện
ABCD
:
;;
4 4 4
A B C D A B C D A B C D
x x x x y y y y z z z z
G



7. Ví dụ minh họa.
Ví d 1. Trong không gian vi
Oxyz
cho ba véc tơ
2 3 5 , 3 4 , 2a i j k b j k c i j
a). Xác định tọa độ các véc tơ
,,abc
,
32x a b
và tính
x
b). Tìm giá tr ca
x
để véc tơ
2 1; ;3 2y x x x
vuông góc với véc tơ
2bc
c). Chng minh rằng các véc
,,abc
không đồng phẳng phân tích véc
3;7; 14u 
qua ba véc tơ
,,abc
.
Li gii.
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II. TÍCH CÓ HƯỚNG CỦA HAI VÉC ỨNG DỤNG.
1. Định nghĩa: Cho
1 1 1
; ; a x y z
2 2 2
; ; b x y z
Tích có hướng của hai véc tơ
1 1 1
; ; a x y z
2 2 2
; ; b x y z
là một véc tơ, kí hiệu
,ab


, được
xác định như sau
1 1 1 1 1 1
1 2 2 1 1 2 2 1 1 2 2 1
2 2 2 2 2 2
, ; ; ; ;
y z z x x y
a b y z y z z x z x x y x y
y z z x x y




2. c tính chất:
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a
cùng phương
,0b a b



,a b a


,a b b


, . .sin( , )a b a b a b


3. c ứng dụng của tích có hướng
Diện tích tam giác:
1
,
2
ABC
S AB AC


.
Thể tích:
Hình hộp:
. ' ' ' '
, . '
ABCD A B C D
V AB AD AA


Tứ diện:
1
,.
6
ABCD
V AB AC AD


.
Điều kiện 3 véctơ đồng phẳng:
,,abc
đồng phẳng
, . 0a b c



, , ,A B C D
đồng phẳng
, . 0AB AC AD



.
4. Ví dụ minh họa.
Ví d 2. Trong không gian
Oxyz
cho bốn điểm
(0;2;0), ( 1;0; 3),AB
(0; 2;0),C
(3;2;1)D
.
a). Chng minh rng bốn điểm
, , ,A B C D
không đồng phng;
b). Tính din tích tam giác
BCD
và đường cao
BH
ca tam giác
BCD
;
c). Tính th tích t din
ABCD
và đường cao ca t din h t
A
;
d). Tìm tọa độ
E
sao cho
ABCE
là hình bình hành;
e). Tính cosin ca góc giữa hai đường thng
AC
BD
;
f). Tìm điểm
M
thuc
Oy
sao cho tam giác
BMC
cân ti
.M
g). Tìm tọa độ trng tâm
G
ca t din
ABCD
chng minh
, , A G A
thng hàng vi
'A
trng tâm tam giác
BCD
.
Li gii.
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B.PHÂN DẠNG VÀ VÍ DỤ MINH HỌA.
DNG 1. XÁC ĐỊNH TA Đ CA ĐIM-TỌA ĐỘ VECTƠ-TÍCH VÔ HƯỚNG
1. Phương pháp:
Dựa vào định nghĩa tọa độ của điểm, tọa độ của véc tơ.
Dựa vào các phép toán véc tơ.
Áp dngc tính cht sau:
Cho các vectơ
1 2 3 1 2 3
( ; ; ), ( ; ; )u u u u v v v v
và s thc
k
tùy ý . Khi đó ta có
a). Hai véc tơ bằng nhau:
11
22
33
uv
u v u v
uv
b).Tng của hai vec tơ
1 1 2 2 3 3
( ; ; )u v u v u v u v
c). Hiu của hai vec tơ
1 1 2 2 3 3
( ; ; )u v u v u v u v
d).Tích của vec tơ với mt s:
1 2 3
( ; ; )ku ku ku ku
2. Bài tp minh ha
Bài tp 1. Trong không gian vi h trc tọa độ
Oxyz
, cho các véc tơ
2 3 , 2a i j k b i k
2 3c j k
a). Xác định tọa độ các véc tơ
,,abc
b). Tìm tọa độ véc tơ
2 3 4u a b c
và tính
u
c). Tìm
x
để véc tơ
(3 1; 2;3 )v x x x
vuông góc vi
b
d). Biu diễn véc tơ
(3;1;7)x
qua ba véc tơ
,,abc
.
Li gii.
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Bài tp 2. Cho hai véc tơ
,ab
tha
0
, 120 , 2, 3a b a b
a). Tính
2ab
b). Tính góc giữa hai véc tơ
a
32x a b
Li gii.
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Bài tp 3. Trong không gian
Oxyz
, cho ba vectơ
(1;0; 2), ( 2;1;3), ( 4;3;5)a b c
a). Tìm to độ vectơ
3. 4. 2a b c
b).Tìm hai s thc
m , n
sao cho
..m a nb c
.
Li gii.
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Bài tp 4. Trong không gian h trc tọa độ
Oxyz
, cho tam giác
ABC
2; 3;1 ,A
1; 1;4B
2;1;6 .C
a). Xác định to độ trng tâm
G
ca tam giác
ABC
;
b). Xác định to độ đim
D
sao cho t giác
ABCD
là hình bình hành và to độ giao điểm hai
đưng chéo ca hình bình hành này;
c). Xác định to độ đim
M
sao cho
2.MA MB
Li gii.
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Bài tp 5. Cho tam giác
ABC
(1;0; 2), ( 1;1;0), ( 2;4; 2).A B C
a). Tìm tọa độ trng tâm
,G
trc tâm
,H
tâm đường tròn ngoi tiếp
I
ca tam giác
.ABC
b). Tìm tọa độ giao điểm ca phân giác trong, phân giác ngoài góc
A
với đường thng
.BC
Li gii.
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Bài tp 6. Trong không gian tọa độ
Oxyz
, cho hình hp
. ' ' ' 'ABCD A B C D
tọa độ các điểm
1;2;3 , 1;4;5 , ' 3;3; 2 , ' 5;3;2A C B D
. Xác định to độ các đỉnh còn li ca hình hp.
Li gii.
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3. Câu hi trc nghim.
Mức độ. Nhận Biết
u 1.(Chuyên Vinh Ln) Trong không gian vi h trc tọa độ
Oxyz
, cho
23a i j k
. Tìm
tọa độ ca
a
.
A.
2; 1; 3
. B.
3;2; 1
. C.
2; 3; 1
. D.
1;2; 3
.
Li gii
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u 2.(Kim Liên Hà Ni 2019)Trong không gian
Oxyz
, cho điểm
M
tha mãn h thc
2OM i j
. Tọa độ đim
M
A.
1;2;0M
. B.
2;1;0M
. C.
2;0;1M
. D.
0;2;1M
.
Li gii
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u 3.(S Đà Nng 2019) Trong không gian
Oxyz
, cho
2 5 3a i k j
. Tọa độ ca
a
là:
A.
2;3; 5
. B.
2;3; 5
. C.
2; 3;5
. D.
2; 5;3
.
Li gii
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u 4.(Lương Thế Vinh Đồng Nai) Trong không gian
Oxyz
với
,,i j k
lần lượt c vecto đơn vị
trên các trục
, , .Ox Oy Oz
Tính tọa độ của vecto
.i j k
A.
( 1; 1;1).i j k
B.
( 1;1;1).i j k
C.
(1;1; 1).i j k
D.
(1; 1;1).i j k
Li gii
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u 5.(Chuyên Lê Qúy Đôn 2019) Trong không gian
Oxyz
vi h tọa độ
; ; ;O i j k
cho
5OA i k
. Tìm tọa độ đim
A
.
A.
1;5
. B.
5; 1;0
. C.
1;0;5
. D.
1;5;0
.
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u 6.(THPT-Chuyên Quang Trung 2019)Trong không gian
Oxyz
cho
B 2;1;0
, điểm
A
tha
mãn
AB 2i j k
vi
,,i j k
là các vectơ đơn vị trên ba trục
Ox
,
Oy
,
Oz
. Tọa độ đim
A
A.
A 0;2; 1
. B.
A 1;2;1
. C.
A 4;0;1
. D.
A 2;1;1
.
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u 7.ng Thành Nam) Trong không gian
Oxyz
, hình chiếu vuông góc của điểm
1;2;3A
trên
mặt phẳng
Oyz
A.
0;2;3M
. B.
1;0;3N
. C.
1;0;0P
. D.
0;2;0Q
.
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u 8.(Chuyên Thái Nguyên) Trong không gian
Oxyz
cho điểm
.
Hình chiếu vuông góc của
A
lên trục
Ox
có tọa độ là:
A.
0;1;0
. B.
2;0;0
. C.
0;0;3
. D.
0;1;3
.
Ligii
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u 9.(KonTum 12 HK2) Trong không gian vi h tọa độ
Oxyz
, hình chiếu vuông góc của điểm
3;2; 4A
lên mt phng
Oxy
có tọa độ
A.
0;2; 4
. B.
0;0; 4
. C.
3;0; 4
. D.
3;2;0
.
Li gii
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u 10.(S Lng Sơn 2019) Trong không gian tọa độ
Oxyz
, cho điểm
(3; 2;5)A
. Hình chiếu
vuông góc của điểm
A
trên mặt phẳng tọa độ
(Ox )z
:
A.
(3; 2;0)M
. B.
(3;0;5)M
. C.
(0; 2;5)M
. D.
(0;2;5)M
.
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u 11.(KonTum 12) Trong không gian
Oxyz
, cho điểm
1; 2;5M
. Khong cách t
M
đến trc
Oz
bng
A.
5
. B.
5
. C.
1
. D.
2
.
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u 12.(Chuyên Thái Bình 2019) Trong không gian
Oxyz
cho điểm
1; 2;3A
. Hình chiếu vuông
góc của điểm
A
trên mt phng
Oyz
là điểm
M
. Tọa độ đim
M
A.
1;0;3M
. B.
0; 2;3M
. C.
1;0;0M
. D.
1; 2;0M
.
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u 13.(Chuyên H Long -2019) Trong không gian với hệ trục tọa độ
Oxyz
cho điểm
;;M x y z
.
Trong các mệnh đề sau, mệnh đề nào đúng?
A. Nếu
M
đối xứng với
M
qua mặt phẳng
Oxz
thì
;;M x y z
.
B. Nếu
M
đối xứng với
M
qua
Oy
thì
;;M x y z
.
C. Nếu
M
đối xứng với
M
qua mặt phẳng
Oxy
thì
;;M x y z
.
D. Nếu
M
đối xứng với
M
qua gốc tọa độ
O
thì
2 ;2 ;0M x y
.
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u 14.(Liên Trường Ngh An) Trong không gian
Oxyz
, cho hai điểm
1; 5; 2A
3; 3; 2B
. Tọa độ trung điểm
M
của đoạn thng
AB
A.
1;1;2M
. B.
2;2;4M
. C.
2; 4;0M
. D.
4; 8;0M
.
Li gii
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u 15.(Đặng Thành Nam Đề 9) Trong không gian Oxyz, cho hai điểm
1;2; 1A
,
B
(1;3;1).AB
Tọa độ của
B
A.
2;5;0
. B.
0; 1; 2
. C.
0;1;2
. D.
2; 5;0
.
Li gii
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u 16.(Chuyên Đại Hc sư phạm2019) Trong không gian vi h trc tọa độ
Oxyz
, cho tam giác
ABC
có ba đỉnh
;0;0Aa
,
0; ;0Bb
,
0;0;Cc
. Tọa độ trng tâm ca tam giác
ABC
A.
;;abc
. B.
;;abc
. C.
;;
3 3 3
abc



. D.
;;
3 3 3
abc



.
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u 17.(THTT ln5) Trong không gian
Oxyz
, cho ba điểm
2; 1;0A
,
1;0 1B
3;0;0C
.
Tọa độ trng tâm
G
ca tam giác
ABC
A.
11
0; ; .
33




B.
0; 1; 1 .
C.
0;1;1 .
D.
11
0; ; .
33



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u 18.ng Thành Nam Đề 10) Trong không gian
Oxyz
, véc nào dưới đây cùng hướng với
véc tơ
3; 1; 2a
?
A.
1
3;1;2u 
. B.
2
1;1;1u
. C.
3
6;2;4u 
. D.
4
12; 4; 8u
.
Li gii
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u 19.(Gia Lc Hải ơng 2019) Trong không gian
Oxyz
, cho các vectơ
1; 1;2a 
,
3;0; 1b 
2;5;1c 
. Tọa độ của vectơ
u a b c
A.
0;6; 6u 
. B.
6;0; 6u 
. C.
6; 6;0u 
. D.
6;6;0u 
.
Li gii
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u 20.(Hoàng Hoa Thám Hưng Yên) Trong không gian vi h trc tọa độ
Oxyz
, cho
2; 3;3a 
,
0;2; 1b 
,
3; 1;5c 
. Tìm tọa độ của vectơ
2 3 2u a b c
.
A.
10; 2;13
. B.
2;2; 7
. C.
2; 2;7
. D.
2;2;7
.
Li gii
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u 21.(Nguyn Du Dak-Lak 2019) Trong không gian tọa độ
; , ,O i j k
, cho ba vectơ
1;2;3a
,
2;0;1b 
,
1;0;1c 
. Tìm tọa độ của vectơ
23n a b c i
.
A.
6;2;6n
. B.
0;2;6n
. C.
6;2; 6n 
. D.
6;2;6n 
.
Li gii
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u 22.(Thăng Long Hà Nội 2019) Trong không gian với hệ tọa độ
, cho
2u i j k
.
Tính
u
.
A.
4u
. B.
5u
. C.
6u
. D.
2u
.
Li gii
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u 23.(THTT S 4-487-2018) Trong không gian vi h tọa độ
Oxyz
, cho ba vectơ
5;7;2a
,
3;0;4b
,
6;1; 1c 
. Tìm tọa độ ca vectơ
32m a b c
.
A.
3;22; 3m
. B.
3;22;3m
. C.
3;22; 3m 
. D.
3; 22;3m
.
Li gii
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u 24.(THPT Chuyên Ti Bình 2018) Cho các vectơ
1;2;3a
;
2;4;1b 
;
1;3;4c 
.
Vectơ
2 3 5v a b c
có tọa độ
A.
7;3;23v
. B.
23;7;3v
. C.
7;23;3v
. D.
3;7;23v
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 25.(THPT Chuyên Hạ Long-2018) Trong không gian với hệ tọa độ
Oxyz
, cho hai véctơ
2; 3; 1a
1;0;4a 
. Tìm tọa độ của véctơ
45u a b
.
A.
13;12; 24u
. B.
13; 12; 24u
. C.
3; 12;16u 
. D.
13; 12; 24u
.
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u 26.(Chuyên Thái Bình 2019) Trong không gian với hệ trục tọa độ
,Oxyz
cho hai điểm
1;3; 1 , 3; 1;5AB
. Tìm tọa độ của điểm
M
thỏa mãn hệ thức
3MA MB
.
A.
5 13
; ;1
33
M



. B.
71
; ; 3
33
M



. C.
71
; ;3
33
M



. D.
4; 3;8M
.
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u 27.(THPT-Yên-Khánh-Ninh 2019)Trong không gian
Oxyz
cho các ctơ
22u i j k
,
;2; 1v m m
với
m
là tham số thực. Có bao nhiêu giá trị của
m
để
uv
?
A. 0. B. 1. C. 2. D. 3.
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u 28.(THPT Chuyên Vĩnh Phúc 2018) Trong không gian vi h trc tọa độ
Oxyz
, cho véctơ
1; 2;3a 
. Tìm tọa độ của véctơ
b
biết rằng véctơ
b
ngược hướng với véctơ
a
2ba
.
A.
2; 2;3b 
. B.
2; 4;6b 
. C.
2;4; 6b
. D.
2; 2;3b
.
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Chn C.
véctơ
b
ngược hướng với véctơ
a
2ba
nên ta có
2 2;4; 6ba
.
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u 29.(Qunh Lưu Ngh An) Trong không gian
Oxyz
, góc giữa hai vectơ
i
3; 0;1u 
A.
120
. B.
60
. C.
150
. D.
30
.
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u 30.(THPT Trần Hưng Đạo2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho hai vectơ
0;3;1a
,
3;0; 1b 
. Tính
cos ,ab
.
A.
1
cos ,
100
ab 
. B.
1
cos ,
100
ab
. C.
1
cos ,
10
ab 
. D.
1
cos ,
10
ab
.
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u 31.(THPT Mộ Đức-Quãng Ngãi 2018) Trong không gian
Oxyz
, cho ba điểm
1; 2;3A 
,
0;3;1B
,
4;2;2C
. Côsin của góc
BAC
bằng
A.
9
35
. B.
9
2 35
. C.
9
2 35
. D.
9
35
.
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u 32.Trong không gian
Oxyz
cho vec-
1;1;2u
2;0;vm
. Tìm giá trị của tham số
m
biết
4
cos ;
30
uv
A.
1m
. B.
1; 11mm
. C.
11m 
. D.
0m
.
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u 33.(Thun Thành BcNinh) Trong không gian vi h tọa độ
Oxyz
, cho ba điểm
1;1;1M
,
2;3;4N
,
7;7;5P
. Để t giác
MNPQ
là hình bình hành thì tọa độ đim
Q
A.
6; 5; 2
. B.
6; 5;2
. C.
6;5;2
. D.
6;5;2
.
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u 34.(Kinh Môn Hải Dương2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho
0; 1;1A
,
2;1; 1B 
,
1; 3; 2C
. Biết rằng
ABCD
là hình bình hành, khi đó tọa độ điểm
D
A.
1;1; 4D
. B.
2
1;1;
3
D



. C.
1; 3; 4D
. D.
1; 3; 2D
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u 35.(HK2 Sở Đồng Tháp)Trong không gian với hệ toạn độ
Oxyz
, cho
1;1;2 , 2; 1;1AB
, 3;2; 3C
. Tìm tọa độ điểm
D
để tứ giác
ABCD
là hình bình hành.
A.
4;2; 4
. B.
0; 2;6
. C.
2;4; 2
. D.
4;0; 4
.
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u 36.(THPT Chuyên Quý Đôn 2018) Trong không gian với hệ tọa độ
Oxyz
cho hình hộp
.ABCD A B C D
. Biết
2;4;0A
,
4;0;0B
,
1;4; 7C 
6;8;10D
. Tọa độ điểm
B
A.
8;4;10B
. B.
6;12;0B
. C.
10;8;6B
. D.
13;0;17B
.
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u 37.(Toán Học Tuổi Trẻ) Trong không gian với hệ trục tọa độ
Oxyz
, cho
2 2 2OA i j k
,
2; 2;0B
4;1; 1C
. Trên mặt phẳng
Oxz
, điểm nào ới đây cách đều ba điểm
A
,
B
,
C
.
A.
31
; 0;
42
M



. B.
31
; 0;
42
N




. C.
31
; 0;
42
P



. D.
31
; 0;
42
Q



.
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u 38.(THPT Tây Thụy Anh 2018)Trong không gian
Oxyz
cho ba điểm:
1; 1;1 ,A
0;1;2 ,B
1;0;1 .C
Trong các mệnh đề sau hãy chọn mệnh đề đúng?
A. Tam giác
ABC
vuông tại
.A
B. Ba điểm
A
,
B
,
C
thẳng hàng.
C. Ba điểm
A
,
B
,
C
không thẳng hàng. D.
B
là trung điểm của
.AC
Li gii
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Mức độ 2. Thông hiểu
u 39.(Hàm Rồng) Cho vectơ
1;3;4u
, tìm vectơ cùng phương với vectơ
u
.
A.
2;6;8n 
. B.
2; 6; 8n
. C.
2; 6;8n
. D.
2; 6; 8n
Li gii
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u 40.(Hùng Vương Bình Phước)Trong không gian với hệ trục
Oxyz
cho ba điểm
1;2; 3A
, 1;0;2 , ; ; 2B C x y
thẳng hàng. Khi đó
xy
bằng
A.
1xy
. B.
17xy
. C.
11
5
xy
. D.
11
5
xy
.
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u 41.(Qunh Lưu Ngh An) Trong không gian
Oxyz
, cho hai điểm
2; 2;1A
,
0;1;2B
. Ta
độ đim
M
thuc mt phng
Oxy
sao cho ba điểm
A
,
B
,
M
thng hàng là
A.
4; 5;0M
. B.
2; 3;0M
. C.
0;0;1M
. D.
4;5;0M
.
Li gii
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u 42.(THPT Nghèn Lần)Trong không gian
Oxyz
cho ba điểm
1;1;2A
,
0;1; 1B
,
2; ; 2C x y
thẳng hàng. Tổng
xy
bằng
A.
7
3
. B.
8
3
. C.
2
3
. D.
1
3
.
Li gii
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u 43.(S GD và ĐT Thanh Hóa 2019) Trong không gian
Oxyz
, cho bốn điểm
1;2;0A
,
3;1;0B
,
0;2;1C
1;2;2D
. Trong đó có ba điểm thẳng hàng là
A.
A
,
C
,
D
. B.
A
,
B
,
D
. C.
B
,
C
,
D
. D.
A
,
B
,
C
.
Li gii
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u 44.(HSG Bc Ninh 2019)Trong không gian với hệ tọa độ
Oxyz
, cho các vectơ
2; 1;3am
, 1;3; 2bn
. Tìm
,mn
để các vectơ
,ab
cùng hướng.
A.
3
7;
4
mn
. B.
4; 3mn
. C.
1; 0mn
. D.
4
7;
3
mn
.
Li gii
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u 45.(THPT Hồng Quang-Hải Dương 2018) Trong không gian với hệ tọa độ
Oxyz
, cho các vectơ
2; 1;3am
,
1;3; 2bn
. Tìm
m
,
n
để các vectơ
a
,
b
cùng hướng.
A.
7,m
3
.
4
n 
B.
7m
;
4
3
n 
. C.
4m
;
3n 
. D.
1m
;
0n
.
Li gii
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u 46.( Nhân Tông)Trong không gian vi h tọa độ
Oxyz
cho
;;3A x y
;
6; 2;4B
;
( 3;7; 5)C 
. Giá tr
x; y
để
;;A B C
thng hàng là
A.
1; 5xy
. B.
1; 5xy
. C.
1; 5xy
. D.
1; 5xy
.
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u 47.(THPT Lương Văn Chánh Phú Yên 2018) Phép tnh tiến biến gc tọa độ
O
thành điểm
1;2A
s biến điểm
A
thành điểm
A
có tọa độ là:
A.
2;4A
. B.
1; 2A

. C.
4;2A
. D.
3;3A
.
Li gii
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u 48.(Chuyên KHTN)Trong không gian
Oxyz
, cho
3;1;2A
, tọa độ điểm
'A
đối xứng với
điểm
A
qua trục
Oy
A.
3; 1; 2
. B.
3; 1;2
. C.
3;1; 2
. D.
3; 1;2
.
Li gii
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u 49.(THPT Chuyên H Long-2018) Trong không gian
Oxyz
, cho điểm
2; 3;5A
. Tìm tọa độ
A
là điểm đối xng vi
A
qua trc
Oy
.
A.
2;3;5A
. B.
2; 3; 5A

. C.
2; 3;5A

. D.
2; 3; 5A
.
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u 50.(Hu Lc Thanh Hóa)Trong không gian vi h trc tọa độ
Oxyz
, cho ba điểm
1;2; 1 ; 2; 1;3 ; 3;5;1A B C
. Tìm tọa độ đim
D
sao cho t giác
ABCD
là hình bình hành.
A.
4; 8; 5D 
B.
4; 8; 3D 
. C.
2;8; 3D 
. D.
2;2;5D
.
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u 51.(Kim Liên)Trong không gian
Oxyz
, cho hai điểm
2;4;1A
4;5;2B
. Điểm
C
tha
mãn
OC BA
có tọa độ
A.
6; 1; 1
. B.
2; 9; 3
. C.
6; 1;1
. D.
2; 9;3
.
Li gii
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u 52.(Liên Trường Ngh An) Trong không gian
Oxyz
, cho hai điểm
3;1; 2A
,
2; 3;5B
.
Đim
M
thuộc đoạn
AB
sao cho
2MA MB
, tọa độ đim
M
A.
7 5 8
;;
3 3 3
M



. B.
4;5; 9M
. C.
3 17
; 5;
22
M



. D.
1; 7;12M
.
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u 53.(Chuyên Đại Hc Vinh 2019)Trong không gian vi h tọa độ
Oxyz
, gi
a
,
b
,
c
lần lượt
khong cách t đim
1;3;2M
đến ba mt phng tọa độ
Oxy
,
Oyz
,
Oxz
. Tính
23
P a b c
?
A.
32P
. B.
18P
. C.
30P
. D.
12P
.
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u 54.(SGD Bà Rịa Vũng Tàu 2018) Trong không gian vi h trc tọa độ
Oxyz
, điểm thuc trc
Oy
và cách đều hai điểm
3;4;1A
1;2;1B
A.
0;4;0 .M
B.
5;0;0M
. C.
0;5;0M
. D.
0; 5;0 .M
Li gii
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u 55.(Sở Nam Định) Trong không gian vói htrục tọa độ
Oxyz
, cho hình thang cân
ABCD
hai đáy
AB
,
CD
thỏa mãn
2CD AB
diện ch bằng
27
, đỉnh
1; 1;0A 
, phương trình
đường thẳng chứa cạnh
CD
2 1 3
2 2 1
x y z

. Tìm tọa độ điểm
D
biết hoành độ điểm
B
lớn hơn hoành độ điểm
A
.
A.
2; 5;1D 
. B.
3; 5;1D 
. C.
2; 5;1D
. D.
3; 5;1D
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 56.(THPT Đô Lương)Trong không gian
Oxyz
, cho điểm
3;4;3A
. Tổng khoảng cách từ
A
đến ba trục tọa độ bằng
A.
34
. B.
10
. C.
34
2
. D.
10 3 2
.
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u 57.(HSG 12 Bc Giang) Trong không gian
Oxyz
, cho ba điểm
3;1;0A
,
0; 1;0B
,
0;0; 6C
. Nếu tam giác
ABC
các đnh tha mãn h thc
0A A B B C C
thì tam giác
ABC
có tọa độ trng tâm là
A.
3; 2;0
. B.
2; 3;0
. C.
1;0; 2
. D.
3; 2;1
.
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u 58.(THPT Hùng ơng 2019)Trong không gian vi h trục tọa độ
Oxyz
cho hai điểm
2;3;2A
,
2; 1;4B 
. Tìm tọa độ đim
E
thuc trc
Oz
sao cho
E
cách đều hai điểm
,AB
.
A.
1
0;0;
2



. B.
1
0;0;
3



. C.
0;0; 1
. D.
0;0;1
.
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u 59.(Cầu Giấy Nội 2019) Trong không gian
Oxyz
, cho điểm
2; 5;4M
. Trong các phát
biểu sau, phát biểu nào sai?
A. Khoảng cách từ
M
đến mặt phẳng tọa độ
xOz
bằng
5
.
B. Khoảng cách từ
M
đến trục
Oz
bằng
29
.
C. Tọa độ điểm
M
đối xứng với
M
qua mặt phẳng
yOz
2;5; 4M
.
D.Tọa độ điểm
M
đối xứng với
M
qua trục
Oy
2; 5; 4M
.
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u 60.(THPT n Lc2018) Trong không gian vi h trc tọa độ
Oxyz
cho
1; 2;3a 
2; 1; 1b
. Khng định nào sau đây đúng?
A.
, 5; 7; 3ab


. B. Vectơ
a
không cùng phương với vectơ
b
.
C. Vectơ
a
không vuông góc với vectơ
b
.
D.
14a
.
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u 61.(THPT Quãng Xương 2018) Trong không gian
Oxyz
, cho
a
,
b
tạo với nhau
1
góc
120
3a
;
5b
. Tìm
T a b
.
A.
5T
. B.
6T
. C.
7T
. D.
4T
.
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u62.(THPT Tứ KỳHải Dương 2018) Trong không gian với hệ tọa độ
Oxyz
, cho vectơ
1;1; 2u 
,
1;0;vm
. Tìm
m
để góc giữa hai vectơ
,uv
bằng
45
.
A.
26m 
B.
26m 
. C.
26m 
. D.
2m
.
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u 63.(THPT Lương Văn Chánh 2019) Trong không gian
Oxyz
, cho hai vectơ
u
v
tạo với
nhau một góc
120
2u
,
5v
. Tính
uv
A.
19
. B.
5
. C.
7
. D.
39
.
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u 64.(THPT Trần Nhânng 2018) Trong không gian với hệ tọa độ O
xyz
, cho ba điểm
2;3; 1M
,
1;1;1N
1; 1;2Pm
. Tìm
m
để tam giác
MNP
vuông tại
N
.
A.
6m 
. B.
0m
. C.
4m 
. D.
2m
.
Li gii
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u 65.(THPT Trần Hưng Đạo 2018) Trong không gian với hệ tọa độ
Oxyz
, cho ba điểm
3;2;8M
,
0;1;3N
2; ;4Pm
. Tìm
m
để tam giác
MNP
vuông tại
N
.
A.
25m
. B.
4m
. C.
1m 
. D.
10m 
.
Li gii
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u 66.(TT Diu Hin-Cn Thơ-tháng 10-năm 2017-2018) Trong không gian vi tọa độ
Oxyz
, cho
hai điểm
1;1;2A
,
1;3; 9B 
. Tìm tọa độ đim
M
thuc
Oy
sao cho
ABM
vuông ti
M
.
A.
0;1 2 5;0
0;1 2 5;0
M
M
. B.
0;2 2 5;0
0;2 2 5;0
M
M
. C.
0;1 5;0
0;1 5;0
M
M
. D.
0;2 5;0
0;2 5;0
M
M
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 67.(THPT Chuyên Vĩnh Phúc 2018)Trong không gian
Oxyz
, cho điểm
3; 1;2M
. Tìm ta
độ đim
N
đối xng vi
M
qua mt phng
Oyz
.
A.
0; 1;2N
. B.
3;1; 2N
. C.
3; 1;2N 
. D.
0;1; 2N
.
Li gii
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u 68.(THPT Kim Liên 2017) Trong không gian với hệ tọa độ
Oxyz
, biết
2u
;
1v
góc
giữa hai vectơ
u
v
bằng
2
3
. Tìm
k
để vectơ
p ku v
vuông góc với vectơ
q u v
.
A.
2
5
k
. B.
5
2
k
. C.
2k
. D.
2
5
k 
.
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u 69.(Sở GD-ĐT Kiên Giang-2019) Trong không gian
Oxyz
cho
hình hộp chữ nhật
.OABC EFGH
các cạnh
5OA
,
8OC
,
7OE
(xem hình vẽ). Hãy tìm tọa độ điểm
H
.
A.
0;7;8H
. B.
7;8;0H
.
C.
8;7;0H
. D.
0;8;7H
.
Li gii
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u 70.(Chuyên T Trng Cn Thơ) Trong không gian vi h tọa độ
Oxyz
, cho hình hp
. ' ' ' 'ABCD A B C D
vi
2;1;3 ,A
2;3;5 ,C
' 2;4; 1 , ' 0;2;1BD
. Tìm tọa độ đim
B
.
A.
1; 3;3B
. B.
1;3;3B
. C.
1;3; 3C
. D.
1;3;3B
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 71.(Đặng Thành Nam) Trong không gian
Oxyz
, véctơ
u
vuông góc với hai véctơ
1;1;1a
1; 1;3b 
; đồng thời
u
tạo với tia
Oz
một góc tù và độ dài véctơ
u
bằng 3. Tìm véctơ
u
.
A.
66
6; ;
22





. B.
66
6; ;
22




.
C.
66
6; ;
22




. D.
66
6; ;
22





.
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u 72.(THPT Lương Văn Chánh 2018) Trong không gian
Oxyz
, cho hình hp
.ABCD A B C D
1;0;1A
,
2;1;2B
,
1; 1;1D
,
4;5; 5C
. Tính tọa độ đỉnh
A
ca hình hp.
A.
4;6; 5A
. B.
2;0;2A
. C.
3;5; 6A
. D.
3;4; 6A
.
Li gii
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u 73.(THPT-Ngô Quyền 2019)Trong không gian
Oxyz
, cho hình hộp
.ABCD A B C D
biết
1;0;1A
,
2;1;2B
,
1; 1;1D
,
4;5; 5C
. Tọa độ của điểm
A
là:
A.
4;6; 5A
. B.
3;4; 1A

. C.
3;5; 6A
. D.
3;5;6A
.
Li gii
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u 74.(Chuyên Hng Phong 2018) Trong không gian vi h to độ
Oxyz
, cho hình hp
.ABCD A B C D
0; 0; 0A
,
3; 0; 0B
,
,
0; 3; 3D
. To đ trng tâm tam giác
ABC

A.
1; 1; 2
. B.
2; 1; 2
. C.
1; 2; 1
. D.
2; 1; 1
.
Li gii
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u 75.(Chuyên Hng Phong 2018)Trong không gian vi h tọa độ
Oxyz
, cho bốn điểm
2;0;0 , 0;2;0 , 0;0;2A B C
2;2;2D
. Gi
,MN
lần lượt trung điểm ca
AB
CD
.
Tọa độ trung điểm
I
ca
MN
là:
A.
1; 1;2I
. B.
1;1;0I
. C.
. D.
1;1;1I
.
Li gii
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u 76.(THPT Việt Đc-Hà Ni)Trong hệ trục tọa độ
Oxyz
, cho các điểm
(1; 1;1),N(2;0; 1)M 
. Xét điểm
Q
sao cho tứ giác
MNPQ
là một hình bình hành. Tọa độ
Q
A.
( 2;1;3)
B.
( 2;1;3)
C.
( 2;1; 3)
D.
(4;1;3)
Li gii
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u 77.(Chuyên Nguyn Du Đăk k) Trong không gian
Oxyz
, cho ba điểm
3;5; 1A
,
7; ;1Bx
9;2;Cy
. Để
A
,
B
,
C
thng hàng thì giá tr
xy
bng
A.
5
. B.
6
. C.
4
. D.
7
.
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u 78.(THPT Chuyên Lê Quý Đôn2018) Trong không gian vi h trc to độ
Oxyz
, cho
2;3;1a
,
1;5;2b 
,
4; 1;3c 
3;22;5x 
. Đẳng thức nào đúng trong các đẳng
thc sau ?
A.
23x a b c
. B.
23x a b c
. C.
23x a b c
. D.
23x a b c
.
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u 79.(THTT S 2-485 tng 11 2018) Trong không gian vi h trc tọa độ
Oxyz
, cho ba điểm
1;2;0A
;
2;1;1B
;
0;3; 1C
. Xét 4 khẳng định sau:
I.
2BC AB
. II. Điểm
B
thuộc đoạn
AC
.
III.
ABC
là một tam giác. IV.
A
,
B
,
C
thẳng hàng.
Trong
4
khẳng định trên có bao nhiêu khẳng định đúng?
A.
1
. B.
2
. C.
3
. D.
4
.
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Mức độ 3. Vận dụng
u 80.(THPT-Yên-Khánh-Ninh-Bình)Trong không gian
Oxyz
cho các điểm
5;1;5A
,
4;3;2B
,
3; 2;1C 
. Điểm
;;I a b c
là tâm đường tròn ngoi tiếp tam giác
ABC
. Tính
2a b c
?
A.
1
. B.
3
. C.
6
. D.
9
.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 81.(TT Diu Hin-Cn Thơ 2018) Trong không gian vi h tọa độ
Oxyz
, cho ba đim
1; 2;2A
,
5;6;4B
0;1; 2C
. Độ dài đường phân giác trong ca góc
A
ca
ABC
là:
A.
3 74
2
. B.
3
2 74
. C.
2
3 74
. D.
2 74
3
.
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u 82.(Chuyên Hng phong 2018) Trong không gian vi h tọa độ
Oxyz
, cho ba điểm
1;2; 1A
,
2; 1;3B
,
4;7;5C
. Tọa độ chân đường phân giác trong góc
B
ca tam giác
ABC
A.
2 11
; ;1
33



. B.
11
; 2;1
3



. C.
2 11 1
;;
3 3 3



. D.
2;11;1
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 83.(THPT Việt Đức-Hà Ni)Trong h trc tọa độ Oxyz, cho điểm
2;1;1H
. Gọi các điểm
,,A B C
lần lượt trên các trc tọa độ
,,Ox Oy Oz
sao cho
H
trc m ca tam giác
ABC
. Khi
đó hoành độ đim
A
là:
A.
3
. B.
5
. C. 3. D. 5
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u 84.(Lê Xoay lần1) Trong không gian với hệ trục tọa độ
Oxyz
cho ba điểm
1;2; 1A
,
2;1;1B
0;1;2C
. Gọi
;;H x y z
là trực tâm tam giác
ABC
. Giá trị của
S x y z
A.
7
. B.
6
. C.
5
. D.
4
.
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u 85.(THPT Trần Quốc Tuấn 2018) Trong không gian với hệ tọa độ
Oxyz
cho tam giác vuông
(4;0;2)A
,
(1; 4; 2)B 
(2;1;1)C
. Tính diện tích
S
của tam giác
ABC
.
A.
242
2
S
. B.
246
2
S
. C.
206
2
S
. D.
210
2
S
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 86.(THPT Chuyên Thái Bình 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho tam giác
ABC
với:
1; 2;2AB 
;
3; 4; 6AC 
. Độ dài đường trung tuyến
AM
của tam giác
ABC
A.
29
. B.
29
. C.
29
2
. D.
2 29
.
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u 87.(Đặng Thúc Ha2018) Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
0;2; 2A
,
2;2; 4B
. Gi s
;;I a b c
là tâm đường tròn ngoi tiếp tam giác
OAB
. Tính
2 2 2
T a b c
.
A.
8T
. B.
2T
. C.
6T
. D.
14T
.
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u 88.(THPT Trn Quc Tun năm 2018) Trong không gian vi h trc tọa độ
Oxyz
cho hình
thang
ABCD
vuông ti
A
B
. Ba đỉnh
(1;2;1)A
,
(2;0; 1)B
,
(6;1;0)C
Hình thang din tích
bng
62
. Gi s đỉnh
( ; ; )D a b c
, tìm mệnh đề đúng?
A.
6abc
. B.
5abc
. C.
8abc
. D.
7abc
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 89.(THPT Lê Q Đôn2018) Trong không gian
Oxyz
cho ba điểm
1;2;3A
,
3;4;4B
,
2;6;6C
;;I a b c
là tâm đường tròn ngoi tiếp tam giác
ABC
. Tính
abc
.
A.
63
5
. B.
31
3
. C.
46
5
. D.
10
.
Li gii
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u 90.(THPT Đặng Thúc Hứa 2018) Trong không gian với hệ tọa độ
Oxyz
, cho hai điểm
0;2; 2A
,
2;2; 4B
. Giả sử
;;I a b c
tâm đường tròn ngoại tiếp tam giác
OAB
. Tính
2 2 2
T a b c
.
A.
8T
. B.
2T
. C.
6T
. D.
14T
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
30
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 91.(Chuyên Quốc Học Huế 2018) Trong không gian với hệ tọa độ
Oxyz
, cho tam giác
ABC
với
1;1;1A
,
2;3;0B
. Biết rằng tam giác
ABC
có trực tâm
0;3;2H
tìm tọa độ của điểm
C
.
A.
3;2;3C
. B.
4;2;4C
. C.
1;2;1C
. D.
2;2;2C
.
Li gii
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u 92.(THPT Phan Châu Trinh-2018) Trong không gian tọa độ
Oxyz
cho hai điểm
2;2;1A
,
8 4 8
;;
333
B



. Biết
;;I a b c
là tâm đường tròn nội tiếp ca tam giác
OAB
. Tính
.S a b c
A.
1S
. B.
0S
. C.
1S 
. D.
2S
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Mức độ. Vận dụng cao
u 93.(TH PT Kinh Dương 2019) Trong không gian với hệ tọa
Oxyz
, cho vectơ
1; 2;4a 
,
0 0 0
;;b x y z
cùng phương với vectơ
a
. Biết vectơ
b
tạo với tia
Oy
một góc nhọn
21b
.
Giá trị của tổng
0 0 0
x y z
bằng
A.
3
. B.
6
. C.
6
. D.
3
.
Li gii
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u 94.(SGD Tĩnh 2018) Trong không gian với hệ tọa độ
Oxyz
, mặt phẳng
đi qua điểm
1;2;1M
cắt tia
Ox
,
Oy
,
Oz
lần lượt tại
A
,
B
,
C
sao cho độ dài
OA
,
OB
,
OC
theo thứ tự
tạo thành cấp số nhân có công bội bằng
2
. Tính khoảng cách từ gốc tọa độ
O
tới mặt phẳng
A.
4
21
. B.
21
21
. C.
3 21
7
. D.
9 21
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
32
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 95.(THPT Xoay 2018) Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
2;2; 2A 
;
3; 3;3B
. Điểm
M
trong không gian tha mãn
2
3
MA
MB
. Khi đó độ dài
OM
ln nht bng
A.
63
. B.
12 3
. C.
53
2
. D.
53
.
Li gii
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u 96.(Chuyên Lạng Sơn 2018) Trong không gian vi h trc tọa độ
Oxyz
, cho điểm
1;0;6A
.
Biết rằng hai điểm
M
,
N
phân bit thuc trc
Ox
sao cho các đường thng
AM
,
AN
cùng
to với đường thng cha trc
Ox
mt góc
45
. Tổng các hoành độ hai điểm
M
,
N
tìm được là
A.
4
. B.
2
. C.
1
. D.
5
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
33
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 97.(THPT Qúy Đôn 2018) Trong không gian
Oxyz
cho ba điểm
1;2;3A
,
3;4;4B
,
2;6;6C
;;I a b c
là tâm đường tròn ngoại tiếp tam giác
ABC
. Tính
abc
.
A.
63
5
. B.
31
3
. C.
46
5
. D.
10
.
Li gii
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u 98.(Lương Thế Vinh 2018) Trong không gian
Oxyz
, cho ba điểm
2;3;1A
,
2;1;0B
,
3; 1;1C 
. Tìm tt c các điểm
D
sao cho
ABCD
là hình thang có đáy
AD
3
ABCD ABC
SS
.
A.
8;7; 1D
. B.
8; 7;1
12;1; 3
D
D

. C.
8;7; 1
12; 1;3
D
D

. D.
12; 1;3D 
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 99.(THPT Lương Thế Vinh 2018) Trong không gian
Oxyz
, cho ba điểm
0;0; 1A
,
1;1;0B
,
1;0;1C
. Tìm điểm
M
sao cho
2 2 2
32MA MB MC
đạt giá tr nh nht.
A.
31
; ; 1
42
M



. B.
31
; ;2
42
M



. C.
33
; ; 1
42
M




. D.
31
; ; 1
42
M




.
Li gii
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DNG 2. NG DNG CỦA TÍCH CÓ NG
4. Phương pháp:
Dựa vào định nghĩa tích có hướng
Cho
1 1 1
; ; a x y z
2 2 2
; ; b x y z
. Tích có hướng của hai véc tơ
a
b
là một véc tơ,
kí hiệu
,ab


, được xác định như sau
1 1 1 1 1 1
1 2 2 1 1 2 2 1 1 2 2 1
2 2 2 2 2 2
, ; ; ; ;
y z z x x y
a b y z y z z x z x x y x y
y z z x x y




Da vào các ng dng sau:
a
cùng phương
, 0.b a b



Diện tích tam giác:
1
,
2
ABC
S AB AC


.
Thể hình hộp:
. ' ' ' '
, . '
ABCD A B C D
V AB AD AA


Thể tứ diện:
1
,.
6
ABCD
V AB AC AD


.
Điều kiện 3ctơ đồng phẳng:
,,abc
đồng phẳng
, . 0a b c



, , ,A B C D
đồng phẳng
, . 0AB AC AD



.
5. Bài tp minh ha
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Bài tp 7. Cho hai véc tơ
,ab
0
2 3, 3,( , ) 30 .a b a b
Tính
a). Độ dài các véc tơ
,5 2 ,3 2 ,a b a b a b
b). Độ dài véc tơ
, , ,3 , 5 , 2 .a b a b a b
Lời giải.
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Bài tp 8. Tìm điều kin ca tham s
m
sao cho
a). Ba véc tơ
(2;1; ), ( 1; 2;0), (1; 1;2)u m v m w
đồng phng. `
b). Bn điểm
(1; 1; ), ( ;3;2 1), (4;3;1), ( 3; ;2 )A m B m m C D m m m
thuc mt mt phng.
c). Góc gia hai véc tơ
(2; ;2 1), ( ;2; 1)a m m b m
0
60 .
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 9. Cho hình chóp
.S ABCD
với điểm
(4; 1;2),A
( 1;0; 1)B 
(0;0; 2),C
(10; 2;4).D
Gi
M
trung điểm ca
CD
. Biết
SM
vuông góc vi mt phng
()ABCD
th
tích khi chóp
.
66
S ABCD
V
(đvtt). Tìm tọa độ đỉnh
S
.
Lời giải.
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Bài tp10.Trong không gian vi h to độ
Oxyz
, cho tam giác
ABC
có tọa độ các điểm như
sau
2; 1;3 , 3;0; 2 , 5; 1; 6A B C
a). Tính cos
BAC
, suy ra s đo của
BAC
;
b). Xác định to độ hình chiếu vuông góc H ca A trên BC và to độ điểm A’ đối xng ca A
qua đường thng BC.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
37
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 11. Trong không gian vi h tọa độ Oxyz ,cho tam giác ABC có A(4;2;0) , B(2;4;0) và
C(2;2;1). Xác định tọa độ trực tâm và tâm đường tròn ngoi tiếp ca tam giác ABC.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
38
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 12. Cho tam giác
ABC
( 1;1; 1), (2;3;5).BC
Đim
A
có tung độ
1
,
3
hình chiếu
của điểm
A
trên
BC
7
1; ;3
3
K



và din tích tam giác
ABC
49
.
3
S
a). Tìm tọa độ đnh
A
biết
A
có hoành độ dương.
b). Tìm tọa độ chân đường vuông góc h t
B
đến
.AC
c). Tìm tọa độ tâm
I
của đường tròn ngoi tiếp và tọa độ trc tâm
H
ca tam giác
.ABC
d). Chng minh
vi
G
là trng tâm tam giác
.ABC
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
39
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 13. Cho t din
ABCD
có các cp cạnh đối bng nhau. Cho đim
(2;4;1), (0;4;4),AB
(0;0;1)C
D
có hoành độ dương.
a). Xác định tọa độ đim
.D
b). Gi
G
là trng tâm ca t din
.ABCD
Chng minh
G
cách đều các đỉnh ca t din.
c). Gi
,MN
lần lượt là trung điểm ca
,.AB CD
Chng minh rng
MN
là đường vuông góc
chung của hai đường thng
AB
.CD
d). Tính độ dài các đường trng tuyến ca t din
.ABCD
Tính tng các góc phng mi
đỉnh ca t din
.ABCD
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
40
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
i tp 14. Cho tam giác
ABC
(2;3;1), ( 1;2;0), (1;1; 2).A B C
a). Tìm tọa độ chân đường vuông góc k t
A
xung
BC
.
b). Tìm tọa độ
H
là trc tâm ca tam giác
ABC
.
c). Tìm tọa độ
I
là tâm đường tròn ngoi tiếp ca tam giác
ABC
.
d). Gi
G
trng m ca tam giác
ABC
. Chng minh rằng các điểm
,,G H I
nm trên mt
đưng thng.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
41
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 15. Trong không gian vi h tọa độ Đề Các vuông góc
Oxyz
cho tam giác đều
ABC
(5;3; 1), (2;3; 4)AB
và điểm
C
nm trong mt phng
()Oxy
có tung độ nh hơn
3
.
a). Tìm tọa độ đim
D
biết
ABCD
là t diện đều.
b). Tìm tọa độ đim
S
biết
,,SA SB SC
đôi một vuông góc.
Lời giải.
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i tp 16. Trong không gian vi h tọa độ
Oxyz
cho điểm
3; 2;4A
a). Tìm tọa độ các hình chiếu ca
A
lên các trc tọa độ và các mt phng tọa độ
b). Tìm
,M Ox N Oy
sao cho tam giác
AMN
vuông cân ti
A
c). Tìm tọa độ đim
E
thuc mt phng
()Oyz
sao cho tam giác
AEB
cân ti
E
din
tích bng
3 29
vi
1;4; 4B 
.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
42
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 17. Trong không gian vi h trc
Oxyz
cho
00
(4;0;0), ( ; ;0)A B x y
vi
00
,0xy
tha
mãn
2 10AB
0
45AOB
.
a). Tìm
C
trên tia
Oz
sao cho th tích t din
OABC
bng
8
.
b). Gi
G
là trng tâm
ABO
M
trên cnh
AC
sao cho
AM x
. Tìm
x
để
OM GM
.
Lời giải.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
43
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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6. Câu hi trc nghim.
Mức độ. Nhận Biết
u 100.(THPT Kinh Môn 2019) Trong không gian vi h tọa độ
Oxyz
, cho hai vectơ
1; 2;1u 
2;1; 1v 
. Vectơ nào dưới đây vuông góc với c hai vectơ
u
v
?
A.
2
1;3;5w 
. B.
4
1;4;7w
. C.
3
1; 4;7w 
. D.
1
2; 6; 10w
.
Li gii
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u101.(THPT Chuyên Quý Đôn-2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho hình
bình hành
ABCD
. Biết
2;1; 3A
,
0; 2;5B
1;1;3C
. Diện tích hình bình hành
ABCD
A.
2 87
. B.
349
2
. C.
349
. D.
87
.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
44
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii
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u 102.(THPT M Đức-Quãng Ngãi 2018) Trong không gian
Oxyz
, cho
1;2; 1A
,
0; 2;3B
.
Tính din tích tam giác
OAB
.
A.
29
6
. B.
29
2
. C.
78
2
. D.
7
2
.
Li gii
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u 103.(THPT Hoàng Hoa Thám-ng Yên-ln 1 năm 2017-2018)Trong không gian vi h tọa độ
Oxyz
, cho
0; 2;2Aa
;
3; 1;1Ba
;
4; 3;0C 
;
1; 2; 1Da
. Tp hp các giá tr ca
a
để bốn điểm
A
,
B
,
C
,
D
đồng phng là tp con ca tp nào sau?
A.
7; 2
. B.
3;6
. C.
5;8
. D.
2;2
.
Li gii
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u 104.(Sở GD ĐT Đồng Tháp-2018) Trong không gian với hệ tọa độ
Oxyz
, cho bốn điểm
1; 2;0A
,
3;3;2B
,
1;2;2C
3;3;1D
. Độ dài đường cao của tứ diện
ABCD
hạ từ đỉnh
D
xuống mặt phẳng
ABC
bằng
A.
9
72
. B.
9
7
. C.
9
14
. D.
9
2
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 105.(Sở GD-ĐT Gia Lai -2018) Trong không gian
Oxyz
, cho bốn điểm
1;1;4A
,
5; 1;3B
,
2;2;Cm
,
3;1;5D
. Tìm tất cả giá trị thực của tham số
m
để
A
,
B
,
C
,
D
bốn đỉnh của một
hình tứ diện.
A.
6m
. B.
6m
. C.
6m
. D.
6m
.
Li gii
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u 106.(THPT NEWTON Hà Nội 2018) Trong không gian với hệ tọa độ
Oxyz
, cho 3 điểm
( ;0;0)Aa
,
(1; ;0)Bb
,
(1;0; )Cc
với
a
,
b
,
c
các số thực thay đổi sao cho
(3;2;1)H
trực tâm
của tam giác
ABC
. Tính
. S a b c
A.
2S
. B.
19S
. C.
11S
. D.
9S
.
Li gii
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u 107.(THPT Chuyên Quý Đôn 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho hình
bình hành
ABCD
. Biết
2;1; 3A
,
0; 2;5B
1;1;3C
. Diện tích hình bình hành
ABCD
A.
2 87
. B.
349
2
. C.
349
. D.
87
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 108. Trong không gian
Oxyz
cho hai điểm
1;5;0A
,
3;3;6B
và đường thẳng
11
:
2 1 2
x y z
d


. Điểm
;;M a b c
thuộc đường thng
d
sao cho chu vi tam giác
MAB
nhỏ
nhất. Khi đó biểu thức
23a b c
bằng
A.
5
. B.
7
. C.
9
. D.
3
.
Li gii
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u 109.(Chuyên Quý Đôn) Trong không gian
Oxyz
, cho mt phng
P
:
20xy
hai
đim
1;2;3A
,
1;0;1B
. Điểm
; ; 2C a b P
sao cho tam giác
ABC
din tích nh nht.
Tính
ab
A. 0. B.
3
. C. 1. D. 2.
Li gii
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u 110.(THPT Nguyn Khuyến)Trong không gian
Oxyz
, cho tam giác
ABC
vi
1;2;5A
,
3;4;1B
,
2;3; 3C
. Gi
G
là trng tâm tam giác
ABC
và
M
là điểm thay đổi trên
mp Oxz
. Độ dài
GM
ngn nht bng
A.
2
. B.
3
. C.
4
. D.
1
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 111.(THPT Nguyễn Trung Thiên) Trong không gian với hệ tọa độ
Oxyz
, cho tam giác
ABC
1;0;0A
,
0;0;1B
,
2;1;1C
. Din tích tam giác
ABC
bằng:
A.
11
2
. B.
7
2
. C.
6
2
. D.
5
2
.
Li gii
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u 112.(THPT Kim Liên2017) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:2 3 3 0P x y z
. Gọi
M
,
N
lần lượt là giao điểm của mặt phẳng
P
vi các trục
Ox
,
Oz
. Tính diện tích tam giác
OMN
.
A.
9
4
. B.
9
2
. C.
3
2
. D.
3
4
.
Li gii
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u 113.(THPT Việt ĐứcNội)Trong hệ trục tọa đ
Oxyz
, cho bốn điểm
0; 2;1 ;A
1;0; 2 ; 3;1; 2 ; 2; 2; 1B C D
. Câu nào sau đây sai?
A. Bốn điểm
, , ,A B C D
không đồng phẳng. B. Tam giác
ACD
là tam giác vuông tại
A
.
C. Góc giữa hai véctơ
AB
CD
là góc tù. D. Tam giác
ABD
là tam giác cân tại
B
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 114.(THPT Nghèn)Trong không gian
Oxyz
, cho các điểm
1;1;2 ; 0; 1; 3AB
. Xét điểm
M
thay đổi trên mặt phẳng
Oxz
, giá trị nhỏ nhất của
23OM MA MB
bằng?
A.
1
. B.
3
2
. C.
1
2
. D.
1
4
.
Li gii
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u 115. Trong không gian vi h trc
Oxyz
, cho tam giác
ABC
vi
2;0; 3A
;
1; 2; 4B 
;
2; 1;2C
. Biết điểm
;;E a b c
điểm để biu thc
P EA EB EC
đạt giá tr nh nht.
Tính
T a b c
A.
3T
. B.
1T
. C.
0T
. D.
1T 
.
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u 116.(Nam Tiền Hải Thái Bình)Trong không gian với hệ trục toạ độ
Oxyz
, cho hai điểm
,
9; 7;2B
. Tìm trên trục
Ox
toạ độ điểm
M
thỏa
22
MA MB
đạt giá trị nhỏ nhất.
A.
5;0;0M
. B.
2;0;0M
. C.
4;0;0M
. D.
9;0;0M
.
Li gii
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u 117.(Triu Thái Vĩnh Phúc)Trong không gian
Oxyz
, cho hai điểm
1;2;1 , 2; 1;3AB
đim
; ;0M a b
sao cho
22
MA MB
nh nht. Giá tr ca
ab
bng
A.
3.
B.
2.
C.
1.
D.
2.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Hệ Tọa Độ Trong Không Gian
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 118.(Gang Thép Thái Nguyên) Trong không gian vi h trc tọa độ
Oxyz
, cho
4
đim
2;4; 1A
,
1;4; 1B
,
2;4;3C
,
2;2; 1D
, biết
;;M x y z
để
2 2 2 2
MA MB MC MD
đạt
giá tr nh nht thì
x y z
bng
A.
6
. B.
21
4
. C.
8
. D.
9
.
Li gii
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u 119.Trong không gian
Oxyz
, cho ba điểm
1;2;2A
,
3; 1; 2B 
,
4;0;3C
. Tìm ta
độ đim
I
trên mt phng
Oxz
sao cho biu thc
25IA IB IC
đạt giá tr nh nht.
A.
37 19
;0;
44
I



. B.
27 21
;0 ;
44
I



. C.
37 23
;0 ;
44
I



. D.
25 19
;0 ;
44
I



.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Tnh Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
A. LÍ THUYẾT GIÁO KHOA.
I. VÉC PHÁP TUYẾN:
1. Định nghĩa:
Cho mặt phẳng
. Véc tơ
0n
gọi là véc tơ pháp tuyến
(VTPT) của
mp
nếu giá của
n
vuông góc với
,
Kí hiệu
n
.
2. C ý:
Nếu
n
là VTPT của
thì
. ( 0)k n k
cũng là VTPT của
. Vậy
mp
có vô số VTPT.
Nếu hai véc tơ
,ab
(không cùng phương) có giá song song (hoặc nằm trên)
mp
thì
là một VTPT của
mp
.
Nếu ba điểm
,,A B C
phân biệt không thẳng hàng thì
véc tơ
,n AB AC


là một VTPT của
mp ABC
.
II. Phương trình tổng qt của mặt phẳng :
1. Phương trình tổng quát.
Cho
mp
đi qua
0 0 0
;;M x y z
, có
;;n A B C
là một VTPT .
Khi đó phương trình tổng quát của () có dạng:
0 0 0
0A x x B y y C z z
.
Nếu
:0Ax By Cz D
thì
;;n A B C
là một VTPT của ().
Nếu
;0;0 , 0; ;0 , 0;0;A a B b C c
;
0abc
thì phương
trình của
ABC
có dạng:
1
x y z
a b c
và được gọi là
phương trình theo đoạn chắn của ().
Ví d 1. Lập phương trình mặt phng
P
biết:
a).
P
đi qua
1;2;3 , 4; 2; 1 , 3; 1;2A B C
;
b).
P
là mt phng trung trực đoạn
AC
( Vi
,AC
câu 1);
c).
P
đi qua
0;0;1 , 0;2;0MN
và song song vi
AB
;
d).
P
đi qua các hình chiếu ca
A
lên các trc tọa độ.
Lời giải
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z
x
y
(
α
)
n
H
M
0
O
C
B
P
n
P
A
C
0;0;c
( )
A
a;0;0
( )
B
0;b;0
( )
z
y
x
O
§BI 2. PHƯƠNG TRÌNH MT PHNG
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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III. Vị trí tương đối của hai mặt phẳng :
Cho hai
:0mp P Ax By Cz D
: ' ' ' ' 0Q A x B y C z D
P
cắt
Q
: : ': ': 'A B C A B C
.
//
' ' ' '
A B C D
PQ
A B C D
' ' ' '
A B C D
PQ
A B C D
' ' ' 0P Q AA BB CC
.
Ví d 2. Xét v trí tương đối ca mi cp mt phng sau cho bởi các phương trình sau.
a).
2 4 0x y z
10 10 20 40 0.x y z
b).
3 2 3 5 0x y z
9 6 9 5 0.x y z
c).
10x y z
2 2 2 2 0.x y z
d).
2 3 0x y z
2 4 2 0.x y z
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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IV. Khoảng cách từ một điểm đến một mặt phẳng:
Khoảng cách từ
0 0 0
;;M x y z
đến mp
:0P Ax By Cz D
là:
0 0 0
2 2 2
,
Ax By Cz D
d M P
A B C

.
Ví d 3. Lập phương trình
P
biết
P
song song vi
:2 3 6 14 0Q x y z
và khong
cách t
O
đến
P
bng
5
.
Lời giải
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B. PHÂN DNG D MINH HA.
DNG 1. Lập phương trình mt phng khi biết mt điểm
0 0 0
;;M x y z
và một vécpháp tuyến
1. Phương pháp chung.
Để lập phương trình của mt
P
ta cn tìm một điểm mà
P
đi qua và mt VTPT ca
P
.
Khi tìm VTPT ca
P
chúng ta cần lưu ý một s tính cht sau :
Nếu giá của hai véc tơ không cùng phương
,ab
có giá song song hoc nm trên
P
thì
,n a b


là mt VTPT ca
P
.
Nếu hai mt phng song song vi nhau thì VTPT ca mt phẳng này cũng VTPT ca mt
phng kia.
Nếu
P
cha (hoc song song) vi
AB
thì giá của véc tơ
AB
s nm trên (hoc song
song) vi
P
.
Nếu
PQ
thì VTPT ca mt phng y s giá nm trên hoc song song vi mt
phng kia.
Nếu
P AB
thì
AB
là mt VTPT ca
P
.
2. Các trường hợp đặc biệt
Mặt phẳng () đi qua ba điểm không trùng với gốc tọa độ
;0;0 , 0; ;0 ,A a B b
0;0;Cc
phương trình
1.
x y z
a b c
Các mặt phẳng tọa độ
: 0, : 0, : 0.Oyz x Ozx y Oxy z
Mặt phẳng () qua gốc tọa độ
0.Ax By Cz
Mặt phẳng () song song
( 0)D
hoặc chứa
( 0)D
trục
Ox
có dạng :
0.By Cz D
Mặt phẳng () song song
( 0)D
hoặc chứa
( 0)D
trục
Oy
có dạng :
0.Ax Cz D
Mặt phẳng () song song
( 0)D
hoặc chứa
( 0)D
trục
Oz
có dạng :
0.Ax By D
Mặt phẳng () song song
( 0)D
với mặt phẳng
Oxy
có phương trình là:
0.Cz D
Mặt phẳng () song song
( 0)D
với mặt phẳng
Oyz
có phương trình là:
0.Ax D
Mặt phẳng () song song
( 0)D
với mặt phẳng
Ozx
có phương trình là:
0.By D
3. Bài toán tng quát và bài tp minh ha.
Bài toán 1. Phương trình mt phng
P
đi qua điểm
M
song song vi mt phng
cho trước.
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54
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Phương pháp.
Mt phng
P
song song vi mt phng
nên VTPT
ca
P
chính là VTPT ca mt phng
.
T đó viết phương trình mặt phng
P
qua
M
VTPT
nn
Bài tp 1. Viết phương trình mặt phng
P
qua
1;2;3M
song song vi mt phng
:2 3 2 1 0Q x y z
.
Lời giải
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Bài toán 2. Phương trình mặt phng
P
đi qua đim
M
vuông góc vi 2
mp Q
mp R
.
Phương pháp.
Mt phng
P
vuông góc vi mt phng
Q
và mt
phng
R
nên
,
PQ
QR
PR
nn
n n n
nn


vi
,,
P Q P
n n n
ln
t là VTPT ca mt phng
,,P Q R
Phương trình mặt phng
P
qua
M
có VTPT
P
n
Bài tp 2. Viết phương trình mặt phng
P
qua
1; 1;2
và vuông góc vi 2 mt phng
: 3 1 0; :2 1 0Q x z R x y z
Lời giải
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Bài toán 3. Phương trình mặt phng
P
đi qua hai điểm
,AB
vuông góc vi mt phng
Q
1. Phương pháp.
Gi
,
Q
nn
lần lượt là VTPT ca mp
P
và mt phng
Q
Vì mt phng
P
đi qua A, B và mp
P
vuông góc vi mt
phng
Q
nên
,
Q
Q
nn
n AB
n AB
.
T đó viết phương trình mặt phng
P
Bài tp 3. Viết phương trình mặt phng
P
đi qua hai điểm
0;1;0A
1;2; 2B
vuông
góc vi mt phng
:2 3 13 0Q x y x
Lời giải
n
P
M
x
0
;
y
0
;
z
0
n
α
P
α
n
P
n
Q
R
Q
P
M
x
0
;
y
0
;
z
0
n
R
P
n
Q
Q
A
B
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Bài toán 4. Viết phương trình mặt phng
P
đi qua 3 điểm
,,A B C
cho trước
1. Phương pháp.
Gi
n
là VTPT ca mt phng
P
.
Vì mp
P
đi qua
,,A B C
nên
,
n AB
n AB AC
n AC


.
Phương trình mặt phng
P
qua
M
có VTPT là
n
Bài tp 4. Viết phương trình mặt phng
P
qua
1;0;1 , 0;2;0 , 0;1;2A B C
Lời giải
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Bài toán 5. Viết phương trình mặt phng
P
đi qua giao tuyến ca 2 mt phng
,QR
dng
:0
: ' ' ' ' 0
Q Ax By Cz D
R A x B y C z D
tha mãn c gi thiết đi qua đim
M
hoc song song
vi mt phng hoc vuông góc vi mt phng.
1. Phương pháp.
Trường hp 1:
mp P
đi qua giao tuyến
điểm
.M
Mọi điểm thuc giao tuyến có tọa độ là nghim ca h
gm
2
phương trình của mt phng
Q
R
:0
: ' ' '
1
'0
Q Ax By Cz D
R A x B y C z D
T h
1
chn ra
2
đim
,AB
thuc giao tuyến sau đó
viết phương trình mặt phẳng qua điểm
,,A B M
như dạng 4.
Trường hp 2:
mp P
đi qua giao tuyến
song song vi
.mp
Nếu
mp P
song song vi
1 1 1 1
:0mp A x By Cx D
và đi qua hai giao tuyến mt phng thì
1 1 2
;;
p
n n A B C

Khi đó,
1 1 1
:0mp P A x By Cx d
,
1
.dD
Tìm
d
bằng cách thay hai điểm
,AB
vào phương trình
mp P
và gii h.
P
n
P
A
B
C
B
A
R
Q
n
P
P
M
n
α
α
P
n
P
Q
R
A
B
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Trường hp 3:
mp P
đi qua giao tuyến
song song vi
.mp
Nếu
mp P
vuông góc vi
1 1 1 1
:0mp A x By Cx D
và đi qua hai giao tuyến mt phng thì
mp P
nhận véctơ
1 1 2
;;n A B C
làm một véc tơ có giá song song hoặc nm
trên
mp P
.
Mà
mp P
đi qua giao tuyến
nên đi qua hai điểm
,AB
Suy ra
mp P
có 1 cặp véctơ nên
,
p
n n AB


Bài tp 5.
a). Viết phương trình mặt phng
P
qua
2;0;1M
và giao tuyến 2 mt phng
: 2 4 0; :2 4 0R x y z Q x y z
b). Viết phương trình mặt phng
P
qua giao tuyến 2 mt phng
: 2 4 0;R y z
: 3 0Q x y z
và song song vi mt phng
: 2 0.x y z
c). Viết phương trình mặt phng
P
qua giao tuyến 2 mt phng
:3 2 0;R x y z
: 4 5 0Q x y
và vuông góc vi mt phng
:2 7 0.xz
Lời giải
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n
α
α
B
A
R
Q
P
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Bài toán 6. Viết phương trình mặt phng
P
đi qua 3 điểm
;0;0 , 0; ;0 , 0;0;b cBCaA
tha
mãn điều kiện cho trước .
1. Phương pháp.
S dụng phương pháp mặt phẳng đoạn chn :
Mặt phẳng
P
đi qua các điểm
;0;0 , 0; ;0 , 0;0;A a B b C c
;
0abc
thì phương trình của
ABC
có dạng:
1
x y z
a b c
S dụng điều kin ca gi thiết để tìm
,,abc
.
Bài tp 6. Lập phương trình mặt phng
đi qua điểm
1;9;4M
ct các trc tọa độ ti
các điểm
,,A B C
(khác gc tọa độ) sao cho
1).
M
là trc tâm ca tam giác
.ABC
2). Khong cách t gc tọa độ
O
đến mt phng
là ln nht.
3).
.OA OB OC
4).
8 12 16 37OA OB OC
0, 0.
AC
xz
Lời giải.
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C
0;0;c
( )
A
a;0;0
( )
B
0;b;0
( )
z
y
x
O
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Bài tp 7. Lập phương trình mặt phng
đi qua
1;4;9M
sao cho
ct các tia
,Ox Oy
,Oz
lần lượt tại 3 điểm
,,A B C
tha:
1).
M
là trng tâm tam giác
ABC
,
2). T din
OABC
có th tích nh nht,
3). Khong cách t
O
đến
ABC
ln nht,
4).
4OA OC OB
9OA OB
.
Lời giải.
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4. Bài tpn luyn.
Bài 1. Lập phương trình của
P
trong các trương hợp sau:
1).
P
đi qua
1;2;1A
và song song vi
: 3 1 0Q x y z
;
2).
P
đi qua
0;1;2 , 0;1;1 , P 2;0;0MN
;
3).
P
là mt phng trung trc của đoạn
MN
(vi
,MN
ý 2) ;
4).
P
đi qua các hình chiếu ca
(1;2;3)A
lên các trc tọa độ ;
5).
P
đi qua
1;2;0 , 0;2;0BC
và vuông góc vi
: 1 0R x y z
;
6).
P
đi qua
1;2;3D
và vuông góc vi hai mt phng :
: 2 0x

;
: 1 0yz
.
Lời giải
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Bài 2. Lập phương trình mặt phng
, biết:
1).
đi qua
2;3;1M
và song song vi mt phng
: 2 3 1 0P x y z
;
2).
đi qua
2;1;1 , 1; 2; 3AB
và () vuông góc vi
:0x y z
;
3).
cha trc
Ox
và vuông góc vi
:2 3 2 0Q x y z
.
4).
đi qua giao tuyến ca hai mt phng
P
Q
, đồng thi
vuông góc vi mt
phng
:3 2 5 0x y z
.
Lời giải
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5. Câu hi trc nghim:
Mức độ 1. Nhận biết
u 1.(THPT Nguyễn Đức Cảnh) Trong không gian
Oxyz
,
( ): 3 0mp P x y z
.
Hỏi
()mp P
đi qua điểm nào dưới đây?
A.
1;1; 1M
. B.
1; 1;1N 
. C.
1;1;1P
. D.
1;1;1Q
.
Lời giải
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u 2.(Đặng Thành Nam) Trong không gian
Oxyz
, mặt phẳng
( ): 3 0P x y z
đi qua điểm
nào dưới đây?
A.
1; 1; 1 .M 
. B.
1;1;1 .N
C.
3;0;0P
. D.
0;0; 3Q
.
Lời giải
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u 3.(THPT Kim Liên 2018) Trong không gian vi h tọa độ
Oxyz
, cho
:1
2 1 3
x y z
mp P
,
véc tơ nào dưới đây là một véc tơ pháp tuyến ca mt phng
P
.
A.
1
3;6;2n
. B.
3
3;6;2n 
. C.
2
2;1;3n
. D.
4
3;6; 2n
Li gii
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u 4.(THPT Mê Linh Hà Nội) Một véc-tơ pháp tuyến của mặt phẳng
:2 2 3 0x y z
A.
4;2; 4n 
. B.
2;1; 2n
. C.
1; 2;1n 
. D.
2;1;2n
.
Li gii
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u 5.(THPT Ngô Sỹ Liên 2019) Mặt phẳng
:1
2 3 2
x y z
P
có một vectơ pháp tuyến là:
A.
3;2;3n
. B.
2;3; 2n 
. C.
2;3;2n
. D.
3;2; 3n 
.
Li gii
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u 6.(Chuyên Thánh Tôn 2019) Vectơ
1; 4;1n
là một vectơ pháp tuyến ca mt phng
nào dưới đây?
A.
4 3 0x y z
. B.
4 1 0x y z
. C.
4 2 0x y z
. D.
4 1 0x y z
.
Li gii
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u 7.(THPT Nghĩa ng Nam Định) Trong không gian
Oxyz
, một vectơ pháp tuyến của mặt
phẳng
1
2 1 3
x y z

A.
(3;6; 2)n 
B.
(2; 1;3)n 
C.
( 3; 6; 2)n
D.
( 2; 1;3)n
Li gii
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u 8.(THPT Chuyên Trần Đại Nghĩa) Trong không gian với hệ trục độ
Oxyz
, cho ba điểm
1; 2;1A
,
1;3;3B
,
2; 4;2C
. Một véc tơ pháp tuyến
n
của mặt phẳng
ABC
là:
A.
1
( 1;9;4)n 
. B.
4
(9;4; 1)n
. C.
3
(4;9; 1)n
. D.
2
(9;4;11)n
.
Li gii
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u 9.(THPT Thuận Thành Bắc Ninh 2019) Trong không gian vi h trc tọa độ
Oxyz
, phương
trình mt phng đi qua điểm
1;2; 3A
có vectơ pháp tuyến
2; 1;3n 
là :
A.
2 3 9 0x y z
. B.
2 3 4 0x y z
. C.
2 4 0xy
. D.
2 3 4 0x y z
.
Li gii
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u 10.(Chuyên Đại Hc Vinh) Trong không gian , cho hai điểm .
Gi là mt phng trung trc ca . Một vectơ pháp tuyến ca có tọa độ
A. . B. . C. . D. .
Li gii
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u 11.(THPT Chuyên Lê Hng Phong 2019) Trong không gian vi h tọa độ
Oxyz
, cho điểm
3 1 3A ;;
,
1 3 1B ;;
P
là mt phng trung trc của đoạn thng
AB
. Một vectơ pháp
tuyến ca
P
có tọa độ là:
A.
1 3 1 ;;
. B.
1 1 2 ;;
. C.
3 1 3;;
. D.
1 2 1;;
.
Li gii
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Oxyz
2; 1;3A 
0;3;1B
AB
2;4; 1
1;2; 1
1;1;2
1;0;1
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u 12.(Sở GD ĐT Quãng Bình 2019) Trong không gian
Oxyz
, cho hai điểm
2; 1; 3A 
0; 3; 1B
. Gọi
mặt phẳng trung trực của đoạn
AB
. Một vectơ pháp tuyến của
tọa
độ là:
A.
2; 4; 1 .n 
B.
1; 0; 1 .n
C.
1; 1; 2 .n 
D.
1; 2; 1 .n 
Lời giải
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u 13.(Sở GD & ĐT Cà Mau) Trong không gian
Oxyz
, cho hai điểm
1;5; 2A
,
3;1;2B
. Viết
phương trình mặt phẳng trung trực của đoạn thẳng
AB
.
A.
2 3 4 0xy
. B.
2 2 8 0x y x
. C.
2 2 8 0x y z
. D.
2 2 4 0x y z
.
Ligii
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u 14.(THPT Chuyên Ti Bình 2019) Trong không gian
Oxyz
, cho hai điểm
1;3; 4A
1;2;2B
. Viết phương trình mặt phẳng trung trực
của đoạn thẳng
AB
.
A.
: 4 2 12 7 0x y z
. B.
: 4 2 12 17 0x y z
.
C.
: 4 2 12 17 0x y z
. D.
: 4 2 12 7 0x y z
.
Li gii
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u 15.(Gang Thép Ti Nguyên 2019) Trong không gian vi h trc tọa độ
Oxyz
, cho điểm
0;1; 1A
điểm
2;1; 3B
. Phương trình nào sau đây là phương trình mt phng trung trc
của đoạn thng
AB
?
A.
2 3 0xy
. B.
2 3 0xy
. C.
30x y z
. D.
2 3 0xz
.
Li gii
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u 16.(THPT Chuyên Quc Hc Huế 2019) Trong không gian hệ tọa độ
Oxyz
, cho hai điểm
1;3; 4 , 1;2;2AB
. Phương trình mặt phẳng trung trực của đoạn
AB
là?
A.
4 2 12z 7 0xy
. B.
4 2 12z 7 0xy
.
C.
4 2 12z 17 0xy
. D.
4 2 12z 17 0xy
.
Li gii
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u 17.(THPT Nông Cng 2019) Trong không gian
Oxyz
, cho điểm
(1; 2;3), (3;0; 1)AB
.
Mt phng trung trc của đoạn thng
AB
có phương trình
A.
10x y z
. B.
2 1 0x y z
. C.
2 1 0x y z
. D.
2 7 0x y z
.
Li gii
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u 18.(Sở GD và ĐT Kiên Giang 2019) Trong không gian
Oxyz
, cho hai điểm
1;1;0A
,
2; 1;1B
. Một vectơ pháp tuyến
n
của mặt phẳng
OAB
(Với
O
là gốc tọa độ) là
A.
3;1; 1n
. B.
1; 1; 3n
. C.
1; 1;3n 
. D.
1;1;3n
.
Li gii
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u 19.(THPT Ngô Quyền Nội) Toạ đmột vectơ pháp tuyến của mặt phẳng
đi qua ba
điểm
2;0;0M
,
0; 3;0N
,
0;0;4P
A.
2; 3;4
. B.
6;4; 3
. C.
6; 4;3
. D.
6;4;3
.
Li gii
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u 20.(S GD ĐT Đin Biên) Cho không gian
Oxyz
, viết phương trình đoạn chắn mặt phẳng
đi qua điểm
2,0,0 ; 0, 3,0 ; 0,0,2A B C
A.
1
2 3 2
x y z
. B.
1
2 3 2
x y z
. C.
1
3 2 2
x y z
. D.
1
2 2 3
x y z
.
Li gii
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u 21.(Gang Thép Thái Nguyên) Trong không gian vi h trc tọa độ
Oxyz
, mt phng qua các
đim
1;0;0A
,
0;3;0B
,
0;0;5C
có phương trình là
A.
15 5 3 15 0.x y z
B.
1 0.
1 3 5
x y z
C.
3 5 1.x y z
D.
1.
1 3 5
x y z
Li gii
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u 22.(THPT Thanh Chương 2019) Trong không gian
Oxyz
, mt phẳng đi qua ba điểm
(0; 2;0)A
,
(0;0;3)B
( 1;0;0)C
có phương trình là
A.
3 6 2 6 0x y z
. B.
6 3 2 6 0x y z
.
C.
2 6 3 6 0x y z
. D.
6 3 2 6 0x y z
.
Li gii
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u 23.(THPT Chuyên Sơn-La 2019)Trong không gian
Oxyz
, phương trình mặt phẳng đi qua ba
đim
1;0;0A
,
0; 2;0B
0;0;3C
A.
1
1 2 3
x y z
. B.
1
1 2 3
x y z
. C.
0
1 2 3
x y z
. D.
1
1 2 3
x y z
.
Li gii
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u 24.(THPT m Rồng) Trong mặt phẳng tọa độ
Oxyz
, cho ba điểm
2;0;0M
,
0;1;0N
0;0;2P
. Mặt phẳng
MNP
có phương trình là
A.
1
2 1 2
x y z
. B.
1
2 1 2
x y z
. C.
1
2 1 2
x y z
. D.
0
2 1 2
x y z
.
Li gii
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u 25.(THPT Kinh Dương 2019) Trong không gian với hệ trục tọa độ
Oxyz
.Mặt phẳng
P
đi
qua các điểm
A 1;0;0
,
0;2;0B
,
0;0; 2C
có phương trình là:
A.
2 2 0x y z
. B.
2 2 0x y z
.
C.
2 2 0x y z
. D.
2 2 0x y z
.
Li gii
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u 26.(THPT Chuyên Thái Bình) Trong không gian
Oxyz
, cho điểm
1;2;3M
. Gọi
,,ABC
lần
lượt là hình chiếu vuông góc của điểm
M
lên các trục
,,Ox Oy Oz
. Viết phương trình
.mp ABC
.
A.
1
1 2 3
x y z
. B.
1
1 2 3
x y z
. C.
0
1 2 3
x y z
. D..
1
1 2 3
x y z
.
Li gii
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u 61.(THPT Lý Nhân Tông 2019) Trong không gian vi h tọa độ
Oxyz
, cho điểm
(3;5;2)A
,
phương trình nào dưới đây là phương trình mặt phẳng đi qua các đim là hình chiếu ca
A
trên
các mt phng tọa độ?
A.
3 5 2 60 0x y z
. B.
10 6 15 60 0x y z
.
C.
10 6 15 90 0x y z
. D.
1
3 5 2
x y z
.
Li gii
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u 27.(Chuyên Đại Học Vinh) Trong không gian
Oxyz
, mặt phẳng nào trong các mặt phẳng sau
song song với trục
Oz
?
A.
( ): 0z
. B.
( ): 0P x y
. C.
( ): 11 1 0Q x y
. D.
( ): 1z
.
Li gii
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u 28.(THPT Nguyn Công Tr 2019)Trong không gian vi h trc tọa độ
Oxyz
, mt phng
Oyz
có phương trình là
A.
0z
. B.
0y
. C.
0yz
. D.
0x
.
Li gii
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u 29.(Chuyên Hưng Yên Lần 3) Trong không gian vi h ta đ
,Oxyz
phương trình nào sau đây là
phương trình của mặt phng
Ozx
?
A.
0.x
B.
1 0.y 
C.
0.y
D.
0.z
Li gii
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u 30.(THPT n y Nội 2019) Trong không gian với hệ trục tọa độ
Oxyz
, mặt phẳng
Oxy
có phương trình là
A.
0xy
. B.
0x
. C.
0z
. D.
0y
.
Lời giải
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u 31.(THPT Phú Dc) Trong không gian
Oxyz
, mt phng
Oxz
có phương trình là
A.
0x y z
. B.
0y
. C.
0x
. D.
0z
.
Li gii
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u 32.(THPT Thch Thành 2019) Trong không gian
Oxyz
, mt phng
Oxy
có phương trình:
A.
0x
. B.
0x y z
. C.
0y
. D.
0z
.
Li gii
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u 33.(THPT Nguyễn Đức Cảnh 2019) Trong không gian
Oxyz
trục
Ox
song song với mặt
phẳng có phương trình nào ?
A.
z0x by c d
với
22
( 0)bc
. B.
z =0y
.
C.
z 1 0by c
với
22
( 0)bc
. D.
10x
.
Li gii
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Mức độ 2. Thông Hiểu
u 34.(THPT Nguyn Khuyến)Trong không gian
Oxyz
, mt phng
()P
đi qua điểm
(1;0;2)A
vuông góc với đường thng
12
:
2 1 3


x y z
d
có phương trình là
A.
2 3 8 0 x y z
. B.
2 3 8 0 x y z
. C.
2 3 8 0 x y z
. D.
2 3 8 0 x y z
.
Li gii
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u 35.(THPT Trần Kim Hưng 2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho đường
thẳng
2 2 3
:
1 1 2
x y z
d

điểm
1; 2;3A
.
Mặt phẳng qua
A
vuông góc với đường
thẳng
d
phương trình là:
A.
2 9 0x y z
. B.
2 3 9 0x y z
. C.
2 9 0x y z
. D.
2 3 14 0x y z
Lời giải
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u 36.(THPT Yên Dũng 2019) Mặt phẳng
P
đi qua điểm
1 ; 2 ; 0A
và vuông góc với đường
thẳng
11
:
2 1 1
x y z
d


có phương trình là
A.
2 4 0x y z
. B.
2 4 0x y z
. C.
2 4 0x y z
. D.
2 4 0x y z
.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Li gii
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u 37.(Chuyên ĐH Vinh) Trong không gian , mt phng đi qua đim ,
đồng thi vuông góc vi giá của vectơ có phương trình là
A. . B. . C. . D. .
Li gii
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u 38.(Thanh Chương Ngh An) Trong không gian
Oxyz
, mt phng
P
song song vi mt
phng
Oyz
và đi qua điểm
1;2;3A
có phương trình
A.
1x
. B.
3z
. C.
2y
. D.
60x y z
.
Li gii
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u 39.(Chuyên ĐH Vinh 2019) Trong không gian , mt phng đi qua điểm
đồng thi song song vi mt phng có phương trình là
A. . B. . C. . D. .
Li gii
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u 40.(THPT SỐ Tư Nghĩa 2019) Trong không gian với hệ tọa độ
Oxyz
, gọi
là mặt phẳng đi
qua điểm
2; 1;1A
song song với mặt phẳng
:2 3 2 0Q x y z
. Phương trình mặt
phẳng
là:
A.
4 2 6 8 0x y z
. B.
2 3 8 0x y z
. C.
2 3 8 0x y z
. D.
4 2 6 8 0x y z
Li gii
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u 41.(THPTn Ninh Bình 2019) Trong không gian
Oxyz
, viết phương trình mặt phng
P
đi qua điểm
1;2;3M
và song song vi mt phng
: 2 3 1 0Q x y z
A.
2 3 6 0x y z
. B.
2 3 16 0x y z
. C.
2 3 6 0x y z
. D.
2 3 16 0x y z
.
Li gii
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u 42.(THPT Cm Giàng) Trong không gian vi h trc tọa độ
,Oxyz
mt phẳng đi qua điểm
Oxyz
P
3; 1; 4M
1; 1; 2a
3 4 12 0x y z
3 4 12 0x y z
2 12 0x y z
2 12 0x y z
Oxyz
P
1; 1; 2M
: 2 3 5 0Q x y z
2 3 3 0x y z
2 3 0x y z
2 3 3 0x y z
2 3 0x y z
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
1;3; 2A
và song song vi mt phng
:2 3 4 0P x y z
là:
A.
2 3 7 0x y z
. B.
2 3 7 0x y z
.
C.
2 3 7 0x y z
. D.
2 3 7 0x y z
.
Li gii
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u 43.(THPT Thun Thành 2019) Trong không gian vi h tọa độ
Oxyz
, phương trình nào
ới đây là phương trình của mt phng cha trc
Oy
và điểm
(2;1; 1)K
?
A.
20xz
. B.
20xz
. C.
20xy
. D.
10y 
Li gii
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u 44.(S GD và ĐT Cần Thơ 2019) Trong không gian
Oxyz
, cho hai điểm
3;1; 1A
2; 1;4B
. Phương trình mặt phng
OAB
vi
O
là gc tọa độ
A.
3 14 5 0x y z
. B.
3 14 5 0x y z
. C.
3 14 5 0x y z
. D.
3 14 5 0x y z
.
Li gii
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u 45.(THPT Nghĩang 2019) Trong không gian với hệ tọa độ
Oxyz
, cho ba điểm
0;1;2A
,B 2; 2;1 , 2;1;0C
. Khi đó, phương trình mặt phẳng
ABC
0ax y z d
. Hãy xác định
a
d
.
A.
1, 1ad
. B.
6, 6ad
. C.
1, 6ad
. D.
6, 6ad
.
Li gii
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u 46.(Chuyên Nguyn Du-Đăk Lăk) Trong không gian h tọa độ
Oxyz
, mt phẳng qua ba điểm
1;3;2A
,
2;5;9B
,
3;7; 2C 
có phương trình là
30x ay bz c
. Giá tr
abc
bng
A.
6
. B.
3
. C.
3
. D.
6
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 47.(THPT Ngô Quyn Hi Phòng 2019) Trong không gian
Oxyz
, cho ba điểm
2;0;0A
,
0;3;0B
0;0; 1C
. Phương trình của mt phng
P
đi qua đim
1;1;1D
song
song vi mt phng
ABC
A.
2 3 6 1 0xyz
. B.
3 2 6 1 0x y z
. C.
3 2 5 0x y z
. D.
6 2 3 5 0x y z
.
Li gii
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u 48.(Chuyên T Trng 2019) Trong không gian vi h tọa độ
Oxyz
, cho hai mt phng
( ):3 2 2 7 0P x y z
( ):5 4 3 1 0Q x y z
. Viết phương trình mặt phng
()R
qua điểm
(3;1;5)M
và vuông góc vi c hai mt phng
()P
()Q
.
A.
2 2 4 0x y z
. B.
2 2 5 0x y z
.
C.
2 2 3 0x y z
. D.
2 2 3 0x y z
.
Li gii
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u 49.(THPT Xoay 2019) Trong không gian hệ tọa độ
Oxyz
, phương trình mặt phẳng
P
đi
qua điểm
2;1; 3B
, đồng thời vuông góc với hai mặt phẳng
: 3 0Q x y z
mặt phẳng
:2 0R x y z
là:
A.
4 5 3 22 0x y z
. B.
4 5 3 12 0x y z
.
C.
2 3 14 0x y z
. D.
4 5 3 22 0x y z
.
Lời giải
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u 50.(Chuyên Lam n 2019) Trong không gian với hệ tọa độ
Oxyz
cho hai mặt phẳng
:3 2 2 7 0x y z
:5 4 3 1 0x y z
. Phương trình mặt phẳng đi qua
O
đồng
thời vuông góc với cả
có phương trình là
A.
2 2 1 0x y z
. B.
2 2 0.x y z
C.
2 2 0x y z
. D.
2 2 0x y z
.
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Tnh Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 51.(Chuyên Lý T Trng Cần Thơ)Trong không gian vi h trc tọa độ
Oxyz
cho bn đim
5;1;3 , 1;6;2 , 5;0;4 , 4;0;6A B C D
. Viết phương trình mặt phng
P
đi qua hai điểm
,AB
và song song với đường thng
CD
A.
:10 9 5 70 0P x y z
. B.
:10 9 5 74 0P x y z
C.
:10 9 5 74 0P x y z
D.
:10 9 5 70 0.P x y z
Li gii
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u 52.(THPT Gia Lc Hi Dương 2019) Trong không gian vi h trc tọa độ
Oxyz
, cho hai đim
2;4;1 1;1;3A ,B
mt phng
: 3 2 5 0P x y z
. Lập phương trình mặt phng
Q
đi
qua hai điểm
A
,
B
và vuông góc vi mt phng
P
.
A.
2 3 11 0yz
. B.
2 3 11 0xy
. C.
3 2 5 0x y z
. D.
3 2 11 0yz
.
Li gii
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u 53.(THPT Chuyên Sơn La Ln 2019) Trong không gian h tọa độ
Oxyz
, mt phng
P
đi
qua hai điểm
0;1;0A
,
2;3;1B
và vuông góc vi mt phng
: 2 0xyQ z
có phương
trình là
A.
4 3 2 3 0x y z
. B.
4 3 2 3 0x y z
. C.
2 3 1 0x y z
. D.
4 2 1 0x y z
.
Li gii
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u 54.(Đại Học KHTN Hà Ni) Trong không gian
Oxyz
, mặt phẳng
R
qua
1;2; 1A
và vuông
góc với mặt phẳng
:2 3 2 0P x y z
;
: 2 0Q x y z
có phương trình là
A.
2 1 0x y z
. B.
4 3 5 0x y z
. C.
4 1 0x y z
. D.
20x y z
.
Li gii
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u 55.(THPT Gia Lc 2019) Trong không gian với hệ tọa độ
Oxyz
, viết phương trình của mặt
phẳng
P
đi qua điểm
2;1; 3M
, đồng thời vuông góc với hai mặt phẳng
: 3 0Q x y z
,
:2 0R x y z
A.
2 3 14 0x y z
. B.
4 5 3 22 0x y z
.
C.
4 5 3 22 0x y z
. D.
4 5 3 12 0x y z
.
Li gii
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u 56.(HSG S GD Và ĐT Bc Ninh) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:P
10x y z
hai điểm
1; 1;2 ; 2;1;1AB
. Mặt phẳng
Q
chứa
,AB
vuông góc với
mặt phẳng
P
, mặt phẳng
Q
có phương trình là:
A.
3 2 3 0x y z
. B.
20x y z
. C.
3 2 3 0x y z
. D.
0xy
.
Li gii
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u 57.(S GD Đào Tạo Hưng Yên) Trong không gian vi h tọa độ
Oxyz
, viết phương trình
mt phng
P
đi qua hai điểm
2;1;1A
,
1; 2; 3B
vuông góc vi mt phng
Q
:
0x y z
.
A.
0x y z
. B.
30xy
. C.
10xy
. D.
40x y z
.
Li gii
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u 58.(Sở GD và Đào Tạo Bình Thuận 2019) Trong không gian hệ tọa độ
,Oxyz
cho hai điểm
3; 1;1 , 1;2;4 .AB
Viết phương trình
mp P
đi qua
A
và vuông góc với đường thẳng
.AB
A.
:2 3 3 16 0.P x y z
B.
:2 3 3 6 0.P x y z
C.
: 2 3 3 6 0.P x y z
D.
: 2 3 3 16 0.P x y z
Li gii
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u 59.(THPT Nguyn Trãi Hi Dương) Mt phng
P
đi qua
3;0;0 , 0;0;4AB
và song song
vi trc
Oy
có phương trình là
A.
4 3 12 0xz
. B.
3 4 12 0xz
. C.
4 3 12 0xz
. D.
4 3 0xz
.
Li gii
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u 60.(THPT Lương Thế Vinh 2019) Viết phương trình mặt phẳng
()P
đi qua điểm
(0; 1;2)A
,
song song với trục
Ox
và vuông góc với mặt phẳng
( ): 2 2 1 0Q x y z
.
A.
( ): 2 2 1 0.P y z
B.
( ): 1 0P y z
. C.
( ): 3 0P y z
. D.
( ):2 2 0P x z
.
Li gii
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u 61.(Chuyên Nguyn Du) Trong không gian
Oxyz
, biết mt phng
50ax by cz
qua hai
đim
3;1; 1A
,
2; 1;4B
và vuông góc vi
:2 3 4 0P x y z
. Giá tr ca
a b c
bng
A.
9
. B.
12
. C.
10
. D.
8
.
Li gii
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u 62.(THPT Phan Đình Tùng Tĩnh) Trong không gian
Oxyz
, cho ba điểm
1;2;4M
;
0;1;2N
;
2;1;3P
và mặt phẳng
:0x Ay Bz C
. Biết
song song với
OP
và đi qua
hai điểm
M
,
N
. Giá trị của biểu thức
A B C
A.
1
. B.
1
. C.
5
. D.
0
.
Li gii
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u 63.(THPT Chuyên Lê Hng Phong) Trong không gian vi h tọa độ
Oxyz
, biết mt phng
24 0ax by cz
qua
1;2;3A
vuông góc vi hai mt phng
:3 2 4 0P x y z
,
:5 4 3 1 0Q x y z
. Giá tr
abc
bng
A.
8
. B.
9
. C.
10
. D.
12
.
Li gii
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u 64.(Chuyên ĐH Vinh) Trong không gian vi h tọa độ , cho hai mt phng có phương
trình , . Mt phng vuông góc vi c đồng
thi ct trc tại điểm có hoành độ bằng 5. Phương trình của mp là:
A.
. B.
. C.
. D. .
Li gii
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Oxyz
( ): 2 3 2 0P x y z
( ): 3 0Q x y
()P
()Q
Ox
3 3 15 0x y z
30x y z
2 6 0xz
2 6 0xz
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Tnh Mặt Phẳng
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u 65.(Chuyên ĐH Vinh 2019) Trong không gian vi h tọa độ , cho hai mt phng ln
ợt có phương trình , . Mt phng vuông góc vi c
đồng thi ct trc tại điểm có hoành độ bằng 3. Phương trình của mp là:
A.
. B.
. C.
. D. .
Li gii
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DNG 2. Lập phương trình mt phng
khi biết một điểm
0 0 0
;;M x y z
, khong cách , góc và
chưa có véc tơ pp tuyến .
1. Phương pháp:
Gọi
;;n A B C
là véc tơ pháp tuyến của mặt phẳng
2 2 2
, 0.A B C
Phương trình mặt phẳng
đi qua điểm
0 0 0
;;M x y z
và có véc tơ pháp tuyến
;;n A B C
Có dạng :
0 0 0
0 (1)A x x B y y C z z
Căn cứ vào giả thiết
n
ẩn
,,A B C
…thì có
1n
phương trình.
Khoảng cách hai điểm
, , , , ,
A A A B B B
A x y z A x y z
2 2 2
B A B A B A
AB x x y y z z
Khoảng cách từ
0 0 0
;;M x y z
đến
:0mp P Ax By Cz D
là:
0 0 0
2 2 2
,
Ax By Cz D
d M P
A B C

.
Diện tích tam giác
ABC
1
,
2
S AB AC


Góc của hai mặt phẳng
,PQ
lần lượt có véctơ pháp tuyến
1
;;n a b c
2
;;n a b c
12
12
2 2 2 2 2 2
12
.
. . .
cos ,
.
.
nn
a a b b c c
nn
nn
a b c a b c

Đặc biệt:
PQ
12
. 0 . . . 0n n a a bb c c
Nếu
1
n
song song
2
n
hoặc cùng phương với nhau thì
12
:.
abc
k n k n
abc
2. Bài tp minh ha:
Bài tp 8. Lập phương trình mặt phng
, biết:
1).
đi qua
1;1;1 , 3;0;2AB
và khong cách t
1;0; 2C
đến () bng
2
;
2).
cách đều hai mt phng
:2 2 1 0, : 2 2 4 0P x y z Q x y z
3). Viết phương trình mặt phng
P
cha trc
Oz
to vi mp
:2 11 3 0Q x y z
mt
góc
60

.
Oxyz
( ): 3 2 1 0P x y z
( ): 2 0Q x z
()P
()Q
Ox
30x y z
30x y z
2 6 0xz
2 6 0xz
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii.
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Bài tp 9. Trong không gian
Oxyz
cho bốn điểm
1;2;3 , 2;3; 1 , 0;1;1A B C
,
4; 3;5D 
.
Lập phương trình mặt phng
biết:
1). () đi qua
A
và cha
Ox
2). () đi qua
,AB
và cách đều hai điểm
,CD
.
Li gii.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 10. Lập phương trình (P) biết (P):
1). Song song vi
:2 3 6 14 0Q x y z
và khong cách t
1; 2;3I
đến
P
bng
2
.
2). Đi qua giao tuyến ca hai mp
: 3 2 0xz
;
( ): 2 1 0yz
và khong cách t
1
0;0;
2
M



đến
P
bng
7
63
.
3). Lập phương trình mặt phng
P
đi qua
O
, vuông góc vi
:0Q x y z
và cách điểm
1;2; 1M
mt khong bng
2
.
Li gii.
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Bài tp 11. Viết phương trình mặt phng
()
biết:
1). () đi qua
1; 1;1 , 2;0;3AB
và () song song vi
Ox
,
2). () đi qua
3;0;1 , 6; 2;1MN
và () to vi
Oyz
mt góc
tha
2
cos
7
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii.
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Bài tp 12. Lập phương trình mặt phng () biết
1). () qua hai điểm
1;2; 1 , 0; 3;2AB
và vuông góc vi
: 2 1 0.P x y z
2). () cách đều hai mt phng
: 2 2 2 0, : 2 2 3 0.x y z x y z

3). () qua hai điểm
1;0;2 , 1; 2;3CD
và khong cách t gc tọa độ ti mp () là
2
.
4). () đi qua
0;1;1E
11
, 2; , ,
7
d A d B


trong đó
1;2; 1 , 0; 3;2 .AB
Li gii.
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Bài tp 13. Tìm
,mn
để 3 mt phẳng sau cùng đi qua một đường thng:
: 2 0P x my nz
,
: 3 1 0Q x y z
: 2 3 1 0R x y z
.
Khi đó hãy viết phương trình mặt phng () đi qua đường thng chung đó tạo vi
()P
mt
góc
sao cho
23
cos
679
.
Li gii.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 14. Lập phương trình mặt phng
biết
1). () đi qua
1;0;2 , 2; 3;3AB
và to vi mt phng
:4 3 0x y z
mt góc
0
60
.
2).
đi qua
2; 3;5 ,C
vuông góc vi
: 5 1 0P x y z
và to vi mt phng
:2 2 3 0Q x y z
góc
0
45
.
Li gii.
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3. Câu hi trc nghim:
Mức độ 3. Vận dụng
u 66.(THPT Nguyễn Du 2019) Trong không gian hệ trục tọa độ
,Oxyz
cho các điểm
0;1;2A
, 2; 2;1 , 2;0;1BC
. Phương trình mặt phẳng đi qua 3 điểm
,,A B C
0ax by cz d
với
22
21a b c
0.a
Khi đó
a b c d
bằng:
A.
2
. B.
4
. C.
5
. D.
3
.
Lời giải
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u 67.(THPT C Loa Hà Ni 2019) Trong không gian
Oxyz
, mt phng
: 27 0P ax by cz
qua hai điểm
3;2;1A
,
3;5;2B
vuông góc vi mt phng
:3 4 0Q x y z
. Tính tng
S a b c
.
A.
2S
. B.
12S 
. C.
4S 
. D.
2S 
.
Lời giải
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u 68.Trong không gian với hệ tọa độ
Oxyz
, cho hain điểm
2;4;1 ; 1;1;3AB
và mặt phẳng
: 3 2 5 0P x y z
. Một mặt phẳng
Q
đi qua hai điểm
,AB
và vuông góc với mặt phẳng
P
có dạng
11 0ax by cz
. Khẳng định nào sau đây là đúng?
A.
5abc
. B.
15abc
. C.
5abc
. D.
15abc
.
Lời giải
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
84
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 69.(THPT Kim Liên 2018) Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
3;0;0M
,
2;2;2N
. Mt phng
P
thay đổi đi qua hai điểm
M,N
ct trc
Oy, Oz
lần lượt ti
0; ;0Bb
,
0;0;Cc
,
0b
,
0c
. H thức nào dưới đây là đúng?
A.
6b+c=
. B.
3bc= b+c
. C.
bc=b+c
. D.
1 1 1
6
+=
bc
.
Lời giải
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u 70.(Kênh Truynnh Giáo Dc Quc Gia 2019) Trong không gian h tọa độ
Oxyz
cho đim
1;2;3M
. Viết phương trình mt phng
P
đi qua
M
ct các trc
,,Ox Oy Oz
ln t ti
,,A B C
sao cho
M
trng tâm tam giác
ABC
.
A.
:6 3 2 18 0P x y z
. B.
:6 3 2 6 0P x y z
.
C.
:6 3 2 18 0P x y z
. D.
:6 3 2 6 0P x y z
.
Lời giải
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u 71.(THPT Chuyên Quc Hc Huế 2019) Trong không gian
Oxyz
, cho điểm
1;4;3G
. Viết
phương trình mặt phng ct các trc tọa độ
, ,Ox Oy Oz
lần lượt ti
, , A B C
sao cho
G
trng
tâm t din
OABC
.
A.
1
3 12 9
x y z
. B.
1.
4 16 12
x y z
C.
3 12 9 78 0x y z
. D.
4 16 12 104 0x y z
Lời giải
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
85
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 72.(THPT Nguyến Hu Huế) Trong không gian
Oxyz
, viết phương trình mặt phẳng
P
đi
qua điểm
1;2;3M
và cắt ba trục tọa độ
Ox
,
Oy
,
Oz
lần lượt tại
A
,
B
,
C
sao cho
M
là trọng
tâm của tam giác
.ABC
A.
: 2 3 14 0P x y z
. B.
: 6 3 2 18 0P x y z
.
C.
: 6 2 2 2 0P x y z
. D.
: 3 2 10 0P x y z
.
Lời giải
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u 73.(THPT Chuyên Nguyn Du 2019) Trong không gian
Oxyz
, mt phng
z 18 0ax by c
ct ba trc to độ ti
,,A B C
sao cho tam giác
ABC
có trng tâm
1; 3;2G 
. Giá tr
ac
bng
A.
3
. B.
5
. C.
5
. D.
3
.
Li gii
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u 74.(THPT Chuyên y Đôn 2019) Cho điểm
1;2;5M
. Mặt phẳng
P
đi qua
M
cắt các
trục
,,Ox Oy Oz
lần lượt tại
,,A B C
sao cho
M
trực tâm tam giác
ABC
. Phương trình mặt
phẳng
P
A.
80x y z
. B.
2 5 30 0x y z
. C.
0
5 2 1
x y z
. D.
1
5 2 1
x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
86
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 75.(Chuyên Thánh Tông 2019) Trong không gian
Oxyz
, cho điểm
2;2;3M
. Mt phng
P
đi qua
M
ct các trc tọa độ
Ox
,
Oy
,
Oz
ln lượt ti các điểm
A
,
B
,
C
không trùng vi
gc tọa độ sao cho
M
trc tâm ca tam giác
ABC
. Trong các mt phng sau, tìm mt phng
song song vi mt phng
P
.
A.
2 3 9 0.x y z
B.
2 2 3 14 0x y z
.
C.
2 9 0x y z
. D.
3 2 14 0x y z
.
Li gii
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u 76.(THPT chuyên H Long 2019)Viết phương trình mặt phẳng
đi qua
2;1; 3M
, biết
cắt trục
,,Ox Oy Oz
lần lượt tại
,,ABC
sao cho tam giác
ABC
nhận
M
làm trực tâm
A.
2 5 6 0.x y z
B.
2 6 23 0.x y z
C.
2 3 14 0.x y z
D.
3 4 3 1 0.x y z
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
87
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 77.(Chuyên Quý Đôn Điện Biên) Trong không gian với hệ tọa độ
Oxyz
, cho
2;1;1H
. Gọi
P
mặt phẳng đi qua
H
cắt các trục tọa độ tại
A
,
B
,
C
sao cho
H
trực tâm tam giác
ABC
. Hãy viết phương trình mặt phẳng
P
.
A.
2 6 0x y z
. B.
2 6 0x y z
. C.
2 2 6 0x y z
. D.
2 6 0x y z
.
Li gii
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u 78.(THPT Chuyên Thái Bình 2019) Trong không gian với hệ tọa độ
Oxyz
cho mặt phẳng
P
chứa điểm
1;2;2H
cắt
Ox
,
Oy
,
Oz
lần lượt tại
A
,
B
,
C
sao cho
H
trực tâm tam giác
ABC
. Phương trình mặt phẳng
P
A.
2 2 9 0x y z
. B.
2 6 0x y z
. C.
2 2 0x y z
. D.
2 2 9 0x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
88
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 79.(THPT Lương Thế Vinh) Cho mặt phẳng
: 2 2 0Q x y z
. Viết phương trình mặt
phẳng
P
song song với mặt phẳng
Q
, đồng thời cắt các trục
Ox
,
Oy
lần lượt tại các điểm
M
,
N
sao cho
22MN
.
A.
: 2 2 0P x y z
. B.
: 2 0P x y z
.
C.
: 2 2 0P x y z
. D.
: 2 2 0P x y z
.
Li gii
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u 80.(S GD & Đào Tạo Hưng n) Trong không gian h tọa độ
Oxyz
, lập phương trình của các
mt phng song song vi mt phng
: 3 0x y z
và cách
mp
mt khong bng
3
.
A.
60x y z
;
0x y z
. B.
60x y z
.
C.
60x y z
;
0x y z
. D.
60x y z
;
0x y z
.
Li gii
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u 81.(S GD & ĐT Tha Thiên Huế 2019) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:2 2 1 0Q x y z
. Viết phương trình
mp P
song song với
mp Q
khoảng cách giữa hai
mặt phẳng
P
Q
bằng
2
.
3
A.
2 2 1 0x y z
hoặc
2 2 3 0x y z
. B.
2 2 3 0x y z
hoặc
2 2 3 0x y z
.
C.
2 2 1 0x y z
hoặc
2 2 3 0x y z
. D.
2 2 4 0x y z
hoặc
2 2 2 0x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
89
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 82.(THPT Chuyên Phan Bội Châu 2019) Trong không gian hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 2 3 0Q x y z
mặt phẳng
P
không qua
O
, song song mặt phẳng
Q
; 1.d P Q
Phương trình mặt phẳng
P
A.
2 2 3 0x y z
. B.
2 2 0x y z
. C.
2 2 1 0x y z
. D.
2 2 6 0x y z
Li gii.
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u 83.(THPT Nguyn Trãi Hi Dương 2019) Trong không gian tọa độ
Oxyz
, cho
2;0;0A
,
0;4;0B
,
0;0;6C
,
2;4;6D
. Gi
P
mt phng song song vi
mp ABC
,
P
cách đều
D
và mt phng
ABC
. Phương trình của
P
A.
6 3 2 24 0x y z
. B.
6 3 2 12 0x y z
.
C.
6 3 2 0x y z
. D.
6 3 2 36 0x y z
.
Li gii
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u 84.(Đặng Thành Nam) Trong không gian
Oxyz
, phương trình mặt phẳng
P
song song
cách mặt phẳng
: 2 2z 3 0Q x y
một khoảng bằng
1
; đồng thời
P
không qua
O
A.
2 2 1 0x y z
. B.
220x y z
.
C.
2 2 6 0x y z
. D.
2 2 3 0x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
90
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 85.(S GD và Đào Tạo Phú Th 2019) Trog không gian vi tọa độ
zOxy
, cho hai mt phng
( ): 3 2 0P x z
,
( ): 3 4 0Q x z
. Mt phẳng song song cách đều
()P
()Q
phương
trình là
A.
3 1 0.xz
B.
3 2 0xz
. C.
3 6 0.xz
D.
3 6 0.xz
Li gii
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u 86.(THPT Toàn Thắng Hải Phòng 2109) Trong không gian với hệ trục tọa độ
Oxyz
, cho hai
mặt phẳng
1
:3 4 2 0Q x y z
2
:3 4 8 0Q x y z
. Phương trình mặt phẳng
P
song
song và cách đều hai mặt phẳng
1
Q
2
Q
là:
A.
:3 4 10 0P x y z
. B.
:3 4 5 0P x y z
.
C.
:3 4 10 0P x y z
. D.
:3 4 5 0P x y z
.
Li gii
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u 87.(THPT Chuyên Quang Trung 2019)Trong không gian
Oxyz
, cho
0;1;1 , 1;0;0AB
mt
phng
: 3 0P x y z
.
Q
mt phng song song vi
P
đồng thời đường thng
AB
ct
Q
ti
C
sao cho
2CA CB
. Mt phng
Q
có phương trình là:
A.
4
0
3
x y z
hoc
0x y z
. B.
0x y z
.
C.
4
0
3
x y z
. D.
20x y z
hoc
0x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
91
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 88.(Sở GD & ĐT Vĩnh Phúc) Trong không gian
Oxyz
, cho điểm
1; 3;2M
. Hỏi bao nhiêu
mặt phẳng đi qua
M
và cắt các trục tọa độ tại
A
,
B
,
C
0OA OB OC
?
A.
3
. B.
1
. C.
4
. D.
2
.
Li gii
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u 89.(THPT Lý Thường Kit 2019) Trong không gian h tọa độ
Oxyz
cho ba điểm
2;0;1A
, 1;0;0 , 1;1;1BC
mt phng
: 2 0P x y z
. Điểm
;;M a b c
nm trên mt phng
P
tha mãn
MA MB MC
. Tính
23T a b c
.
A.
5T
. B.
4T
. C.
3T
. D.
2T
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
92
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Câu 90. Minh Ha BGD) Trong không gian
Oxyz
cho ba điểm
0;1;1A
;
1;1;0B
;
1;0;1C
mặt phẳng
: 1 0P x y z
. Điểm
M
thuc
P
sao cho
MA MB MC
. Thtích khi chóp
.M ABC
A.
1
4
. B.
1
2
. C.
1
6
. D.
1
.
3
Li gii
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u 91.(Tp Chí Toán Hc 2019) Trong không gian
Oxyz
, cho hai điểm
(1;2;1)A
(3; 1;5)B
. Mt
phng
()P
vuông góc với đường thng
AB
ct các trc
Ox
,
Oy
Oz
lần lượt tại các điểm
D
,
E
F
. Biết th tích ca t din
ODEF
bng
3
2
, phương trình mặt phng
()P
A.
3
2 3 4 36 0x y z
. B.
3
2 3 4 0
2
x y z
.
C.
2 3 4 12 0x y z
. D.
2 3 4 6 0x y z
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
93
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 92.ng Thành Nam 2019) Trong không gian
Oxyz
, bao nhiêu mt phẳng qua điểm
4; 4;1M
chn trên ba trc tọa độ
Ox
,
Oy
,
Oz
theo ba đoạn thng độ dài theo th t
lp thành cp s nhân có công bi bng
1
2
?
A. 1. B. 2. C. 3. D. 4.
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u 93.(THPT Gia Lc Hi Dương 2019) Trong không gian
Oxyz
, cho tam giác
ABC
vi
1;0;0A
,
0;0;1B
2;1;1C
. Gi
;;I a b c
là tâm đường tròn ngoi tiếp tam giác. Khi đó
2a b c
bng
A.
2
. B.
4
. C.
3
. D.
5
.
Li gii
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u 94.(THPT Hoàng Hoa Thám 2019) Trong không gian vi h trc tọa độ
Oxyz
, cho hai điểm
3;1;7A
,
5;5;1B
mt phng
:2 4 0P x y z
. Điểm
M
thuc
P
sao cho
35MA MB
. Biết
M
có hoành độ nguyên, ta có
OM
bng
A.
22
. B.
23
. C.
32
. D.
4
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
94
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 95.(S GD & ĐT Đà Nẵng 2019) Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
1;0;0A
,
0;0;1B
mt phng
:2 2 5 0P x y z
. Tìm tọa độ đim
C
trên trc
Oy
sao cho mt
phng
ABC
hp vi mt phng
P
mt góc
45
A.
22
0; ;0
0
C




. B.
1
0; ;0
4
C



. C.
22
0; ;0
2
C




. D.
1
0; ;0
4
C



.
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u 96.(THPT Chuyên Qúy Đôn 2019) Trong không gian
Oxyz
cho
1; 1;0A 
,
0;1;0B
,
vi
0b
thuc mt phng
: 2 0P x y z
sao cho
2AM
mt phng
ABM
vuông góc vi mt phng
.P
Khi đó
2
24T a b c
bng
A.
8
. B.
7
. C.
28
. D.
17
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
95
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 97.(THPT Bình Minh Ninh Bình 2018) Trong không gian với hệ trục tọa độ
Oxyz
, viết phương
trình mặt phẳng
P
đi qua điểm
1;2;3M
cắt các tia
Ox
,
Oy
,
Oz
lần lượt tại các điểm
A
,
B
,
C
khác với gốc tọa độ
O
sao cho biểu thức
6 3 2OA OB OC
có giá trị nhỏ nhất.
A.
6 2 3 19 0x y z
. B.
2 3 14 0x y z
.
C.
6 3 2 18 0x y z
. D.
3 2 13 0x y z
.
Li gii
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u 98.(THPT Nguyễn Đức Cảnh 2019) Trong không gian hệ tọa độ
,Oxyz
cho hai mt phng
: 2 3 0P x y z
,
: 2 3 0Q x y z
có bao nhiêu điểm
M
có hoành độ nguyên thuc
Ox
sao cho tng khong cách t
M
đến hai mt phng
P
,
Q
bng khong cách gia
P
và
Q
.
A.
2
. B.
4
. C.
6
. D.
7
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 99.(THPT Thun Thành 2019) Trong không gian vi h tọa độ
Oxyz
, cho các điểm
(1;0;0), (0;1;0)AB
. Mt phẳng đi qua các điểm
,AB
đồng thi ct tia
Oz
ti
C
sao cho t din
OABC
có th tích bng
1
6
có phương trình dạng
0x ay bz c
. Tính giá tr
32a b c
.
A.
16
. B.
1
. C.
10
. D.
6
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u 100.(S GD & ĐT Bc Ninh 2019) Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
1;2;1 , 3;4;0AB
, mt phng
: 46 0P ax by cz
. Biết rng khong cách t
,AB
đến mt
phng
P
lần lượt bng
6
3
. Giá tr ca biu thc
T a b c
bng
A.
3
. B.
6
. C.
3
. D.
6
.
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u 101.(THPT Gia Lc Hi ơng 2019) Mt phng
P
đi qua điểm
1;1;1M
ct các tia
Ox
,
Oy
,
Oz
lần lượt ti
;0;0Aa
,
0; ;0Bb
,
0;0;Cc
sao cho th tích khi t din
OABC
nh nht.
Khi đó
23a b c
bng
A.
12
. B.
21
. C.
15
. D.
18
.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 102.(Kênh Truyn Hình GD Quc Gia 2019) Trong không gian vi h tọa độ
Oxyz
. Viết
phương trình mặt phng
P
đi qua điểm
1;2;3M
ct các trc
,,Ox Oy Oz
lần lượt ti ba
đim
,,A B C
khác vi gc tọa độ
O
sao cho biu thc
2 2 2
1 1 1
OA OB OC

có giá tr nh nht.
A.
: 2 14 0P x y z
. B.
: 2 3 14 0P x y z
.
C.
: 2 3 11 0P x y z
. D.
: 3 14 0P x y z
.
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u 103.(THPT Hàm Rồng 2019) Trong không gian với hệ tọa độ
Oxyz
, cho
2;0;0A
,
1;1;1M
.
Mặt phẳng
P
thay đổi qua
AM
cắt các tia
Oy
,
Oz
lần lượt tại
B
,
C
. Khi mặt phẳng
P
thay
đổi thì diện tích tam giác
ABC
đạt giá trị nhỏ nhất bằng bao nhiêu?
A.
56
. B.
26
. C.
46
. D.
36
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 104.(THPT Chuyên Bc Giang) Trong không gian vi h tọa độ
Oxyz
, cho điểm
4;1;9M
. Gi
P
mt phẳng đi qua
M
và ct 3 tia
,,Ox Oy Oz
lần lượt ti các điểm
,,A B C
(khác
O
) sao
cho
OA OB OC
đạt giá tr nh nht. Tính khong cách
d
t đim
0;1;3I
đến
mp P
.
A.
34
5
d
. B.
36
5
d
. C.
24
7
d
. D.
30
7
d
.
Li gii
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u 105.(THPT Thch Thành 2019) Trong không gian h tọa độ
Oxyz
, cho mt phng
:P
3 3 2 37 0x y z
các điểm
4;1;5A
,
3;0;1B
,
1;2;0C
. Biết rằng điểm
;;M a b c
thuc mt phng
P
để biu thc
. . .MAMB MB MC MC MA
đạt giá tr nh nht. Biu thc
2 2 2
a b c
có giá tr
A.
69
. B.
61
. C.
18
. D.
22
.
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
99
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 106.(Chuyên Khoa Hc T Nhiên 2019) Trong không gian với hệ trục toạ độ
Oxyz
, cho tứ
diện
ABCD
điểm
1;1;1 , 2;0;2AB
,
1; 1;0 , 0;3;4CD
. Trên các cạnh
, , AB AC AD
lần
lượt lấy các điểm
', ', 'B C D
thoả mãn
4
' ' '
AB AC AD
AB AC AD
. Viết phương trình mặt phẳng
' ' 'B C D
, biết tứ diện
' ' 'AB C D
có thể tích nhỏ nhất?
A.
16 40 44 39 0 x y z
. B.
16 40 44 39 0 x y z
.
C.
16 40 44 39 0 x y z
. D.
16 40 44 39 0 x y z
.
Li gii
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u 107.(Tạp Chí Toán Học 2019) Trong không gian với hệ tọa độ
,Oxyz
cho hai điểm
9; 3;4 ,A
;;B a b c
. Gọi
,,M N P
lần lượt giao điểm của đường thẳng
AB
với các
mp
,,Oxy mp Oxz
mp Oyz
. Biết các điểm
,,M N P
đều nằm trên đoạn AB sao cho
AM MN NP PB
. Giá trị của
ab bc ca
bằng
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
A.
17
. B.
17
. C.
9
. D.
12
.
Li gii
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u 108. Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 0P x y z
hai điểm
3;4;1 ; 7; 4; 3AB
. Điểm
; ; 2M a b c a
thuộc
P
sao cho tam giác
ABM
vuông tại
M
có diện tích nhỏ nhất. Khi đó giá trị biểu thức
T a b c
bằng:
A.
6T
. B.
8T
. C.
4T
. D.
0T
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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DNG 3. V trí tương đối ca hai mt phng, khong cách và góc ca hai mt phng .
1. Phương pháp:
Cho hai
:0mp P Ax By Cz D
: ' ' ' ' 0Q A x B y C z D
P
cắt
Q
: : ': ': 'A B C A B C
.
//
' ' ' '
A B C D
PQ
A B C D
' ' ' '
A B C D
PQ
A B C D
' ' ' 0P Q AA BB CC
.
Khoảng cách từ
0 0 0
;;M x y z
đến mp
:0P Ax By Cz D
là:
0 0 0
2 2 2
,
Ax By Cz D
d M P
A B C

.
Chú ý: nếu hai mặt
P
Q
song song với nhau thì chọn điểm
0 0 0
;;M x y z mp Q
Khi đó
0 0 0
2 2 2
,,
Ax By Cz D
d Q P d M P
A B C


Cho hai
:0mp P Ax By Cz D
: ' ' ' ' 0Q A x B y C z D
lần lượt một véctơ
pháp tuyến
;;
P
n A B C
;;
P
n A B C
. Khi đó, góc của hai mặt phẳng
2 2 2 2 2 2
. . .
cos ;
.
pQ
A A B B C C
nn
A B C A B B


2. Câu hi trc nghim:
Mức độ 1. Nhận biết
u 109.Trong không gian với hệ tọa độ
,Oxyz
cho hai mặt phẳng
P
Q
có các véc tơ pháp
tuyến
1 1 1 2 2 2
; ; ; ; ;a a b c b a b c
. Góc
góc giữa hai mặt phẳng đó
osc
biểu thức nào
sau đây
A.
1 2 1 2 1 2
a a b b c c
ab

. B.
1 2 1 2 1 2
2 2 2 2 2 2
1 2 3 1 2 3
.
a a b b c c
a a a b b b

.
C.
1 2 1 2 1 2
;
a a b b c c
ab



. D.
1 2 1 2 1 2
a a b b c c
ab

.
Li gii
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u 110. Trong không gian với hệ trục toạ độ
Oxyz
, cho điểm
2;1;2H
,
H
hình chiếu vuông
góc của gốc toạ độ
O
lên mặt phẳng
P
, số đo góc của mặt phẳng
P
mặt phẳng
: 11 0Q x y
.
A.
0
60
. B.
0
30
. C.
0
45
. D.
0
90
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 111.(Gia Bình I Bắc Ninh 2018) Trong không gian
Oxyz
, cho điim
(3; 1;1)A
. Tính khong
cách t
A
đến mt phng
Oyz
.
A.
1.
B.
3.
C.
0.
D.
2.
Li gii
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u 112.(THPT Xoay Vĩnh phúc 2018) Trong không gian vi h tọa độ
Oxyz
, khong cách t
2;1; 6A 
đến mt phng
Oxy
A.
6
. B.
2
. C.
1
. D.
7
41
.
Li gii
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u 113.(THPT Thuận Thành 2019) Trong không gian
Oxyz
, mặt phẳng
:6 3 2 6 0P x y z
.
Tính khoảng cách từ điểm
1; 2;3M
đến mặt phẳng
P
.
A.
31
7
d
. B.
12 85
85
d
. C.
12
7
d
. D.
18
7
d
.
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u 114.(THPT Lê Xoay-Vĩnh pc 2018) Trong không gian vi h tọa độ
Oxyz
, khong cách t
2;1; 6A 
đến mt phng
Oxy
A.
6
. B.
2
. C.
1
. D.
7
41
.
Li gii
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u 115.(THPT Ngc To Hà Ni 2018) Trong không gian h tọa độ
Oxyz
, cho ba điểm
2; 1;1A
,
4;4;5B
,
0;0;3C
. Trng tâm
G
ca tam giác
ABC
cách mt phng tọa độ
Oxy
mt
khong bng
A.
2
. B.
3
. C.
5
. D.
1
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 116. Trong không gian hệ tọa độ
Oxyz
, cho
:16 12 15 4 0mp P x y z
tọa độ điểm
2; 1; 1A 
. Gọi
H
là hình chiếu của điểm
A
lên mặt phẳng
P
. Tính độ dài đoạn thẳng
AH
.
A.
5
. B.
11
5
. C.
11
25
. D.
22
5
.
Li gii
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u 117.(THPT Thuận Thành 2018)Trong không gian
Oxyz
, cho
1;0;0A
,
0; 2;0B
,
0;0;3C
,
1; 1; 2D 
. Khoảng cách từ điểm
D
đến mặt phẳng
ABC
bằng
A.
1
7
. B.
1
7
. C.
7
. D.
2
7
Li gii.
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u 118.(Đề Minh Họa BGD 2019) Trong không gian với hệ tọa độ
Oxyz
, cho ba điểm
,
0; 2 ; 0B
,
0; 0; 4C
. Tính khoảng cách từ gốc tọa độ
O
đến mặt phẳng
ABC
.
A.
4 21
21
. B.
2 21
21
. C.
21
21
. D.
3 21
21
.
Li gii
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u 119.(Đề minh hoạ BGD 2019) Trong không gian với hệ tọa độ
Oxyz
, cho hai mặt phẳng
:P
5 5 5 1 0x y z
: 1 0Q x y z
. Khoảng cách giữa hai mặt phẳng
P
Q
bằng
A.
23
15
. B.
2
5
. C.
2
15
. D.
23
5
Li gii
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u 120. Trong không gian hệ trục tọa độ
Oxyz
, khoảng cách giữa
: 2 2 10 0mp P x y z
: 2 2 3 0mp Q x y z
bằng
A.
8
3
. B.
7
3
. C.
3
. D.
4
3
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 121. Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 1 0P x z
. Chọn câu đúng
nhất trong các nhận xét sau:
A.
P
đi qua gốc tọa độ
O
. B.
P
song song với
Oxy
.
C.
P
vuông góc với trục
Oz
. D.
P
song song với trục
Oy
.
Li gii
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u 122. Ba mặt phẳng
2 6 0x y z
,
2 3 13 0x y z
,
3 2 3 16 0x y z
cắt nhau tại
điểm
M
. Tọa độ của
M
là :
A.
1;2; 3M 
. B.
1; 2;3M
. C.
1; 2;3M 
. D.
1;2;3M
.
Li gii
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u 123.(THPT Chuyên Hà Tĩnh 2019)Trong không gian h tọa độ
Oxyz
, cho mt phng
:
2 3 0xy
. Mệnh đề nào dưới đây đúng?
A.
// Oxy
. B.
//Oz
. C.
Oz
. D.
Oz
.
Li gii
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u 124.(THPT Chuyên Hà Tĩnh 2019)Trong không gian h tọa độ
Oxyz
, cho mt phng
:
2 3 0z 
. Mệnh đề nào dưới đây đúng?
A.
Oxy
. B.
//Oz
. C.
Oz
. D.
Oz
.
Li gii
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u 125.(Tp Chí Toán Hc 2019) Trong không gian vi h tọa độ
Oxyz
, mt phẳng nào dưới đây
song song vi
(O )xz
?
A.
( ): 3 0Px
. B.
( ): 2 0Qy
. C.
( ): 1 0Rz
. D.
( ): 3 0S x z
.
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
105
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 126.(THPT Chuyên Hà Tĩnh 2019) Trong không gian h tọa độ
Oxyz
, cho mt phng
:
20xy
. Mệnh đề nào dưới đây đúng?
A.
// Oxy
. B.
//Oz
. C.
Oz
. D.
Oy
.
Li gii
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u 127.(THPT Gia Lc Hải Dương 2019)Trong không gian vi h tọa độ
Oxyz
, mt phng
:P
2 2 0x y z
vuông góc vi mt phẳng nào dưới đây ?
A.
2 2 0x y z
. B.
20x y z
. C.
20x y z
. D.
2 2 0x y z
.
Li gii
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u 128.(THPT Chuyên Thái Bình 2018) Trong không gian hệ trục tọa độ
Oxyz
, cho mặt phẳng
: 2 2 3 0P x y z
, mặt phẳng
: 3 5 2 0Q x y z
. Cosin của góc giữa hai mặt phẳng
P
,
Q
A.
35
7
. B.
35
7
. C.
5
7
. D.
5
7
.
Li gii
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u 129.(THPT Chuyên Nguyễn Quang Diệu 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho
điểm
2; 1; 2H 
hình chiếu vuông góc của gốc tọa độ
O
xuống mặt phẳng
P
, số đo góc
giữa mặt
P
và mặt phẳng
Q
:
11 0xy
bằng bao nhiêu?
A.
45
. B.
30
. C.
90
. D.
60
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
106
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 130.(Sở GD & ĐT Kiên Giang 2018) Trong không gian hệ tọa độ
Oxyz
cho điểm
4;2; 3B
mặt phẳng
: 2 4 7 0Q x y z
. Gọi
B
điểm đối xứng của
B
qua mặt phẳng
Q
. Tính
khoảng cách từ
B
đến
Q
.
A.
2 21
7
. B.
6 13
13
. C.
10 13
13
. D.
10 21
21
.
Li gii
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Mức độ 2. Thông hiểu
u 131.(THPT Chuyên Lào Cai 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho điểm
1;2;3M
gọi
,,A B C
lần lượt hình chiếu vuông góc của điểm
M
lên các trục
,,Ox Oy Oz
. Khi
đó khoảng cách từ điểm
0;0;0O
đến mặt phẳng
ABC
có giá trị bằng
A.
1
2
. B.
6
. C.
6
7
. D.
1
14
.
Li gii
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u 132.(Pt triển đề minh họa 2019) Trong không gian tọa độ
Oxyz
, cho tứ diện
ABCD
với
1;2;3 , 3;0;0 , 0; 3;0 , 0;0;6 .A B C D
Độ dài đường cao hạ từ đỉnh
A
của tứ diện
ABCD
A.
9
. B.
1
. C.
6
. D.
3
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
107
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 133.(THPT Gia Bình 2018) Trong không gian với hệ toạ độ
Oxyz
, cho
1;0;0A
,
0; ;0Bb
,
0;0;Cc
,
0, 0bc
mặt phẳng
: 1 0P y z
. Tính
S b c
biết mặt phẳng
ABC
vuông góc với mặt phẳng
P
và khoảng cách từ
O
đến
ABC
bằng
1
3
.
A.
1S
. B.
2S
. C.
0S
. D.
3
2
S
.
Li gii
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u 134.(THPT Chuyên Thái Bình 2019) Trong không gian với hệ tọa độ
Oxyz
cho hai mặt phẳng
:2 1 0P x my z
: 3 2 3 2 0Q x y m z
. Giá trị của
m
để
PQ
A.
1m 
. B.
1m
. C.
0m
. D.
2m
.
Li gii
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u 135.(Sở GD & ĐT Đồng Tháp 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
10x my z
m
, mặt phẳng
Q
chứa trục
Ox
qua điểm
1; 3;1A
. Tìm số thực
m
để hai mặt phẳng
P
,
Q
vuông góc.
A.
3m 
. B.
1
3
m 
. C.
1
3
m
. D.
3m
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
108
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 136.(THPT Lương Thế Vinh 2019) Gọi
m
,
n
là hai giá trị thực thỏa mãn giao tuyến của hai
mặt phẳng
: 2 1 0
m
P mx y nz
: 2 0
m
Q x my nz
vuông góc với mặt phẳng
:
4 6 3 0x y z
. Tính
mn
.
A.
0mn
. B.
2mn
. C.
1mn
. D.
3mn
.
Li gii
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u 137.(THPT Hùng Vương 2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho hai mặt phẳng
: 2 1 0x y z
:2 4 2 0x y mz
. Tìm
m
để
song song với nhau.
A.
1m
. B.
2m 
. C.
2m
. D. Không tồn tại
m
Li gii
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u 138.(THPT Th Quãng Tr 2019) Trong không gian h tọa độ
Oxyz
, cho hai mt phng
: 1 0x y z
:2 1 0x y mz m
, vi
m
là tham s thc.
Giá tr ca
m
để

A.
1
. B.
0
. C.
1
. D.
4
.
Li gii
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u 139.ng Thành Nam) Trong không gian h tọa độ
Oxyz
, có bao nhiêu s thc
m
để mt
phng
: 2 2 1 0P x y z
song song vi mt phng
:2 ( 2) 2 0Q x m y mz m
?
A.
1
. B.
0
. C. Vô s. D.
2
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
109
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 140.(THPT Kim Liên 2017) Trong không gian với hệ trục tọa độ
Oxyz
, cho hai mặt phẳng
P
:
2 4 3 0x by z
Q
:
3 2 1 0ax y z
,
,ab
. Với giá trị nào của
a
b
thì hai
mặt phẳng
P
Q
song song với nhau.
A.
1a
;
6b 
. B.
1a 
;
6b 
. C.
3
2
a 
;
9b
. D.
1a 
;
6b
.
Li gii
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u 141.(THPT Đoàn Thượng 2019) Trong không gian vi h trc tọa độ
Oxyz
, cho hai mt
phng
: 1 0x y z
và
: 2 2 2 0x my+ z
. Tìm
m
để
song song vi
.
A.
2.m=
B. không tn ti
.m
C.
2.m=
D.
1
.
2
m=
Li gii
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u 142.(THPT Thuận Thành 2019) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:P
1 10 0mx m y z
mặt phẳng
:2 2z 3 0Q x y
. Với giá trị nào của
m
thì
P
Q
vuông góc với nhau?
A.
2m 
. B.
2m
. C.
1m
. D.
1m 
.
Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
110
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 143.(THPT Chuyên Thái Bình 2019) Trong không gian
Oxyz
cho điểm
2;1;5M
. Mặt phẳng
P
đi qua điểm
M
cắt các trục
Ox
,
Oy
,
Oz
lần lượt tại các điểm
A
,
B
,
C
sao cho
M
trực
tâm của tam giác
ABC
. Tính khoảng cách từ điểm
1;2;3I
đến mặt phẳng
P
A.
17 30
30
. B.
13 30
30
. C.
19 30
30
. D.
11 30
30
.
Li gii
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Mức độ 3. Vận dung
u 144.(Chuyên KHTN 2019) Biết rằng trong không gian với hệ tọa độ
Oxyz
có hai mặt phẳng
P
Q
cùng thỏa mãn các điều kiện đi qua hai điểm
1;1;1A
0; 2;2B
, đồng thời cắt
các trục tọa độ
,Ox Oy
tại hai điểm cách đều
O
. Giả sử
P
phương trình
1 1 1
0x b y c z d
Q
có phương trình
2 2 2
0x b y c z d
. Tính giá trị biểu thức
1 2 1 2
bb c c
.
A.
7
. B.
9
. C.
7
. D.
9
.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 145.(Toán Học Tuổi Trẻ số 6-2018) Trong không gian với hệ tọa độ
Oxyz
, biết mặt phẳng
:0P ax by cz d
với
0c
đi qua hai điểm
0;1;0A
,
1;0;0B
tạo với mặt phẳng
yOz
một góc
60
. Khi đó giá trị
abc
thuộc khoảng nào dưới đây?
A.
0;3
. B.
3;5
. C.
5;8
. D.
8;11
.
Li gii
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u 146.(THPT Chuyên ĐHSP 2018) Trong không gian tọa độ
Oxyz
, cho điểm
1;2;2 .A
Các số
a
,
b
khác
0
thỏa mãn khoảng cách từ điểm
A
đến mặt phẳng
:0P ay bz
bằng
2 2.
Khẳng
định nào sau đây là đúng?
A.
ab
. B.
2ab
. C.
2ba
. D.
ab
.
Li gii
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u 147.(S GD & ĐT Hậu Giang 2018) Trong không gian vi h tọa độ
Oxyz
, cho
1; 2; 3A
,
3; 4; 4B
. Tìm tt c các giá tr ca tham s
m
sao cho khong cách t đim
A
đến mt phng
2 1 0x y mz
bằng độ dài đoạn thng
AB
.
A.
2m
. B.
2m 
. C.
3m 
. D.
2m 
.
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
112
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii
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u 148.(THPT Trần Hưng Đạo 2018) Trong không gian với hệ tọa độ
Oxyz
, cho bốn điểm
0;0; 6A
,
0;1; 8B
,
1;2; 5C
4;3;8D
. Hỏi tất cả bao nhiêu mặt phẳng cách đều bốn
điểm đó?
A. Có vô số mặt phẳng. B. 1 mặt phẳng. C. 7 mặt phẳng. D. 4 mặt phẳng.
Li gii
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u 149.(THPT Chuyên Hà Tĩnh 2018) Trong không gian
Oxyz
cho ba điểm
1;2;3A
,
1;0; 1B
,
2; 1;2C
. Điểm
D
thuc tia
Oz
sao cho độ dài đường cao xut phát t đỉnh
D
ca t din
ABCD
bng
3 30
10
có tọa đọ
A.
0;0;1
. B.
0;0;3
. C.
0;0;2
. D.
0;0;4
.
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u 150.(THPT Chuyên Thái Bình 2018) Trong không gian
Oxyz
, cho ba điểm
;0;0Aa
,
0; ;0Bb
,
0;0;Cc
vi
,,abc
các s thực dương thay đổi tùy ý sao cho
2 2 2
1abc
.
Khong cách t
O
đến mt phng
ABC
ln nht là
A.
1
3
. B.
1
. C.
1
3
. D.
3
.
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u 151.(Cụm Đồng Bằng Sông Cửu Long 2018) Trong không gian với hệ tọa độ
Oxyz
bao
nhiêu mặt phẳng song song với mặt phẳng
: 3 0Q x y z
, cách điểm
3;2;1M
một khoảng
bằng
33
biết rằng tồn tại một điểm
;;X a b c
trên mặt phẳng đó thỏa mãn
2abc
?
A.
1
. B. Vô s. C.
2
. D.
0
.
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u 152.(THPT Chuyên Trn Phú 2018) Trong không gian vi h tọa độ , cho các điểm
, , . Phương trình mặt phng nào dưới đây đi qua , gc ta
độ và cách đều hai điểm ?
A. . B. . C. . D.
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u 153.(THPT Đặng Thúc Ha 2019) Trong không gian vi h trc to độ , cho ba điểm có
, , . Đường thng qua trc m ca tam giác nm trong
mt phng cùng to với các đường thng , mt góc một véctơ chỉ
phương là vi là mt s nguyên t. Giá tr ca biu thc bng
A. . B. . C. . D. .
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u 154.(THPT Chuyên Lam Sơn 2018) Trong không gian vi h trc tọa đ cho các điểm
, , , . tt c bao nhiêu mt phng phân biệt đi qua trong
đim , , , , ?
A. . B. . C. . D. .
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Oxyz
1; 2;0A 
0; 4;0B
0;0; 3C
P
A
O
B
C
:2 3 0P x y z
:6 3 5 0P x y z
:2 3 0P x y z
: 6 3 4 0P x y z
Oxyz
1;2; 1A
2;0;1B
2;2;3C
H
ABC
ABC
AB
AC
o
45
;;u a b c
c
ab bc ca
67
23
33
37
Oxyz
1;0;0A
0;2;0B
0;0;3C
2; 2;0D
3
5
O
A
B
C
D
7
5
6
10
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DNG 3. m hình chiếu của điểm
0 0 0
;;M x y z
xung
mp
, tìm điểm đối xng
'.M
Bài toán 1. Tìm hình chiếu của điểm
0 0 0
;;M x y z
xung mt phng
mp P
.
1. Phương pháp.
ch 1. Vn dng khi
, , 0abc
.
H
là hình chiếu vuông góc ca
M
lên
mp P
Gi s
1 1 1 1 1 1
; ; 0 1H x y z ax by cz d
1 0 1 0 1 0
; ;z 2MH x x y y z
MH
cùng phương với VTPT
;;n a b c
ca
mp P
:.
P
t MH t n
1 0 1 0 1 0
; ; ; ;x x y y z z ta tb tc
1 0 1 0
1 0 1 0
1 1 0
x x ta x x ta
y y tb y y tb
z z tc z z tc





Thay
1 1 1
;;x y z
vào
1 1 1
:0mp P ax by cz d t
1
1
1
x
y
z
Cách 2:
Gi
H
là hình chiếu vuông góc ca
M
lên mt phng
PM
là giao điểm ca mt
phng
P
với đường thng
qua
M
và vuông góc vi mt phng
P
.
Viết phương trình tham số của đưng thng
đi qua điểm
M
và nhận véc tơ pháp tuyến
;;n a b c
làm véctơ chỉ phương.
H
thuc mt phng
P
thay vào phương trình mặt phng
1 1 1
2; 8; 1 2;8;1P t H x y z H
Bài tp 15.
a). Tìm hình chiếu vuông góc ca
3;6;2M
lên
:5 2 25 0.mp P x y z
b). Tọa độ hình chiếu
H
ca
2;1;0A
trên mt phng
P
là:
: 2 2 9 0P x y z
.
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n
P
P
M
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Bài tp 16. Trong không gian
Oxyz
, cho
2; 0; 1 , 1; 2; 3 , 0;1; 2A B C
.
Tọa độ hình chiếu vuông góc ca gc tọa độ
O
lên mt phng
ABC
điểm
H
, khi đó ta độ
đim
H
là:
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Bài toán 2. Tìm điểm đi xng
'M
của điểm
0 0 0
;;M x y z
qua mt phng
mp P
.
1. Phương pháp.
c 1. Tìm
H
là hình chiếu vuông góc ca
M
lên
.mp P
c 2.
'M
đối xng vi
M
qua
mp P
H
là trung điểm ca
1
MM
'
'
'
2
2'
2
M H M
M H M
M H M
x x x
y y y M
z z z


2. Bài tp Minh ha:
Bài tp 17. Cho điểm
2;3;5A
và mt phng
:2 3 17 0P x y z
.
Gi
A
đim đối xng ca
A
qua
.P
Tọa độ đim
A
là:
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3. Câu hi trc nghim:
Mức độ 3. Vận dụng
H
P
n
P
M'
M
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u 156. (THPT Hoàng Hoa Thám 2017) Trong không gian với hệ tọa độ
Oxyz
, hình chiếu của điểm
1; 3; 5M 
trên mặt phẳng
Oxy
có tọa độ
;;abc
. Khi đó
abc
bằng ?
A.
0
. B.
3
. C.
2
. D.
1
.
Lời giải
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u 157. (THPT Thanh Thy 2017) Trong không gian vi h tọa độ
Oxyz
, tìm tọa độ hình chiếu
vuông góc
N
của điểm
1;2;3M
trên mt phng
Oxz
.
A.
0;2;3N
. B.
1;2;0N
. C.
0;2;0N
. D.
1;0;3N
.
Lời giải
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u 158. (Sở Bình Phước 2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho điểm
4;1; 2A
.
Tọa độ điểm đối xứng với
A
qua mặt phẳng
Oxz
là?
A.
4; 1; 2A

. B.
4; 1;2A
. C.
4; 1;2A

. D.
4;1;2A
.
Lời giải
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u 159.(THPT chuyên Nguyn trãi) Cho điểm
3;5;0A
và mt phng
7:2 3 0xyP z
. Tìm
tọa độ đim
M
là điểm đối xng với điểm
A
qua
P
.
A.
1; 1;2M 
. B.
2; 1;1M
. C.
0; 1; 2M 
. D.
7;1; 2M
.
Lời giải
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u 160.(THPT chuyênThánh Tông 2019) Trong không gian h trc tọa độ
,Oxyz
cho điểm
0;1;2M
mt phng
:0P x y z
. Tìm tọa độ đim
N
hình chiếu vuông góc của điểm
M
trên mt phng
P
.
A.
2;0;2N
. B.
1;1;0N
. C.
. D.
2;2;0N
.
Lời giải
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u 161.(THPT ơng i 2019) Trong mặt phẳng
,Oxyz
cho mặt phẳng
:3 2 6 0x y z
điểm
2; 1;0A
. Hình chiếu vuông góc của
2; 1;0A
lên mặt phẳng
là.
A.
1;1; 1
. B.
1; 1;1
.
C.
3; 2;1
. D.
5; 3;1
.
Lời giải
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u 162. Tọa độ hình chiếu của điểm
5; 1; 2A 
lên mặt phẳng
3 2 9 0x y z
là.
A.
2; 0; 1
. B.
2; 0; 1
. C.
1; 1; 2
. D.
1; 5; 0
.
Lời giải
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u 163.(THPT chuyên Phan Bi Châu 2019) Trong không gian vi h tọa độ
Oxyz
cho hai điểm
1;2;1A
,
3;0; 1B
và mt phng
: 1 0 P x y z
. Gi
M
N
lần lượt là hình chiếu ca
A
B
trên mt phng
P
. Tính độ dài đoạn
MN
.
A.
2
3
. B.
23
. C.
42
3
. D.
4
.
Lời giải
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u 164.(THPT chuyên Lương Thế Vinh 2019) Trong không gian vi h tọa độ
Oxyz
, tọa độ hình
chiếu vuông góc của điểm
6;5;4A
lên mt phng
:9 6 2 29 0P x y z
là:
A.
3; 1;2
. B.
5;3; 1
. C.
5;2;2
. D.
1; 3; 1
.
Lời giải
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u 165. Cho mặt phẳng
: 2 2 9 0P x y z
điểm
2;1;0A
. Tọa độ hình chiếu
H
của
A
trên mặt phẳng
P
là:
A.
1; 3; 2H 
. B.
1;3;2H
. C.
1;3; 2H
. D.
1;3; 2H 
.
Lời giải
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u 166.(THPT Chuyên Sơn La 2019) Trong không gian với hệ tọa độ
,Oxyz
cho mặt phẳng
:P
2 2 3 0x y z
điểm
1; 2;4M
. Tìm tọa độ hình chiếu vuông góc của điểm
M
trên mặt
phẳng
.P
.
A.
0;0; 3
. B.
1;1;3
. C.
3;0;3
.
D.
5;2;2
.
Lời giải
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u 167. (THPT chuyên KHTN) Trong không gian
Oxyz
, cho
3; 5; 0A
,
2; 0; 3B
,
0;1; 4C
2; 1; 6D 
. Tọa độ của điểm
A
đối xng vi
A
qua mt phng
BCD
là.
A.
1; 1; 2
. B.
1;1; 2
. C.
1;1; 2
. D.
1; 1; 2
.
Lời giải
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u 168. Cho mặt phẳng
: 2 2 9 0P x y z
điểm
2;1;0A
. Tọa độ hình chiếu
H
của
A
trên mặt phẳng
P
là:
A.
1; 3; 2H 
. B.
1;3;2H
. C.
1;3; 2H
. D.
1;3; 2H 
.
Lời giải
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u 169. Trong không gian tọa độ
Oxyz
, cho
2; 0; 1 , 1; 2; 3 , 0; 1; 2A B C
. Tọa độ hình chiếu
vuông góc của gốc tọa độ
O
lên mặt phẳng
ABC
là điểm
H
, khi đó tọa độ điểm
H
là:
A.
11
1; ;
32
H



.
B.
11
1; ;
22
H



.
C.
31
1; ;
22
H



. D.
11
1; ;
23
H



.
Lời giải
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u 170. Cho điểm
2;3;5A
và mặt phẳng
:2 3 17 0P x y z
. Gọi
A
điểm đối xứng của
A
qua
.P
Tọa độ điểm
A
là:
A.
12 18 34
;;
7 7 7
A



. B.
12 18 34
;;
7 7 7
A




. C.
12 18 34
;;
7 7 7
A




. D.
12 18 34
;;
7 7 7
A



.
Lời giải
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DNG 4. i toán cc tr (giá tr ln nht nh nht ).
Bài toán 1. Tìm đim
M
sao cho tng hoc hiệu c véc tơ đạt giá tr ln nht, nh nht.
Trong không gian cho
n
đim
12
, ,...,
n
A A A
.
Loi 1. Tìm
M
sao cho
2 2 2
1 1 2 2
...
nn
P MA MA MA
a). Nh nht khi
12
... 0
n
b). Ln nht khi
12
... 0
n
Loi 2. Tìm
M
sao cho
1 1 2 2
...
nn
P MA MA MA
nh nht hoc ln nhất , trong đó
1
0
n
i
i
.
1. Phương pháp.
Gọi
I
là điểm thỏa mãn:
1 1 2 2
... 0
nn
IA IA IA
điểm
I
tồn tại và duy nhất nếu
1
0
n
i
i
.
Khi đó
Loại 1.
2 2 2
1 1 2 2
...
nn
P MI IA MI IA MI IA
22
1 2 1
1
( ... )
n
ni
i
IM IA
Do
2
1
1
n
i
i
IA
không đổi nên:
Nếu
12
... 0
n
thì
P
nhỏ nhất
MI
nhỏ nhất
Nếu
12
... 0
n
thì
P
lớn nhất
MI
nhỏ nhất
Bài toán 2.
1 1 2 2
1
... .
n
n n i
i
P MI IA MI IA MI IA MI
Do đó
P
nhỏ nhất hoặc lớn nhất
MI
nhỏ nhất hoặc lớn nhất.
Nếu
M
thuộc đường thẳng
(hoặc mặt phẳng
()P
) thì
MI
lớn nhất khi chỉ khi
M
hình chiếu của
I
lên
(hoặc
()P
).
Nếu
M
thuộc mặt cầu
S
đường thẳng đi qua
I
tâm của
S
, cắt
S
tại hai điểm
,AB
(
)IA IB
thì
MI
nhỏ nhất (lớn nhất)
MB
(
MA
).
2. Bài toán minh ha.
Bài tp 18. Trong không gian cho ba điểm
(1;2;3), ( 1;0; 3),AB
(2; 3; 1)C 
1). Tìm
M
thuộc mặt phẳng
( ): 2 2 1 0x y z
sao cho
2 2 2
3 4 6S MA MB MC
đạt giá
trị nhỏ nhất.
2). Tìm
M
thuộc đường thẳng
1 1 1
2 3 1
x y z
sao cho
75P MA MB MC
đạt giá
trị nhỏ nhất:.
3). Tìm
M
thuộc mặt cầu
2 2 2
( ):( 2) ( 2) ( 8) 36S x y z
thỏa
2 2 2
42F MA MB MC
đạt giá trị lớn nhất, giá trị nhỏ nhất.
n
P
P
M
I
u
M
I
R
n
p
A=M
B=M
K
I
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải.
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Bài tp 19. Trong không gian
Oxyz
cho ba điểm
(2;3;1), ( 1; 2;0),AB
(1;2; 2)C
.
1). Lập phương trình mặt phẳng
()ABC
;
2). Tìm
,ab
để mặt phẳng
( ):(2 ) (3 2 ) 1 1 0a b x a b y z
song song với
()ABC
;
3). Tìm
( ):3 1 0M x y z
sao cho
2 2 2
243S MA MB MC
nhỏ nhất;
4). Tìm
( ):3 3 29 0N x y z
sao cho
3 5 7P NA NB NA
nhỏ nhất.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 20. Cho các điểm
( 2;3;1), (5; 2;7), (1;8; 1)A B C
.
Tìm tp hợp các điểm
M
trong không gian tha
1).
2 2 2
MA MB MC
2).
AM AB BM CM
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 21. Trong không gian
Oxyz
cho các điểm
(1; 4; 5), (0; 3;1),AB
(2; 1; 0)C
và mt
phng
( ) : 3 3 2 15 0.P x y z
Tìm điểm
M
thuc mt phng
()P
sao cho
1).
2 2 2
MA MB MC
có giá tr nh nht.
2).
2 2 2
24MA MB MC
có giá tr ln nht.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 22. Cho
(1; 4; 2), ( 1; 2; 4)AB
12
:.
1 1 2
x y z
Tìm điểm
M
thuộc đường thng
sao cho
1).
22
MA MB
nh nht
2).
3 2 4OM AM BM
nh nht.
3). Din tích tam giác
MAB
nh nht.
Lời giải.
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Bài 23. Cho tam giác
ABC
3; 2;5 , 2;1; 3 , 5;1; 1 .A B C
Đim
M
các thành phn
tọa độ bng nhau.
1). Chng minh rng tam giác
ABC
là tam giác nhn.
2). Tìm tọa độ đim
M
sao cho
3MA BC
đạt giá tr nh nht.
3). Tìm điểm
M
sao cho
2 2 2
24MA MB MC
đạt giá tr ln nht.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải.
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Bài 24. Cho ba điểm
(1;2; 3), (2;4;5), (3;6;7)A B C
và mt phng
( ): 3 0.P x y z
1). Tìm tọa độ hình chiếu trng tâm
G
ca tam giác
ABC
trên mt phng
( ).P
2). Tìm tọa độ đim
G
đối xng với điểm
G
qua mt phng
( ).P
3). Tìm tọa độ đim
M
thuc mt phng
()P
sao cho biu thc
T
có giá tr nh nht vi
2 2 2
.T MA MB MC
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài 25. Cho các điểm
(1; 0; 1), (0; 2; 3), ( 1;1;1)A B C
đường thng
11
:.
1 2 2
x y z
Tìm
đim
M
thuộc đường thng
sao cho
1).
2 2 2
24MA MB MC
ln nht. 2).
AM BC
nh nht.
Lời giải.
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Bài 26. Cho đường thng
12
: (1 ) ( ),
2
m
xt
y m t t
z mt

m
là tham s. Tìm giá tr ca
m
sao cho
1). Khong cách t gc tọa độ đến
m
là ln nht, nh nht.
2.
m
to vi mt phng
()xOy
mt góc ln nht.
3. Khong cách gia
m
và trc
Oy
ln nht.
Lời giải.
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Bài tp 27. Cho và ba điểm
1;1;1 , 0;1;2 , 2;0;1A B C
.
1). Tìm tọa độ đim
()MP
sao cho
MA MB
1
M
y
;
2). Tìm
()NP
sao cho
2 2 2
2S NA NB NC
nh nht.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 28. Trong không gian
Oxyz
cho ba điểm
2;3;1 , 1; 2;0 , 1;2; 2A B C
1). Lập phương trình mặt phng
ABC
,
2). Tìm
,ab
để mt phng
: 2 3 2 1 1 0a b x a b y z
song song vi
()ABC
,
3). Tìm
:3 1 0M x y z
sao cho
2 2 2
243S MA MB MC
nh nht,
4). Tìm
:3 3 29 0N x y z
sao cho
3 5 7P NA NB NA
nh nht.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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4. Câu hi trc nghim:
Mức độ 3. Vận dụng
u 171.(THPT Thị Quảng Trị) Trong không gian
Oxyz
, cho ba điểm
0;1;2A
,
1;1;1B
,
2; 2;3C
và mặt phẳng
: 3 0P x y z
. Gọi
;;M a b c
là điểm thuộc mặt phẳng
P
thỏa
mãn
MA MB MC
đạt giá trị nhỏ nhất. Giá trị của
bằng
A.
7
. B.
5
. C.
3
. D.
2
.
Li gii
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u 172.(THPT Chuyên Lam n 2019) Trong hệ trục
,Oxyz
cho điểm
1;3;5 ,A
2;6; 1 ,B
4; 12;5C
mặt phẳng
: 2 2 5 0. P x y z
Gọi
M
điểm di động trên
.P
Gía trị nhỏ
nhất của biểu thức
S MA MB MC
A.
42.
B.
14.
C.
14 3.
D.
14
.
3
Li gii
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u 173.(Toán Học Tuổi trẻ 2019) Trong không gian hệ tọa độ
Oxyz
, cho ba điểm
1;2;2 ,A
3; 1; 2 , 4;0;3BC
. Tọa độ điểm
I
trên
mp Oxz
sao cho biểu thức
23IA IB IC
đạt
giá trị nhỏ nhất là
A.
19 15
;0;
22
I



. B.
19 15
;0;
22
I




. C.
19 15
;0;
22
I



. D.
19 15
;0;
22
I



.
Li gii
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u 174.(THPT chuyên Hùng Vương 2019) Trong không gian với hệ tọa độ
Oxyz
, cho ba điểm
0; 2; 1A 
,
2; 4;3B 
,
1;3; 1C
và mặt phẳng
: 2 3 0P x y z
.
Biết điểm
;;M a b c P
thỏa mãn
2T MA MB MC
đạt giá trị nhỏ nhất. Tính
S a b c
.
A.
1S 
. B.
1
2
S
. C.
0S
. D.
1
2
S 
.
Li gii
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u 175. Trong không gian h tọa độ
Oxyz
, cho mt phng
:2 2 9 0x y z
và ba điểm
2;1;0 ,A
0;2;1 , 1;3; 1BC
. Điểm
M
sao cho
2 3 4MA MB MC
đạt giá tr nh nht.
Khẳng định nào sau đây đúng?
A.
3
M M M
x y z
. B.
2
M M M
x y z
.
C.
1
M M M
x y z
. D.
4
M M M
x y z
.
Li gii
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u 176.(THPT Thuận Thành 2019) Trong không gian
Oxyz
, cho ba điểm
(1;1;1)A
,
( 2;3;4)B
( 2;5;1)C
. Điểm
( ; ;0)M a b
thuộc mặt phẳng
Oxy
sao cho
2 2 2
MA MB MC
đạt giá trị nhỏ nhất.
Tổng
22
T a b
bằng
A.
10T
. B.
25T
. C.
13T
. D.
17T
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 177.(THPT Ngô Quyn Hi Phòng 2019) Trong không gian
Oxyz
, cho ba điểm
(1; 1; 1)A
,
( 1; 2; 0)B
,
(3; 1; 2)C
M
điểm thuc mt phng
:2 2 7 0x y z
. Tính giá tr nh
nht ca
3 5 7P MA MB MC
.
A.
min
20P
. B.
min
5P
. C.
min
25P
. D.
min
27P
.
Li gii
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u 178.(THPT Kim Liên 2018) Trong không gian vi h trc tọa độ
Oxyz
, cho hai điểm
3;5; 5 , 5; 3;7AB
mt phng
:0P x y z
. Tìm tọa độ đim
M
trên mt phng
P
sao
cho
22
2MA MB
ln nht.
A.
2;1;1M
. B.
2; 1;1M
. C.
6; 18;12M
. D.
6;18;12M
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 179.(THPT Yên Phong 2019) Trong không gian h trc tọa độ
Oxyz
, cho điểm
3;5; 5A 
,
5; 3;7B
mt phng
:0x y z
. Xét điểm
M
thay đổi trên
, giá tr ln nht ca
22
2MA MB
bng
A.
398
. B.
379
. C.
397
. D.
498
.
Li gii
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u 180.(THPT Chuyên T Trng 2019) Trong không gian
Oxyz
, cho hai điểm
(2; 2;4)A
,
( 3;3; 1)B 
mặt phẳng
( ):2 2 8 0P x y z
. Xét
M
điểm thay đổi thuộc
()P
, giá trị nhỏ nhất
của
22
23MA MB
bằng
A.
145
. B.
108
. C.
105
. D.
135
.
Li gii
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u 181.(THPT Nghĩa Hưng Nam Định) Trong không gian với hệ trục tọa độ
Oxyz
, cho tam giác
ABC
với
2;1;3A
,
1; 1;2B
,
3; 6;1C
. Điểm
thuộc mặt phẳng
Oyz
sao cho
2 2 2
MA MB MC
đạt giá trị nhỏ nhất. Tính giá trị biểu thức
P x y z
.
A.
0P
. B.
2P
. C.
6P
. D.
2P 
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 182. Trong không gian
Oxyz
, cho ba điểm
0;0; 1A
,
1;1;0B
,
1;0;1C
.
Tìm điểm
M
sao cho
2 2 2
32MA MB MC
đạt giá trị nhỏ nhất.
A.
31
; ; 1
42
M



. B.
33
; ; 1
42
M




. C.
31
; ; 1
42
M




. D.
31
; ;2
42
M



.
Li gii
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u 183.(S GD & ĐT ngn 2019) Trong không gian vi h tọa độ
Oxyz
, cho ba điểm
1;4;5A
,
3;4;0B
,
2; 1;0C
mt phng
:3 3 2 29 0P x y z
. Gi
;;M a b c
điểm thuc
P
sao cho
2 2 2
3MA MB MC
đạt giá tr nh nht. Tính tng
abc
.
A.
8
. B.
10
. C.
10
. D.
8
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 2. Phương Trình Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 184.(S GD & ĐT Quãng Bình 2019) Trong không gian
Oxyz
, cho
0;1;1A
,
2; 1;1B
,
4;1;1C
: 6 0P x y z
. Xét điểm
;;M a b c
thuộc
mp P
sao cho
2MA MB MC
đạt giá trị nhỏ nhất. Giá trị của
24a b c
bằng:
A.
6
. B.
12
. C.
7
. D
5
.
Li gii
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u 185.(THPT Cm Giàng 2019) Trong không gian
Oxyz
, cho ba điểm
10; 5;8A 
,
2;1; 1B
,
2;3;0C
mt phng
: 2 2 9 0P x y z
. Xét
M
điểm thay đổi trên
P
sao cho
2 2 2
23MA MB MC
đạt giá tr nh nht. Tính
2 2 2
23MA MB MC
.
A.
54
.
B.
282
.
C.
256
.
D.
328
.
Li gii
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u 186.(S GD & ĐT Ninh Bình 2019) Trong không gian
Oxyz
, cho các điểm
1;4;5A
,
0;3;1B
,
2; 1;0C
:3 3 2 15 0mp P x y z
. Gi
;;M a b c
điểm thuc mt phng
P
sao cho
tổng các bình phương khoảng cách t
M
đến
,,A B C
nh nht. Tính
abc
.
A.
5
. B.
5
. C.
3
. D.
3
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Loi 2. i toán tìm điểm
M
sao đ dài c vec tơ đt giá tr ln nht, nh nht.
Bài toán 1. Trong không gian
,Oxyz
cho các điểm
( ; ; ), ( ; ; )
A A A B B B
A x y z B x y z
và mt phng
( ): 0.P ax by cz d
Tìm điểm
()MP
sao cho:
1).
MA MB
nh nht.
2).
ln nht vi
( , ( )) ( , ( )).d A P d B P
1. Phương pháp.
Xét v trí tương đối của các điểm
,AB
so vi mt phng
( ).P
Nếu
( )( ) 0
A A A B B B
ax by cz d ax by cz d
thì điểm
,AB
cùng phía vi mt phng
( ).P
Nếu
( )( ) 0
A A A B B B
ax by cz d ax by cz d
thì hai điểm
,AB
nm khác phía vi
( ).mp P
MA MB
nh nht.
Trưng hp 1. Hai điểm
,AB
khác phía so vi mt phng
( ).P
,AB
khác phía so vi mt phng
()P
nên
MA MB AB
nh nht bng
AB
khi và ch khi
1
( ) .M M P AB
Lập phương trình đường thẳng
.AB
Tọa độ
1
M
là nghiệm hệ phương trình của đường thẳng
AB
mặt phẳng
( ).P
Trường hp 2: Hai đim
,AB
cùng phía so vi mt phng
( ).P
,AB
cùng phía so vi mt phng
()P
nên ta phi là các
c:
Gọi
'A
đối xứng với
A
qua mặt phẳng
( ).P
Khi đó khi đó
'A
B
ở khác phía
()P
.MA MA
Lúc này
.MA MB MA MB A B

MA MB
nh nht bng
AB
khi và ch
1
( ) ' .M M P A B
Vy
MA MB
nh nht bng
AB
khi
( ).M A B P

2. Bài tp minh ha.
Bài tp 29. Trong không gian
,Oxyz
cho hai đim
1;3; 2 , 3;7; 18AB
và phương trình
: 2 1 0mp P x y z
.
1). Viết phương trình mặt phng cha
AB
và vuông góc vi
mp P
.
2). Tìm to độ đim
M
thuc
mp P
sao cho
MA MB
nh nht.
P
M
1
M
A
B
n
p
P
H
M
1
A
M
B
A'
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii.
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2.
ln nht.
Trường hp 1. Hai điểm
,AB
cùng phía so vi mt phng
( ).P
,AB
cùng phía so vi mt phng
()P
nên
MA MB AB
MA MB
ln nht bng
AB
khi
1
( ) .M M P AB
Lập phương trình đường thẳng
AB
Tọa độ
1
M
là nghiệm hệ phương trình của đường thẳng
AB
mặt phẳng
( ).P
Trường hp 2: Hai đim
,AB
khác phía so vi mt phng
( ).P
,AB
khác phía so vi mt phng
()P
nên ta phi là các
c:
Gọi
'A
đối xứng với
A
qua mặt phẳng
( ).P
Khi đó khi đó
'A
B
ở khác phía
()P
.MA MA
Lúc này
.MA MB MA MB A B

MA MB
ln nht bng
AB
khi và ch
1
( ) ' .M M P A B
P
M
1
B
M
A
P
H
M
1
A'
M
B
A
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2. Bài tp minh ha.
Bài tp 30. Trong không gian
Oxyz
cho
:2 2 6 0P x y z
và hai điểm
5; 2;6 ,A
3; 2;1B
. Tìm điểm
M
thuc
()P
sao cho:
1).
MA MB
nh nht 2).
MA MB
ln nht.
Lời giải.
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Bài tp 31. Cho các điểm
1; 1; 2 , 2;1; 0 , 2; 0;1A B C
và mt phng
P
có phương
trình
2 3 0.x y z
Tìm điểm
M
thuc
P
sao cho
1).
MA MB
có giá tr nh nht. 2).
MA MC
có giá tr ln nht.
3).
MA MC
có giá tr nh nht. 4).
có giá tr ln nht.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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3. Câu hi trc nghim:
Mức độ 3. Vận dụng
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III- Phương Tnh Mặt Phẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 187.(Chuyên ĐH Vinh 2019) Trong không gian , cho hai điểm , . Gi
s là điểm thay đổi trong mt phng Tìm giá tr ln nht ca biu
thc
A. . B. . C. . D. .
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u 188.(Chuyên ĐH Vinh 2019) Trong không gian , cho mt phng
. Tìm tọa độ đim sao cho đạt giá tr ln nht.
A. . B. . C. . D. .
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u 189.(Chuyên ĐH Vinh) Trong không gian tọa độ , cho mt phng
hai điểm . Biết sao cho đạt giá tr nh nhất. Khi đó,
hoành độ của điểm
A. . B. . C. . D. .
Li gii
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Oxyz
1;2;3A
4;4;5B
M
( ):2 2 2019 0.P x y z
.P AM BM
17
77
7 2 3
82 5
Oxyz
1;1;0 , 3; 1;4AB
: 1 0x y z
M
MA MB
1;3; 1M
3 5 1
;;
4 4 2
M



1 2 2
;;
3 3 3
M



0;2;1M
Oxyz
: 2 1 0x y z
0; 1;1 , 1;1; 2AB
M
MA MB
M
x
M
1
3
M
x
1
M
x 
2
M
x 
2
7
M
x
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u 190.(Chuyên Nguyn Du 2019) Trong không gian
Oxyz
, cho hai điểm
2;0;1A
,
điểm
;;M a b c
di động trên mt phng
Oxy
. Khi
MA MB
đạt giá tr nh nht thì giá tr
3a b c
bng
A.
2
. B.
3
. C.
5
. D.
4
.
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u 191.(THPT Chuyên H Long 2019) Cho
4;5;6 ; 1;1;2AB
,
M
một điểm di động trên mặt
phẳng
:2 2 1 0P x y z
. Khi đó
nhận giá trị lớn nhất là?
A.
77
. B.
41
. C.
7
. D.
85
.
Li gii
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u 192.(THPT Lương Thế Vinh 2019) Trong không gian tọa độ
Oxyz
, cho hai điểm
1;2; 2A
,
2; 1;2B
. Tìm tọa độ điểm
M
trên mặt phẳng
Oxyz
cho
MA MB
đạt giá trị nhỏ nhất.
A.
1;1;0M
. B.
31
; ;0
22
M



. C.
2;1;0M
. D.
13
; ;0
22
M



.
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u 193.(THPT Thuận Thành 2019) Trong không gian
Oxyz
, cho hai điểm
1; 1;1A
,
0;1; 2B
và điểm
M
thay đổi trên mặt phẳng
Oxy
. Tìm giá trị lớn nhất của
MA MB
.
A.
14
. B.
14
. C.
6
. D.
6
.
Li gii
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u 194.(Sở GDĐT Lâm Đồng 2019). Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt phẳng
( ):2 1 0P x y z
hai điểm
( 1;3;2), ( 9;4;9)AB
. Tìm điểm
M
trên
P
sao cho
MA MB
đạt giá trị nhỏ nhất.
A.
( 1;2;3)M
. B.
( 1;2; 3)M 
. C.
(1; 2;3)M
. D.
( 1;2; 3)M 
.
Li gii
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u 195. Trong không gian vi h tọa độ
Oxyz
, cho hai điểm
1;3;4A
,
3;1;0B
. Gi
M
đim trên mt phng
Oxz
sao cho tng khong cách t
M
đến
A
B
ngn nht. Tìm
hoành độ
0
x
của điểm
M
.
A.
0
4x
. B.
0
3x
. C.
0
2x
. D.
0
1x
.
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Bài toán 3. Tìm mt phng
P
sao cho khong cách t mt điểm đến
P
nh nht.
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1. Phương pháp.
Trong không gian h trc tọa độ
,Oxyz
cho mt phng
:0P Ax By Cz D
đi qua hai điểm c định
,M
.N
Khi đó khoảng cách t một điểm
S mp P
đến
mt phng
P
ln nht khi:
Gi
H
hình chiếu ca
S
lên
P
,
K
hình chiếu
ca
S
lên
MN
.
Khi đó
;d S P SH
;d S MN SK
,
Trong tam giác vuông
SHK
ta có
SH SK
(không đổi) theo quan h đường xiên và đường
vuông góc.
Vy
;d S P
ln nht khi
HK
.
Suy ra mt phng
P
nhn
SK
làm véctơ pháp tuyến.
2. Bài toán minh ha.
u 196.(THPT Chuyên Sơn La ) Trong không gian hệ tọa độ
Oxyz
, gọi
: 3 0P ax by cz
(với
,,abc
là các số nguyên không đồng thời bằng
0
) là mặt phẳng đi qua hai điểm
0; 1;2 ,M
không đi qua điểm
0;0;2H
. Biết rằng khoảng cách từ
H
đến mặt phẳng
P
đạt giá trị lớn nhất. Tổng
2 3 12T a b c
bằng
A.
16
. B.
8
. C.
12
. D.
16
.
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u 197.(THPT Phan Đình Tùng 2019) Trong không gian
Oxyz
, cho điểm
1;2;3M
. Mặt phẳng
:0P x Ay Bz C
chứa trục
Oz
cách điểm
M
một khoảng lớn nhất, khi đó tổng
A B C
bằng
A.
6
. B.
3
. C.
3
. D.
2
.
Li gii
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P
M
N
S
H
K
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u 198.(THPT Chuyên Trần Đại Nghĩa) Trong không gian với hệ trục tọa độ
Oxyz
, cho
(1;2;1)M
. Viết phương trình mặt phẳng
()P
qua
M
cắt các trục
lần lượt tại
,,A B C
sao cho
2 2 2
1 1 1
OA OB OC

đạt giá trị nhỏ nhất.
A.
( ): 2 3 8 0P x y z
. B.
( ): 1
1 2 1
x y z
P
.
C.
( ): 4 0P x y z
. D.
( ): 2 6 0P x y z
.
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u 199.(THPT TX Quãng Tr 2019) Trong không gian
Oxyz
, cho các điểm
(6;0;0)A
,
(0;3;0)B
mt phng
( ): 2 2 0P x y z
. Gi
d
đường thẳng đi qua
(2 ; 2 ; 0)M
, song song vi
()P
tng khong cách t
A
,
B
đến đường thng
d
đạt giá tr nh nhất. Vectơ nào dưới đây
một vectơ chỉ phương của
d
?
A.
1
( 10;3;8)u 
. B.
2
(14; 1; 8)u
. C.
3
(22; 3; 8)u 
. D.
4
( 18; 1;8)u
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A. L THUY󰈸T
I. VÉC CHỈ PHƯƠNG:
1. Định nghĩa:
Cho đường thẳng
. Véc tơ
0u
gọi là véc tơ chỉ phương
(VTCP) của đường thẳng
nếu giá của nó song song hoặc
trùng với
.
2. C ý :
Nếu
u
là VTCP của
thì
. ( 0)k u k
cũng là VTCP của
Nếu đường thẳng
đi qua hai điểm
A
B
thì
AB
là một VTCP.
Nếu
là giao tuyến của hai mặt phẳng
P
Q
thì
,
pQ
nn


là một VTCP của
(Trong đó
,
pQ
nn
lần lượt
là VTPT của
P
Q
).
II. PƠNG TRÌNH CỦA ĐƯỜNG THẲNG.
1. Phương trình tham số của đường thẳng.
Cho đường thẳng
đi qua
0 0 0
;;A x y z
và có VTCP
;;u a b c
.
Khi đó phương trình đường thẳng tham số
có dạng:
0
0
0
(1)
x x at
y y bt t
z z ct


t
gọi là tham số.
Chú ý . Cho đường thẳng
có phương trình
1
;;u a b c
là một VTCP của
.
Nếu điểm
0 0 0
;;M M x at y bt z ct
. Đây kỹ thuật chọn điểm thuộc đường thẳng
(
1 ẩn theo
t
) để giải các bài toán lập hệ dựa vào tính chất: vuông góc, cùng phương, thẳng
hàng, khoảng cách, góc….
2. Phương trình chính tắc:
Cho đường thẳng
đi qua
0 0 0
;;M x y z
và có VTCP
;;u a b c
với
0abc
. Khi đó phương trình
đường thẳng
có dạng:
0 0 0
(2)
x x y y z z
a b c

2
gọi là phương trình chính tắc của đường thẳng
.
3. Ví dụ minh họa.
Ví d 1. Viết phương trình tham số của đưng thng
, biết
1).
đi qua hai điểm
1;2;4A
3;5; 1 .B 
2).
đi qua
A
( ý 1) và song song với đường thng
12
:
2 1 1
x y z
d


3).
là giao tuyến ca hai mt phng
: 3 0x y z
:2 1 0yz
4).
nm trong mt phng
: 3 0x y z
đồng thi
ct vuông góc với đường
thng
12
:
2 1 1
x y z
d


u
A
u
n
β
n
α
M
β
α
§BI 3. PHƯƠNG TRÌNH ĐƯNG THNG
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Lời giải
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III. Vị trí tương đối giữa hai đường thẳng.
Cho hai đường thẳng
0 0 0
:
x x y y z z
d
a b c

đi qua
0 0 0
;;M x y z
có VTCP
;;
d
u a b c
, , ,
0 0 0
':
' ' '
x x y y z z
d
a b c

đi qua
, , ,
0 0 0
' ; ;M x y z
có VTCP
'
'; '; '
d
u a b c
.
Nếu
'
, ' 0
dd
u u MM d



'd
đồng phẳng. Khi đó xảy ra ba trường hợp
d
'd
cắt nhau
, ' 0uu



và tọa độ giao điểm là nghiệm hệ:
0 0 0
, , ,
0 0 0
' ' '
x x y y z z
a b c
x x y y z z
a b c


.
[ , '] 0
/ / '
[ , '] 0
uu
dd
u MM
[ , '] 0
'
[ , ']=0
uu
dd
u MM

Nếu
[ , '] ' 0u u MM 
d
'd
chéo nhau .
Ví d 2. Xét v trí tương đối giữa các đường thng
12
,.
Tính góc giữa hai đường thng và tìm
giao điểm ca chúng (nếu có). Biết
1).
1
1 1 5
:
2 3 1
x y z
2
1 1 1
:.
4 3 5
x y z
2).
1
:3
12
xt
yt
zt


2
0
: 9 .
55
x
yt
zt

3).
1
33
:
1 4 3
x y z

2
là giao tuyến ca hai mp
1
2
:0
: 2 2 0
x y z
x y z
.
Lời giải
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Ví d 3. Xét v trí tương đối gia các cặp đường thng sau
1).
1
13
:
2 1 2
x y z
d



2
1 1 2
:
2 1 3
x y z
d

2).
1
1 2 3
:
1 2 2
x y z
d

2
3 5 6
:
3 1 1
x y z
d

3).
1
1 2 1
:
1 2 2
x y z
d

2
2 1 1 2
:
1 1 1
x y z
d


.
Lời giải
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3. C ý :
Để xét v trí tương đối giữa hai đường thng
1 1 1
1
1 1 1
:
x x y y z z
d
a b c

2 2 2
2
2 2 2
:
x x y y z z
d
a b c

.
Ta làm như sau: Xét hệ phương trình :
1 1 2 2
1 1 2 2
1 1 2 2
'
'
'
x a t x a t
y bt y b t
z c t z c t
Nếu
có nghim duy nht
00
;'tt
thì hai đường thng
1
d
2
d
ct nhau ti
1 1 0 1 1 0 1 1 0
;;A x a t y bt z c t
.
Nếu
có vô s nghiệm thì hai đường thng
1
d
2
d
trùng nhau.
Nếu
vô nghiệm, khi đó ta xét sự cùng phương của hai véc tơ.
1 1 1 1
;;u a b c
2 2 2 2
;;u a b c
.
Nếu
1 2 1 2
//u ku d d
Nếu
12
.u k u
thì
1
d
2
d
chéo nhau.
IV. Vị trí tương đối giữa đường thẳng và mặt phẳng
Cho
:0mp Ax By Cz D
;;n A B C
VTPT đường thẳng .
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Đường thẳng
0 0 0
:
x x y y z z
a b c
;;u a b c
là VTCP và đi qua
0 0 0 0
;;M x y z
.
cắt
n
u
không cùng phương
0Aa Bb Cc
. Khi đó tọa độ giao điểm là
nghiệm của hệ :
0 0 0
0 (a)
(b)
Ax By Cz D
x x y y z z
a b c

T
0 0 0
,,b x x at y y bt z z ct
thế vào
()at
giao điểm
0 0 0
0
0
//
0
Aa Bb Cc
nu
Ax By Cz D
M

0 0 0
0
0
0
Aa Bb Cc
nu
Ax By Cz D
M

v nu
cùng phương
.n k u
.
d 4. Xét v trí tương đối giữa đưng thng
d
mp
. Tìm tọa độ giao đim ca chúng
nếu có.
1).
12 4
: 9 3 , :3 4 2 0
1
xt
d y t t x y z
zt


2).
10 4 1
: : 4 17 0
3 4 1
x y z
d y z

3).
13 1 4
: : 2 4 1 0.
8 2 3
x y z
d x y z
Lời giải.
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d 5. Xét v trí tương đối giữa đường thng
d
mp
. Tìm tọa độ giao điểm ca chúng
nếu có.
1).
12 4
: 9 3 ; :3 4 2 0
1
xt
d y t x y z
zt


2).
10 4 1
: ; : 4 17 0
3 4 1
x y z
d y z

3).
13 1 4
: ; : 2 4 1 0.
8 2 3
x y z
d x y z
Lời giải.
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V. KHOẢNG CH.
1. Khoảngch từ một điểm đến một đường thẳng:
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Cho đường thẳng
đi qua
0
M
, VTCP
u
điểm
M 
.
Khi đó để tính khoảng cách từ
M
đến
ta có các cách sau:
ch 1: Sử dụng công thức:
0
[ , ]
,
M M u
dM
u

.
ch 2: Lập phương trình
mp P
đi qua
M
vuông góc với
.
Tìm giao điểm
H
của
P
với
.
Khi đó độ dài
MH
là khoảng cách cần tìm.
Ví d 6. Tính khong cách t
2;3; 1A
đến đường thng
32
:
1 3 2
x y z
Lời giải.
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2. Khoảngch giữ hai đường thẳng chéo nhau:
Cho hai đường thẳng chéo nhau
đi qua
0
M
có VTCP
u
'
đi qua
0
'M
có VTCP
'u
. Khi đó khoảng cách giữa hai đường
thẳng
'
được tính theo các cách sau:
ch 1: Sử dụng công thức:
00
, ' . '
,'
,'
u u M M
d
uu


.
ch 2: Tìm đoạn vuông góc chung
MN
. Khi đó độ dài
MN
khoảng cách cần tìm.
ch 3: Lập phương trình
mp P
đi qua
và song song với
'
.
Khi đó khoảng cách cần tìm khoảng cách từ một điểm bất
trên
'
đến (P).
3. Ví dụ minh họa.
Ví d 7. Tính khong cách giữa hai đường thng
1
1
: 4 2
3
x
yt
zt

2
3'
: 3 '
2
xt
yt
z


.
Lời giải.
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H
u
M
M
O
u'
u
'
M
M'
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Ví d 8. Tính các khong cách sau
1). Khong cách t
3;2;1A
đến đường thng
12
:
2 3 1
x y z
2). Khong cách giữa hai đường thng
1
1 1 2
:
2 1 3
x y z
2
2 1 3
:
1 2 4
x y z
.
3). Khong cách giữa đường thng
1 1 2
:
2 1 3
x y z
và mt phng
: 4 2 1 0x x z
Lời giải.
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Ví d 9. Cho đường thng
1 2 1
:
2 1 3
x y z
và điểm
2; 5; 6A 
1). Tìm tọa đ hình chiếu ca
A
n đưng thng
.
2). Tìm tọa đ đim
M
nm trên
sao cho
35AM
.
Lời giải.
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VI. GÓC
1. Góc giữa hai đường thẳng:
Cho hai đường thẳng
0 0 0
:
x x y y z z
a b c
có VTCP
;;u a b c
và đường thẳng
0 0 0
' ' '
':
' ' '
x x y y z z
a b c
có VTCP
' '; '; 'u a b c
.
Đặt
,'
, khi đó:
2 2 2 2 2 2
' ' '
cos cos , '
. ' ' '
aa bb cc
uu
a b c a b c


.
Ví d 10. Trong không gian vi h trc tọa độ
Oxyz
, cho hai đường thng
: 5 2
14 3
xt
yt
zt

14
': 2
15
xt
yt
zt

. Xác định góc giữa hai đường thng
'
.
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Lời giải.
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d 11. Trong không gian vi h ta đ
Oxyz
, cho bn điểm
1;0;0A
,
0;1;0B
,
0;0;1C
2;1; 1D 
. Góc gia hai cnh
AB
CD
có s đo là:
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Lời giải.
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Ví d 12. Trong không gian vi h tọa độ
Oxyz
, cho hai đường thng
1
11
:
2 2 1
x y z
d


2
1 2 3
:
1 2 1
x y z
d

.
Tính
cosin
ca góc giữa hai đường thng
1
d
2
d
.
A.
6
3
B.
3
2
C.
6
6
D.
2
2
Lời giải.
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Ví d 13. Trong không gian vi h tọa độ
Oxyz
, cho hai đường thng
1
1
:2
2
xt
d y t
zt


2
2
: 1 2
2
xt
d y t
z mt



.
Để hai đường thng hp vi nhau mt góc bng
0
60
thì giá tr ca
m
bng:
A.
1m
B.
1m 
C.
1
2
m
D.
1
2
m 
Lời giải.
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2. Góc giữa đường thẳng và mặt phẳng
Cho
:0mp Ax By Cz D
;;n A B C
là một véctơ pháp tuyến và đường thẳng
:
o o o
x x y y z z
a b c
;;u a b c
là VTCP.
Gọi
là góc giữa
mp
và đường thẳng
, khi đó ta có:
2 2 2 2 2 2
sin cos ,
Aa Bb Cc
nu
A B C a b c


d 14. Trong không gian vi h tọa độ
Oxyz
, cho mt phng
: 2 1x y z
đường
thng
1
:
1 2 1
x y z
. Góc gia
A.
30
. B.
120
. C.
150
. D.
60
.
Li gii
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d 15. Trong không gian vi h tọa độ
Oxyz
, cho đường thng
65
:2
1
xt
d y t
z


mt phng
:3 2 1 0P x y
. Tính góc hp bi giữa đường thng
d
và mt phng
P
.
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Li gii
..........................................................................................................................................................................................................
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d 16. Trong không gian vi h tọa độ
Oxyz
, cho đưng thng
32
:
2 1 1
x y z
d


mt
phng
:3 4 5 8 0x y z
. Góc giữa đường thng
d
và mt phng
có s đo là:
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Li gii
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d 17. Trong không gian vi h tọa độ
Oxyz
, cho mt phng
: 2 2 3 0P x y z
và
đưng thng
:
2 1 1
x y z
d 
. Tính
sin
ca góc gia đưng thng
d
mt phng
P
.
A.
2
2
B.
3
2
C.
6
6
D.
6
3
Li gii
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3. Góc giữa hai mặt phẳng
Cho hai mặt phẳng
:0Ax By Cz D
có VTPT
1
;;n A B C
: ' ' ' ' 0A x B y C z D
có VTPT
2
'; '; 'n A B C
.
Gọi
là góc giữa hai mặt phẳng (
00
0 90

). Khi đó:
12
2 2 2 2 2 2
' ' '
cos cos ,
' ' '
AA BB CC
nn
A B C A B C


.
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d 18. Trong không gian vi h ta độ
Oxyz
, cho hai mt phng
:2 2 9 0P x y z
: 6 0Q x y
. S đo góc tạo bi hai mt phng bng:
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Lời giải.
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Ví d 19. Trong kng gian vi h trc tọa đ
Oxyz
, cho t din
ABCD
0;2;0A
,
2;0;0B
,
0;0; 2C
0; 2;0D
. S đo góc của hai mt phng
ABC
ACD
:
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Lời giải.
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Ví d 20. Trong không gian vi h ta đ
Oxyz
, cho ba điểm
1;0;0 , 0;1;0 , 0;0;1M N P
. Cosin
ca góc gia hai mt phng
MNP
và mt phng
Oxy
bng:
A.
1
3
B.
2
5
C.
1
3
D.
1
5
Lời giải.
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d 21. Trong không gian vi h tọa độ
Oxyz
, cho hai mt phng
: 6 0P x y
Q
.
Biết rằng điểm
2; 1; 2H 
hình chiếu vuông góc ca gc tọa độ
0;0;0O
xung mt phng
Q
. S đo góc giữa mt phng
P
và mt phng
Q
bng:
A.
0
30
B.
0
45
C.
0
60
D.
0
90
Lời giải.
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d 22. Trong không gian vi h ta đ
Oxyz
, cho c điểm
1;0;0 , 0;2;0 , 0;0;A B C m
. Để
mt phng
ABC
hp vi mt phng
Oxy
mt góc
0
60
thì giá tr ca
m
là:
A.
12
5
m 
B.
2
5
m 
C.
12
5
m 
D.
5
2
m 
Lời giải.
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B. PHÂN DẠNG BÀI TẬP MINH HỌA .
DNG 1. Viết phương trình đường thng.
1. Phương pháp chung.
Phương pháp chung để lập phương trình của đường thng
ta cn đi tìm một điểm đi qua
mt véc tơ ch phương (VTCP). Khi tìm VTCP của đường thng
, ta cần lưu ý:
Nếu giá của hai véc tơ không cùng phương
,ab
cùng vuông góc vi
thì
,ab


mt VTCP
ca
.
Nếu đường thng
đi qua hai đim phân bit
,MN
thì
MN
mt VTCP của đưng thng
.
2. Bài tp minh ha.
Bài tp 1. Lp ptts và ptct của đường thng
d
biết:
1).
d
đi qua
2;0;1A
và có
1; 1; 1u
là VTCP .
2).
d
đi qua
1;2;1A
1;0;0B
.
3).
d
đi qua
2;1;0M
và vuông góc vi
: 2 2 1 0P x y z
.
4).
d
đi qua
1;2; 3N 
và song song vi
13
:
2 2 1
x y z
.
5).
d
là giao tuyến ca hai mt phng
: 3 0x y z
:2 5 4 0x y z
.
a
b
M
x
0
;
y
0
;
z
0
u
u
u
M
N
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Lời giải.
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3. Câu hi trc nghim.
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Mức độ 1. Nhận biết
u 1.(Sở GD & ĐT Điện Biên) Trong không gian
Oxyz
, đường thẳng
2
3
2
xt
yt
zt


đi qua đim nào
sau đây:
A.
1;2; 1A
. B.
. C.
3; 2; 1A 
. D.
3; 2;1A 
.
Li gii
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u 2.(Chuyên KHTN 2019) Trong không gian
Oxyz
, vectơ nào dưới đây là một vectơ chỉ phương
của đường thẳng
12
:
2 1 3
x y z
d


?
A.
2; 1;3
. B.
2;1;3
. C.
1; 2;0
. D.
1;2;0
.
Li gii
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u 3.(THPT Thị Quảng Trị) Trong không gian
Oxyz
, đường thẳng
1 2 2
:
2 3 1
x y z
một vectơ chỉ phương là
A.
1
(1; 2; 2)u
. B.
2
( 2; 3; 1)u
. C.
3
( 1;2;2)u 
. D.
4
(2; 3; 1)u
.
Li gii
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u 4.(THPT Ninh Bình 2019) Trong không gian
Oxyz
, cho đường thẳng
d
song song với trục
Oy
. Đường thẳng
d
có một vectơ chỉ phương là
A.
1
2019; 0; 0u
. B.
2
0; 2019; 0u
. C.
3
0; 0; 2019u
. D.
4
2019; 0; 2019u
Li gii
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u 5.(Chuyên Đại Hc Vinh 2019) Trong không gian
Oxyz
cho đường thng
vuông góc vi mt
phng
: 2 3 0xz
. Một véc tơ chỉ phương của
là:
A.
1;0;2a
. B.
2; 1;0b
. C.
1;2;3v
. D.
2;0; 1u
.
Li gii
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u 6.(THPT Kim Liên 2019) Trong không gian h tọa độ
Oxyz
, vectơ nào sau đây một vectơ
ch phương của đường thng
12
:?
1 1 2
x y z
A.
1; 2;0u 
. B.
2;2; 4u
. C.
1;1;2u
. D.
1;2;0u 
.
Li gii
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u 7.ng Thành Nam) Trong không gian
Oxyz
, đường thẳng qua hai điểm
2;1;2M
,
3; 1;0N
có một vectơ chỉ phương là
A.
1;0;2u
. B.
5; 2; 2u
. C.
1;0;2u 
. D.
5;0;2u
.
Li gii
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u 8.(THPT Chuyên Hà Tĩnh 2019) Trong không gian vi h tọa độ
Oxyz
, cho
2 3 5OA i j k
;
24OB j k
. Tìm một vectơ chỉ phương của đường thng
AB
.
A.
2;5; 1u 
. B.
2;3; 5u 
. C.
2; 5; 1u
. D.
2;5; 9u 
.
Li gii
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u 9.(THPT Chuyên Phan Bội Châu 2019) Trong không gian với hệ tọa độ
,Oxyz
cho đường
thẳng
1 2 1
:
2 1 2
x y z
d

nhận vectơ
;2;u a b
là vectơ chỉ phương. Tính
.ab
A.
8
. B.
8
. C.
4
. D.
4
.
Li gii.
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u 10.(S GD & ĐT Đà Nẵng 2019) Trong không gian vi h tọa độ
Oxyz
, phương trình mt
phng
P
vuông góc với đường thng
22
1 2 3
x y z

và đi qua điểm
3; 4;5A
A.
3 4 5 26 0x y z
. B.
2 3 26 0x y z
.
C.
3 4 5 26 0x y z
. D.
2 3 26 0x y z
.
Li gii
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u 11.(THPT Chuyên ĐH Vinh) Trong không gian tọa độ , cho đường thng đi qua điểm
và có véctơ chỉ phương là . Phương trình nào sau đây không phải là phương
trình của đường thng ?
A. . B. . C. . D. .
Li gii
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u 12.(THPT Lương Thế Vinh) Trong hệ tọa độ
Oxyz
, cho đường thẳng
122
:
1 2 3
x y z
d

.
Phương trình nào sau đây là phương trình tham số của
d
?
A.
1
2
23
x
yt
zt

. B.
1
22
13
xt
yt
zt



. C.
1
22
23
xt
yt
zt


. D.
1
2
1
x
yt
zt


.
Li gii
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u 13.(THPT Kim Liên 2019) Trong không gian
Oxyz
, đường thng
Oz
có phương trình là
A.
0x
yt
zt
. B.
0
0
1
x
y
zt

. C.
0
0
xt
y
z
. D.
0
0
x
yt
z
.
Li gii
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u 14.(THPT Kinh n 2019) Trong không gian cho
1;2;3A
2; 1;2B
. Đường thẳng đi
qua hai điểm
AB
có phương trình là.
A.
1
23
3
xt
yt
zt


. B.
1 2 3
1 3 1
x y z

.C.
2 1 2
1 3 1
x y z


. D.
32
46
12
xt
yt
zt


Li gii
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Oxyz
(1;2;3)M
2;4;6u
52
10 4
15 6
xt
yt
zt
2
42
63
xt
yt
zt
12
24
36
xt
yt
zt
32
64
12 6
xt
yt
zt
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u 15.(THPT Chuyên KHTN 2019) Trong không gian
Oxyz
, phương trình đường thẳng đi qua
điểm
1;2; 3M
và vuông góc với mặt phẳng
: 2 1 0P x y z
là:
A.
1 2 3
1 1 2
x y z


. B.
1 2 3
1 1 2
x y z

.
C.
1 2 3
1 1 2
x y z

. D.
1 2 3
1 1 2
x y z

.
Li gii
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u 16.(S GD & ĐT PThọ 2019) Trong không gian
Oxyz
, cho điểm
(1; 2 ; 3)A
mt phng
( ):3 4 7 2 0P x y z
. Đường thẳng đi qua
A
và vuông góc vi mt phng
()P
có phương trình
A.
3
4 2 ( ).
73
xt
y t t
zt


B.
13
2 4 ( ).
37
xt
y t t
zt


C.
13
2 4 ( ).
37
xt
y t t
zt


D.
14
2 3 ( ).
37
xt
y t t
zt


Li gii
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u 17.(Sở GD & ĐT Bắc Ninh 2019) Trong không gian với hệ tọa độ
Oxyz
, phương trình đường
thẳng
d
đi qua điểm
1;2;1A
và vuông góc với mặt phẳng
: 2 1 0P x y z
có dạng
A.
1 2 1
:
1 2 1
x y z
d

. B.
22
:
1 2 1
x y z
d


.
C.
1 2 1
:
1 2 1
x y z
d

. D.
22
:
2 4 2
x y z
d


.
Li gii
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u 18.(THPT Thuan Thanh 2019) Trong không gian
,Oxyz
cho tam giác
ABC
vi
1;4; 1 ,A
2;4;3 ,B
2;2; 1 .C
Phương trình tham số của đường thng đi qua điểm
A
song song vi
BC
A.
1
4.
12
x
yt
zt

B.
1
4.
12
x
yt
zt


C.
1
4.
12
x
yt
zt

D.
1
4.
12
x
yt
zt

Li gii
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u 19.(THPT Nguyn Du 2019) Trong không gian tọa độ
Oxyz
, gi
d
giao tuyến ca hai mt
phng
: 3 0x y z
: 4 0x y z
. Phương trình tham số của đường thng
d
A.
2
22
xt
yt
zt


. B.
. C.
2
22
xt
yt
zt

. D.
2
22
xt
yt
zt

.
Li gii
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u 20.(Chuyên Nguyn Du 2019)Trong không gian
Oxyz
, phương trình đường thng đi qua hai
đim
3;1;2A
,
1; 1;0B
A.
11
2 1 1
x y z


. B.
3 1 2
2 1 1
x y z

. C.
3 1 2
2 1 1
x y z

. D.
11
2 1 1
x y z


.
Li gii
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u 21.(THPT Chuyên Hùng Vương 2019) Trong không gian
Oxyz
cho điểm
2; 1;1A
mt
phng
:2 2 1 0P x y z
. Viết đường thng
đi qua
A
và vuông góc vi mt phng
P
A.
22
:1
12
xt
yt
zt


. B.
22
:1
1
xt
yt
zt


. C.
22
:1
2
xt
yt
zt


. D.
24
: 1 2
1
xt
yt
zt


Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
169
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 22.(THPT Nguyễn Đình Chiểu 2019) Trong không gian tọa độ
Oxyz
, cho điểm
( 1;2;3)A
(3; 2;1)B
. Mặt phẳng trung trực của đoạn thẳng
AB
có phương trình
A.
2 2 4 0.x y z
. B.
2 2 0.x y z
C.
2 2 4 0x y z
. D.
2 2 0x y z
.
Li gii
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u 23.(THPT Chuyên Lê Hồng Phong 2019) Trong không gian tọa độ
Oxyz
, cho mặt phẳng
: 2 3 0P x y
. Đường thẳng
qua
1;2; 3A
vuông góc với mặt phẳng
P
phương
trình là
A.
1
22
3
xt
yt
z


. B.
1
22
33
xt
yt
zt


. C.
1
22
3
xt
yt
zt



. D.
1
22
3
xt
yt
z



.
Li gii
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u 24.(THPT Thuận Thành 2019) Trong không gian
Oxyz
, phương trình nào dưới đây
phương trình tham số của đường thẳng đi qua hai điểm
2;1;0A
;
1;3;1B
?
A.
23
12
xt
yt
zt



. B.
2
13
xt
yt
zt


. C.
32
2
1
xt
yt
z



. D.
23
12
xt
yt
zt



.
Li gii
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Mức độ 2. Thông Hiểu
u 25.(THPT Lương Thế Vinh 2019) Cho điểm
1;2;3A
và hai mặt phẳng
:2 2 1 0 P x y z
,
:2 2 1 0 Q x y z
. Phương trình đường thẳng
d
đi qua
A
song song với cả
P
Q
A.
1 2 3
1 1 4

x y z
. B.
1 2 3
1 2 6

x y z
. C.
1 2 3
1 6 2

x y z
. D.
1 2 3
5 2 6


x y z
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 26.(S GD & ĐT Mau 2019) Trong không gian
Oxyz
, cho điểm
1;1;1M
hai mt phng
: 2 1 0P x y z
,
: 2 3 0Q x y
. Viết phương trình tham s của đường thng
d
đi qua
đim
M
đồng thi song song vi c hai mt phng
P
Q
.
A.
12
: 1 4
13
xt
d y t
zt



. B.
2
:4
3
xt
d y t
zt


. C.
12
: 1 4
13
xt
d y t
zt



. D.
1
:1
12
xt
d y t
zt



.
Li gii
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u 27.(THPT Nguyn Trãi 2019) Đưng thng
giao ca hai mt phng
50xz
2 3 0x y z
thì có phương trình là
A.
21
1 3 1
x y z

. B.
21
1 2 1
x y z

. C.
2 1 3
1 1 1
x y z

. D.
2 1 3
1 2 1
x y z

Li gii
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u 28.(THPT Thanh Chương 2019) Trong không gian
Oxyz
, cho tam giác
ABC
2;1; 1 ,A
2;3;1B
0; 1;3C
. Gi
d
đường thẳng đi qua tâm đường tròn ngoi tiếp tam giác
ABC
và vuông góc vi mt phng
ABC
. Phương trình đường thng
d
A.
1 1 2
1 1 1
x y z

. B.
1
1 1 1
x y z

. C.
2
2 1 1
x y z

. D.
1
1 1 1
x y z

.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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3. Mt s k thut lập phương trình đường thẳng đc biêt.
K thuật đim
M
thuộc đưng thng
.
1. Phương pháp.
Tìm hai điểm
,AB
thuộc đường thng
.
Đim thuộc đường thng:
0 0 0
:
x x y y z z
Md
a b c
0 0 0
;;M x at y bt z ct
Da vào gi thiết để thiết lập phương trình, hệ phương trình:
Vuông góc : tích vô hướng bng 0.
Song song, thẳng hàng : tích có hướng bng
0
hoc
.
' ' '
x y z
a k b
x y z
Độ dài
2 2 2
a x y z
2. Bài tp minh ha.
Bài tp 2. Lập phương trình chính tắc của đường thng
, biết:
1).
đi qua
1;2;1A
đồng thi
cắt đường thng
1
1
:2
xt
d y t
zt


và vuông góc với đường
thng
2
113
:
2 1 2
x y z
d

.
2).
đi qua
1; 1;1M
, ct c
2
đưng thng
1
22
:1
2
xt
yt
zt


2
2
: 3 3
xt
yt
zt

.
3).
ct c
2
đưng thng
1
1
:
1 2 1
x y z
d

2
114
:
1 2 3
x y z
d

đồng thi song song
với đường thng
4 7 3
':
1 4 2
x y z
.
4).
đi qua
0; 1;2P
, đồng thi
ct
1
1 1 1
:
1 2 2
x y z
d

2
:d
13
1 2 2
x y z


ln
t ti
,AB
khác
I
tha mãn
AI AB
, trong đó
I
là giao điểm ca
1
d
2
d
Lời giải.
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 29.(THPT Lương Thế Vinh 2019) Cho các đường thẳng
1
11
:
1 2 1
x y z
d


và đường thẳng
2
23
:
1 2 2
x y z
d


. Viết phương trình đường thẳng
đi qua
1;0;2A
, cắt
1
d
và vuông góc
2
.d
A.
12
2 2 1
x y z

. B.
12
4 1 1
x y z


C.
12
2 3 4
x y z

. D.
12
2 2 1
x y z

.
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Li gii
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u 30.(THPT Lương Thế Vinh 2019) Cho các đường thẳng
1
11
:
1 2 1
x y z
d


đường thẳng
2
23
:
1 2 2
x y z
d


. Phương trình đường thẳng
đi qua
1;0;2A
, cắt
1
d
và vuông góc với
2
d
A.
12
2 2 1
x y z

. B.
12
4 1 1
x y z


C.
12
2 3 4
x y z

. D.
12
2 2 1
x y z

.
Li gii
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u 31.(THPT Chuyên Hà Tĩnh 2019) Trong không gian
,Oxyz
cho điểm
2;1;0M
và đường thng
11
:
2 1 1
x y z
d
. Viết phương trình đường thng đi qua điểm
M
ct và vuông góc vi
đưng thng
.d
A.
21
.
1 4 1
x y z

B.
21
.
1 4 1
x y z

C.
21
.
2 4 1
x y z

D.
21
.
1 4 2
x y z


Li gii
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u 32.(THPT Kim Liên 2017) Trong không gian hệ tọa độ
Oxyz
, cho hai đường thẳng
1
11
:
1 2 1
x y z
d


2
23
:
1 2 2
x y z
d


. Viết phương trình đường thẳng
đi qua điểm
1;0;2A
cắt
1
d
và vuông góc với
2
d
.
A.
12
:
2 3 4
x y z
. B.
3 3 2
:.
2 3 4
x y z
C.
5 6 2
:
2 3 4
xyz

. D.
12
:
2 3 4
x y z

.
Li gii
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u 33.(THPT Chuyên Hà Tĩnh 2019) Trong không gian
,Oxyz
cho điểm
1; 1;2A
hai đường
thng
1
12
:;
2 1 1
x y z
d
2
1
: 1 2
25
xt
d y t
zt
. Viết phương trình đường thng đi qua
A
vuông
góc vi
1
d
2
.d
A.
45
32
57
xt
yt
zt
B.
17
1 11
23
xt
yt
zt


. C.
1
12
2
x
yt
zt

. D.
7
11
32
xt
yt
zt


Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 34.(THPT Chuyên Tĩnh 2019) Trong không gian
,Oxyz
cho điểm
1; 2;3A
hai đường
thng
1
13
:;
2 1 1
x y z
d
2
1
:2
1
xt
d y t
z
. Viết phương trình đường thng đi qua
A
vuông
góc vi
1
d
2
.d
A.
1
2
3
xt
yt
zt


B.
2
12
33
xt
yt
zt

. C.
1
2
3
xt
yt
zt


. D.
12
2
33
xt
yt
zt


Li gii
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u 35.(S GD & ĐT Cn Thơ 2019) Trong không gian tọa độ
,Oxyz
cho hai đường thng
1
2 2 3
:
2 1 1
x y z
d

,
2
1
: 1 2
1
xt
d y t
zt


điểm
1;2;3A
. Đường thẳng đi qua đim
A
, vuông
góc vi
1
d
và ct
2
d
có phương trình là
A.
1 2 3
1 3 1
x y z


. B.
1 2 3
1 3 1
x y z

. C.
1 2 3
1 3 5
x y z

. D.
1 2 3
1 3 5
x y z


Li gii
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u 36.(Cm Trn Kim Hưng 2019) Trong không gian vi h tọa độ
Oxyz
cho mt phng
( ):P
2 4 0x y z
đường thng
12
:.
2 1 3
x y z
d


Đường thng
nm trong mt phng
()P
đồng thi ct và vuông góc với đường thng
d
có phương trình
A.
1 1 1
5 1 2
x y z

. B.
1 1 1
5 2 3
x y z

. C.
1 3 1
5 1 3
x y z

. D.
1 1 1
5 1 3
x y z


.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 37.(S󰉉 GD & ĐT Mau 2019) Trong không gian
Oxyz
, cho mt phng
:3 0x y z
đưng thng
3 4 1
:
1 2 2
x y z
. Phương trình của đường thng
d
nm trong mt phng
, ct và vuông góc với đường thng là:
A.
22
: 2 5
17
xt
d y t
zt


. B.
14
:5
37
xt
d y t
zt


. C.
4
:5
73
xt
dy
zt


. D.
14
:5
37
xt
d y t
zt

.
Li gii
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u 38.(THPT n y Nội 2019) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:P
2 10 0x y z
, điểm
1;3;2A
đường thẳng
2 1 1
:
2 1 1
x y z
d

. Tìm phương trình
đường thẳng
cắt
P
d
lần lượt tại
M
N
sao cho
A
là trung điểm của
MN
.
A.
6 1 3
7 4 1
x y z

. B.
6 1 3
7 4 1
x y z

.
C.
6 1 3
7 4 1
x y z


. D.
6 1 3
7 4 1
x y z


Li gii
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
177
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 39.(THPT Triu Thái 2019) Trong không gian với hệ tọa độ
Oxyz
, cho đường thng
2 1 1
:
1 1 1
x y z
d


mt phng
:2 2 0P x y z
. Đường thng
nm trong
P
, ct
d
và vuông góc vi
d
có phương trình là:
A.
1
2
xt
y
zt



. B.
1
2
xt
y
zt



. C.
1
2
xt
yt
zt


. D.
1
2
xt
y
zt


.
Li gii
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u 40.(THPT Chuyên Nguyễn Huệ 2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt
phẳng
:2 2 9 0P x y z
đường thẳng
1 3 3
:
1 2 1
x y z
d

. Phương trình tham số của
đường thẳng
đi qua
0; 1;4A
, vuông góc với
d
và nằm trong
P
là:
A.
5
:1
45
xt
yt
zt

. B.
2
:
42
xt
yt
zt


. C.
:1
4

xt
y
zt
. D.
: 1 2
4
xt
yt
zt


.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 41.(THPT ơng Thế Vinh 2019) Trong hệ tọa độ
Oxyz
, lập phương trình đường vuông góc
chung
của hai đường thẳng
1
1 3 2
:
1 1 2
x y z
d

2
3
:
13
xt
d y t
zt

.
A.
2 2 4
1 3 2
x y z


. B.
3 1 2
1 1 1
x y z

. C.
1 3 2
3 1 1
x y z

. D.
1
1 6 1
x y z

.
Li gii
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u 42.(THPT Đô Lương 2019) Trong không gian
Oxyz
, cho đường thng
1
:2
32
xt
d y t
zt


mt
phng
: 2 3 2 0P x y z
. Đường thng
nm trong mt phng
P
đồng thi ct vuông
góc đường thng
d
có phương trình là:
A.
57
: 6 5
5
xt
d y t
zt

. B.
57
: 6 5
5
xt
d y t
zt

. C.
17
: 2 5
3
xt
d y t
zt


. D.
17
:5
1
xt
d y t
zt

.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 43.(Đặng Thành Nam) Trong không gian Oxyz, cho hai đường thẳng
1
22
:
1 1 1
x y z
d


;
2
21
:
1 2 3
x y z
d


Phương trình đường thẳng
cắt
12
,dd
lần lượt tại
A
B
sao cho
AB
nhỏ nhất là
A.
32
2
xt
yt
zt


. B.
2
12
xt
yt
zt

. C.
1
12
2
xt
yt
zt


. D.
2
12
xt
yt
zt



.
Li gii
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u 44.(Chuyên Đại Học KHTN) Trong không gian với hệ tọa độ
Oxyz
, phương trình đường thẳng
đi qua điểm
1; 0;1M
và vuông góc với hai đường thẳng
1
:4
3
xt
d y t
zt

2
12
: 3 2
4
xt
d y t
zt


là:
A.
11
3 3 4
x y z

. B.
11
1 3 4
x y z

. C.
11
1 3 4
x y z

. D.
11
1 3 4
x y z

.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 45.(SGD & ĐT Bình Thuận 2019) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:3 2 0P x y z
hai đường thẳng
1
16
:
1 2 1
x y z
d


2
124
:
3 1 4
x y z
d


. Đường
thẳng vuông góc với
P
cắt cả hai đường thẳng
1
d
2
d
có phương trình là
A.
21
3 1 2
x y z

. B.
54
3 1 2
x y z

. C.
2 8 1
3 1 2
x y z

. D.
1 2 2
3 1 2
x y z

.
Li gii
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Câu 46.(THPT Nguyn Khuyến 2019) Trong không gian h tọa độ
Oxyz
,
cho điểm
1;2;3A
đưng thng
3 1 7
:
2 1 2
x y z
d

. Đường thẳng đi qua
A
, vuông góc vi
d
và ct trc
Ox
phương trình là
A.
1
22
32



xt
yt
zt
. B.
12
2
3
xt
yt
zt
. C.
12
2

xt
yt
zt
. D.
1
22
33



xt
yt
zt
.
Lời gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 47.(THPT Đoàn Thượng 2019) Trong không gian h tọa độ
Oxyz
, viết phương trình tham số
của đường thng đi qua điểm
1;2;3M
và song song vi giao tuyến ca hai mt phng lần lượt
:3 3 0P x y
,
:2 3 0Q x y z
.
A.
1
23
3
xt
yt
zt



. B.
1
23
3
xt
yt
zt



. C.
1
23
3
xt
yt
zt



. D.
1
23
3
xt
yt
zt



.
Li gii
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u 48.(THPT S1 nghĩa 2019) Viết phương trình đường thẳng
d
qua
1;2;3A
cắt đường
thẳng
1
2
:
2 1 1
x y z
d

và song song với mặt phẳng
: 2 0P x y z
.
A.
1
2
3
xt
yt
zt



. B.
1
2
3
xt
yt
z


. C.
1
2
3
xt
yt
z


. D.
1
2
3
xt
yt
zt



.
Li gii
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u 49.(THPT Kim Liên Hà Ni 2019) Trong không gian
Oxyz
, phương trình đưng thng
đi
qua
1;2;4A
song song vi
P
:
2 4 0x y z
và cắt đường thng
:d
2 2 2
3 1 5
x y z

A.
. B.
. C.
12
2
44
xt
y
zt

. D.
.
Li gii
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u 50.(THPT Chuyên Nguyn Du 2019) Trong không gian
Oxyz
, đường thng qua
1;2; 1M
và song song vi hai mt phng
: 8 0P x y z
,
:2 5 3 0Q x y z
có phương trình là
A.
1 2 1
4 7 3
x y z


. B.
1 2 1
4 7 3
x y z

. C.
1 2 1
4 7 3
x y z

. D.
1 2 1
4 7 3
x y z

.
Li gii
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u 51.(THPT Toàn Thng 2019) Trong không gian vi h tọa độ
Oxyz
, cho điểm
1;2;3A
và mt
phng
:2 4 1 0P x y z
. Đường thng
d
đi qua điểm
A
, song song vi mt phng
P
,
đồng thi ct trc
Oz
. Viết phương trình tham s của đường thng
d
.
A.
15
26
3
xt
yt
zt



. B.
2
2
xt
yt
zt

. C.
13
22
3
xt
yt
zt



. D.
1
26
3
xt
yt
zt



.
Li gii
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3.2. K thut lp hai mt phng ct nhau theo giao tuyến là đưng thng
.
1. Phương pháp.
Tìm hai mt phng phân bit chứa đường thng
. Khi đó
chính giao tuyến ca hai mt
phẳng đó. nhiều mt phng cha
nên khi chn mt phng cha
, ta thường da vào
các du hiu sau:
Nếu đường thng
đi qua
M
vuông góc vi
d
thì đường thng
nm trong mt phng
đi qua
M
và vuông góc vi
d
Nếu đường thng
đi qua
M
cắt đường thng
d
thì đường thng
nm trong mt
phẳng đi qua
M
và đường thng
d
.
Nếu đường thng
đi qua
M
và song song vi m
mp P
thì đường thng
nm trong mt
phẳng đi qua
M
và song song vi
P
.
Nếu đường thng
song song với đường thng
d
cắt đường thng
'd
thì đường thng
nm trong mt phng cha
'd
và song song với đường thng
d
.
2. Bài tp minh ha.
Bài tp 3. Lập phương trình chính tắc của đường thng
, biết:
1).
đi qua
1;2;1A
đồng thi
cắt đường thng
1
1
:2
xt
d y t
zt


và vuông góc với đường
thng
2
113
:
2 1 2
x y z
d

,
2).
đi qua
9;0; 1B
, đồng thi
cắt hai đường thng
1
1 3 1
:
2 1 1
x y z
,
2
2 3 4
:
1 1 3
x y z

.
Lời giải.
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Bài tp 4. Lập phương trình của đường thng
biết
đi qua
1;0; 1M
và vuông góc vi hai
đưng thng
12
21
: ; : 1 2
5 8 3
0
xt
x y z
d d y t
z

Lời giải.
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Bài tp 5. Lập phương trình của đường thng
đi qua
1;4; 2M
song song vi hai mt
phng
: 6 6 2 3 0P x y z
:3 5 2 1 0Q x y z
.
Lời giải.
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Bài tp 6. Lập phương trình của đường thng
nm trong
: 2 0P y z
cắt hai đường
thng
11
1 2 '
: ; : 4 2 '
41
x t x t
d y t d y t
z t z






.
Lời giải.
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Bài tp 7. Lập phương trình của đường thng
đi qua
4; 5;3M 
và cắt hai đường thng
1
1 3 2
:
3 2 1
x y z
d


2
2 1 1
:
2 3 5
x y z
d

Lời giải.
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Bài tp 8. Lập phương trình của đường thng
đi qua
, ct c
2
đưng thng
1
22
:1
2
xt
yt
zt


2
2
: 3 3
xt
yt
zt

.
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Lời giải.
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Bài tp 9. Lập phương trình của đường thng
ct c
2
đưng thng
1
1
:
1 2 1
x y z
d

2
114
:
1 2 3
x y z
d

đồng thi song song với đường thng
4 7 3
':
1 4 2
x y z
Lời giải.
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Bài tp 10. Trong không gian
Oxyz
cho ha đường thng
1
1 3 2
:
1 2 3
x y z
d

2
4 2 3
:
1 4 3
x y z
d


Chng minh rằng hai đường thng
12
,dd
chéo nhau. Viết phương trình đường vuông góc chung
và tính khong cách giữa hai đường thẳng đó.
Lời giải.
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Bài tp 11. Viết phương trình đường thng
biết
1).
đi qua
2;2;1A
và ct
Oy
tại điểm
B
sao cho
2OB OA
2).
đi qua
1;1;2B
và cắt đường thng
2 3 1
:
1 2 1
x y z
d

ti
C
sao cho tam giác
OBC
có din tích bng
83
2
.
3).
đi qua
2;3;1M
và to
1
1 1 1
:
1 2 2
x y z
d

,
2
13
:
1 2 2
x y z
d


mt tam giác cân ti
A
. Biết rng
A
là giao điểm
1
d
2
d
.
Lời giải.
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Mức độ 3. Vận dụng
u 52.(Sở GD & ĐT Nam) Trong không gian
Oxyz
cho đường thẳng
31
:
2 1 1
x y z
d


mặt phẳng
:x y 3z 2 0.P
Gọi
d
đường thẳng nằm trong
P
, cắt vuông góc với
d
.
Đường thẳng
'd
có phương trình là:
A.
11
2 5 1
x y z


. B.
11
2 5 1
x y z

. C.
11
2 5 1
x y z

. D.
11
2 5 1
x y z

.
Lời giải
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u 53.(THPT Chuyên Nguyễn Du 2019) Trong không gian tọa độ
Oxyz
, cho
2;3; 1M
đường thẳng
3
:
2 4 1
x y z
d

. Đường thẳng qua
M
vuông góc với
d
và cắt
d
có phương trình là
A.
2 3 1
5 6 32
x y z

. B.
2 3 1
6 5 32
x y z

. C.
2 3 1
5 6 32
x y z

. D.
2 3 1
6 5 32
x y z

Lời giải
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u 54.(THPT Yên A Ninh Bình 2019) Trong không gian hệ tọa độ
Oxyz
, cho đường thẳng
1 1 3
:
1 2 2
x y z
d

mặt phẳng
:2 2 3 0P x y z
, phương trình đường thẳng
nằm
trong mặt phẳng
P
, cắt
d
và vuông góc với
d
A.
22
15
56
zt
yt
zt


. B.
22
15
56
zt
yt
zt

. C.
22
15
56
zt
yt
zt

. D.
22
15
56
zt
yt
zt


.
Lời giải
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u 55.(THPT ISCHOOL Nha Trang) Trong không gian
Oxyz
, cho điểm
1;0;2A
và đường thng
11
:
1 1 2
x y z
d


. Phương trình đường thng
đi qua
A
, vuông góc và ct
d
là:
A.
12
1 1 1
x y z

. B.
12
1 1 1
x y z

. C.
12
2 2 1
x y z

. D.
12
1 3 1
x y z

.
Lời giải
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190
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 56.(THPT Phúc Trch Tĩnh 2019) Trong không gian vi h tọa độ
Oxyz
, cho đường thng
22
:
1 1 1
x y z
mt phng
: 2 3 4 0P x y z
. Phương trình tham s của đường
thng
d
nm trong
P
, cắt và vuông góc đường thng
A.
32
1
1
xt
yt
zt


. B.
13
23
1
xt
yt
zt

. C.
33
12
1
xt
yt
zt


. D.
3
12
1
xt
yt
zt


.
Lời giải
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u 57.(Chuyên Đi Hc Vinh) Trong không gian ,
cho 2 đường thng ,
mt phng . Đưng thng vuông góc vi mt phng ,
ct
có phương trình
A. . B. . C. . D. .
Lời giải
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Oxyz
12
:
13
xt
d y t
zt
2
: 1 2
2
xt
d y t
zt



: 2 0P x y z
P
d
d
3 1 2
1 1 1
x y z

1 1 1
1 1 4
x y z


2 1 1
1 1 1
x y z

1 1 4
2 2 2
x y z

Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
191
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 58. tham kho BGD 2018) Trong không gian h tọa độ
Oxyz
, cho hai đường thng
1
3 3 2
:
1 2 1
x y z
d


;
2
5 1 2
:
3 2 1
x y z
d

mt phng
: 2 3 5 0P x y z
. Đưng
thng vuông góc vi
P
, ct
1
d
2
d
có phương trình là
A.
11
1 2 3
x y z

. B.
2 3 1
1 2 3
x y z

. C.
3 3 2
1 2 3
x y z

. D.
11
3 2 1
x y z

.
Lời giải
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u 59.(THPT Chuyên Hoàng Văn Th 2019) Trong không gian tọa độ
Oxyz
, cho hai đường thng
1
d
,
2
d
mt phng (
) có phương trình
1
13
:2
12
xt
d y t t
zt

,
2
24
:
3 2 2
x y z
d



, mt phng
( ): 2 0x y z
. Phương trình đường thng
nm trong mt phng (
), ct c hai đường
thng
1
d
2
d
A.
2 1 3
8 7 1
x y z

. B.
2 1 3
8 7 1
x y z


. C.
2 1 3
8 7 1
x y z

. D.
2 1 3
8 7 1
x y z

Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 60.(THPT Chuyên ĐH KHTN 2019) Phương trình đường thẳng song song với đường thẳng
12
:
1 1 1
x y z
d


và cắt hai đường thẳng
1
112
:
2 1 1
x y z
d

;
2
1 2 3
:
1 1 3
x y z
d

là:
A.
112
1 1 1
x y z


. B.
11
1 1 1
x y z

. C.
1 2 3
1 1 1
x y z

. D.
11
1 1 1
x y z

.
Lời giải
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u 61.(THPT y Đôn 2019) Trong không gian h tọa độ
Oxyz
, cho điểm
1;3;2A
,
:mp P
20x y z
đường thng
11
:
2 1 1
x y z
d


. Viết phương trình đưng thng
ct
P
d
lần lượt ti
M
,
N
sao cho
A
là trung điểm ca
MN
.
A.
1
:3
22
xt
yt
zt


. B.
1
:3
22
xt
yt
zt


. C.
1
:3
22
xt
yt
zt

. D.
1
:3
22
xt
yt
zt


.
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 62.(THPT Ngô Quyn Nội 2019) Trong không gian với hệ tọa độ
Oxyz
, cho điểm
2;1;1A
và hai đường thẳng
1
3
:1
2
xt
dy
zt


,
2
32
:3
0
xt
d y t
z


. Phương trình đường thẳng đi qua
,A
vuông góc
với
1
d
và cắt
2
d
A.
12
.
2 1 2
x y z

B.
2 1 1
1 1 1
x y z


. C.
2 1 1
2 1 2
x y z

. D.
12
1 1 1
x y z

.
Lời giải
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u 63.(THPT Lương Thế Vinh 2019) Trong hệ tọa độ
Oxyz
, lập phương trình đường vuông góc
chung
của hai đường thẳng
1
1 3 2
:
1 1 2
x y z
d

2
3
:
13
xt
d y t
zt

.
A.
2 2 4
1 3 2
x y z


. B.
3 1 2
1 1 1
x y z

. C.
1 3 2
3 1 1
x y z

. D.
1
1 6 1
x y z

.
Lời giải
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 64.(THPT Thanh Chương 2019) Trong không gian
Oxyz
, cho hai đường thng chéo nhau
1
112
:
3 2 2
x y z
d

,
2
4 4 3
:
2 2 1
x y z
d

. Phương trình đường vuông góc chung ca
hai đường thng
12
,dd
A.
1
41
:
2 1 2
x y z
d


. B.
2 2 2
6 3 2
x y z

. C.
2 2 2
2 1 2
x y z

. D.
41
2 1 2
x y z


Lời giải
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u 65.(THTT S 3-2018) Trong không gian vi h tọa độ
Oxyz
, viết phương trình đường vuông
góc chung của hai đường thng
2 3 4
:
2 3 5

x y z
d
1 4 4
:
3 2 1


x y z
d
.
A.
1
1 1 1

x y z
. B.
2 2 3
2 3 4

x y z
. C.
2 2 3
2 2 2

x y z
. D.
23
2 3 1


x y z
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 66.(THTT s6-2018) Trong không gian với hệ tọa độ
,Oxyz
cho
:2 10 0,mp P x y z
điểm
1;3;2A
đường thẳng
22
:1
1
xt
d y t
zt


. Tìm phương trình đường thẳng
cắt
P
d
lần lượt tại hai điểm
M
N
sao cho
A
là trung điểm cạnh
MN
.
A.
6 1 3
7 4 1
x y z


. B.
6 1 3
7 4 1
x y z

. C.
6 1 3
7 4 1
x y z

. D.
6 1 3
7 4 1
x y z


Lời giải
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Mức độ 3. Vận dụng cao
u 67.(Chuyên ĐH Vinh 2019) Trong không gian cho ba đường thng
. Đường thng vuông góc vi đồng thi ct
tương ứng ti sao cho độ dài nh nht. Biết rng một vectơ chỉ phương
Giá tr bng
A. B. C. D.
Lời giải
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Oxyz
1
:,
1 1 2
x y z
d

1
31
:,
2 1 1
x y z
2
12
:
1 2 1
x y z
d
12
,
,HK
HK
; ;1 .u h k
hk
0.
4.
6.
2.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 68.(THPT Nguyn Trãi 2019) Đưng thng
đi qua điểm
3;1;1M
, nm trong mt phng
: 3 0x y z
to với đường thng
1
: 4 3
32
x
d y t
zt

mt góc nh nhất thì phương trình
ca
A.
1
2
x
yt
zt

. B.
85
34
2
xt
yt
zt


. C.
12
1
32
xt
yt
zt



. D.
15
14
32
xt
yt
zt



.
Lời giải
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u 69.i Học Sư Phạm Hà Ni 2019) Trong không gian to đ
Oxyz
, cho điểm
1;2;4A
hai
đim
,MB
tho mãn
. . 0MA MA MB MB
. Gi s đim
M
thay đổi trên đường thng
3 1 4
:
221
x y z
d

. Khi đó điểm
B
thay đổi trên đường thẳng có phương trình là:
A.
1
7 12
:
2 2 1
x y z
d


. B.
2
1 2 4
:
2 2 1
x y z
d

.
C.
3
:
2 2 1
x y z
d 
. D.
4
5 3 12
:
2 2 1
x y z
d

.
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 70.Trong không gian vi h tọa độ
Oxyz
, cho điểm
2;1;5A
và hai mt phng có phương
trình
:2 3 7 0,P x y z
:3 2 1 0Q x y z
. Gi
M
điểm nm trên mt phng
P
đim
N
nm trên mt phng
Q
tha mãn
2AN AM
. Khi
M
di động trên mt phng
P
thì
qu tích điểm
N
là một đường thng có phương trình là
A.
35
8 11
67
xt
yt
zt

. B. . C. . D. .
Lời giải
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u 71.(S GD & ĐT Phú Th 2019) Trong không gian h trc tọa độ
Oxyz
, cho mt phng
:2 3 2 12 0x y z
. Gi
,,A B C
lần lượt giao điểm ca
vi ba trc tọa độ, đường
thng
d
đi qua tâm đường tròn ngoi tiếp tam giác
ABC
và vuông góc vi
có phương trình
A.
3 2 3
2 3 2
x y z

. B.
3 2 3
2 3 2
x y z

. C.
3 2 3
2 3 2
x y z

. D.
3 2 3
2 3 2
x y z

Lời giải
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7 11
85
67
xt
yt
zt


7 11
85
87
xt
yt
zt

25
3 11
17
xt
yt
zt


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u 72.(Đề tham khảo BGD 2018) Trong không gian
Oxyz
, cho hai điểm
2; 2;1A
,
8 4 8
;;
333
B



.
Đường thẳng đi qua tâm đường tròn nội tiếp tam giác
OAB
vuông góc với mặt phẳng
OAB
có phương trình là
A.
1 3 1
1 2 2
x y z

. B.
1 8 4
1 2 2
x y z

. C.
1 5 11
3 3 6
1 2 2
x y z

. D.
2 2 5
9 9 9
1 2 2
x y z

Lời giải
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Dng 3. Hình chiếu của điểm, ca đường thng lên đưng thng, mt phng.
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Bài toán 1. Tìm hình chiếu của điểm
;;
A A A
A x y z
xung đưng thng
:
0
0
0
:
x x at
y y bt
z z ct


Đim
đối xng
A
ca
A
qua
.
Gi
H
là hình chiếu vuông góc ca
A
lên mt phng
P
H
là giao điểm ca mt phng
P
với đường thng
qua
M
và vuông góc vi mt phng
P
.
Viết phương trình mt phng
P
đi qua điểm
A
và nhn
véc tơ chỉ phương ca
;;
P
u n a b c

làm véc tơ
pháp tuyến.
Gii h phương trình của mt phng
P
.tH
H
là trung điểm ca
AA
'
'
'
2
2
2
A H A
A H A
A H A
x x x
y y y A
z z z


2. Bài tp minh ha.
Bài tp 12. Cho đường thng
và mt phẳng (P) có phương trình
12
: 1 ,
2
xt
y t t R
zt

: 2 2 11 0.P x y z
1). Tìm tọa đ đim
H
là hình chiếu ca
1; 2; 5A 
trên
.
2). Tìm tọa đ đim
A
sao cho
2AA AH
và ba điểm
,,A A H
thng hàng.
3). Tìm tọa đ đim
B
đối xng với điểm
1; 1; 2B
qua
P
.
Lời giải.
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P
A
H
u
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Bài tp 13. Trong không gian
,Oxyz
cho đường thng
112
2 3 1
x y z

và điểm
4;3;2A
1). Tìm tọa đ đim
M
thuộc đường thng
sao cho
105AM
,
2). Tìm tọa đ đim
'A
đối xng vi
A
qua
.
3). Tìm tọa đ đim
D
thuc
sao cho khong cách t
D
đến
: 2 2 2 0x y z
bng
1
.
Lời giải.
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Bài tp 14. Trong không gian
,Oxyz
cho điểm
5;5;0A
và đường thng
d
:
1 1 7
2 3 4
x y z

1). Tìm tọa đ đim
'A
đối xng với điểm
A
qua đường thng
d
.
2). Tìm to đ đim
,BC
thuc
d
sao cho tam giác
ABC
vuông ti
C
29BC
.
Lời giải.
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3. Câu hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
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u 73.(THPT Lê Qúy Đôn 2019) Trong không gian
,Oxyz
đưng thng
12
:
2 3 1
x y z
d


đi qua
điểm nào dưới đây?
A.
1; 0; 2M
. B.
2; 3; 1N
. C.
1; 0; 2P
. D.
1; 0; 2Q
.
Lời giải
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u 74.(Cụm Trần Kim Hưng 2019) Trong không gian với hệ tọa độ
Oxyz
, cho đường thẳng
3 2 1
:
2 1 4
x y z
d

. Điểm nào sau đây không thuộc đường thẳng
d
.
A.
1; 1; 5M 
. B.
1; 1;3M
. C.
3; 2; 1M 
. D.
5; 3;3M
.
Lời giải
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u 75.(Nguyn Tt Thành Yên Bái) Trong không gian vi h tọa độ
Oxyz
, cho đường thng
d
phương trình
1 2 3
3 2 4
x y z

. Điểm nào sau đây không thuộc đường thng
d
?
A.
7;2;1P
. B.
2; 4;7Q 
. C.
4;0; 1N
. D.
1; 2;3M
.
Lời giải
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u 76.(THPT Chuyên Hưng Yên 2019) Trong không gian với hệ tọa độ
,Oxyz
đường thẳng
:
2
1
23
xt
y
zt

không đi qua điểm nào sau đây?
A.
2;1; 2 .M
B.
4;1; 4 .P
C.
3;1; 5 .Q
D.
0;1;4 .N
Lời giải
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u 77.(Toán Hc Tui Tr 2019) Trong không gian
Oxyz
, cho điểm
3; 2 ; 1M
. Hình chiếu
vuông góc của điểm
M
lên trc
Oz
là điểm:
A.
3
3; 0 ; 0M
. B.
4
0; 2 ; 0M
. C.
1
0; 0; 1M
. D.
2
3; 2 ; 0M
.
Lời giải
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u 78.(Tn Hc Tui Tr 2019) Trong không gian
Oxyz
, cho điểm
3; 2;1M 
.
Hình chiếu vuông góc của điểm
M
lên mt phng
Oxy
là điểm:
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A.
3
3; 0; 0M
. B.
4
0; 2 ;1M
. C.
1
0; 0;1M
. D.
2
3; 2; 0M 
.
Lời giải
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u 79.(THPT Chuyên Hùng Vương 2019) Trong không gian vi h tọa độ
Oxyz
, cho tam giác
ABC
1;1;2 , 2;3;1 , 3; 1;4A B C
. Viết phương trình đường cao ca tam giác
ABC
k t
đỉnh
B
A.
2
3
1
xt
yt
zt


. B.
2
3
1
xt
y
zt

. C.
2
3
1
xt
yt
zt


. D.
2
3
1
xt
yt
zt


.
Lời giải
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Mức độ 3. Vận dụng
u 80.(THPT Thăng Long 2019) Trong không gian
Oxyz
cho đường thẳng
12
:
2 1 1
x y z
điểm
4;1;1A
. Gọi
'A
là hình chiếu của
A
trên
. Mặt phẳng nào sau đây vuông góc với
'?AA
A.
2 2 0xy
. B.
4 7 1 0x y z
. C.
3 3 0x y z
. D.
4 1 0x y z
.
Lời giải
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u 81.(THPT Chuyên Ngoi Ng Ni) Trong không gian tọa độ
Oxyz
, cho mt phng
( ):2 2 7 0P x y z
điểm
(1;1; 2)A
. Điểm
( ; ; 1)H a b
hình chiếu vuông góc ca
()A
trên
()P
. Tng
ab
bng
A.
2
. B.
3
. C.
1
. D.
3
.
Lời giải
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u 82.(THPT Đô Lương 2019) Trong không gian
Oxyz
, cho điểm
và đường thng
2
: 1 2
2
xt
yt
zt

. Hình chiếu vuông góc của điểm
A
lên đường thng
A.
3; 1;2M
. B.
11; 17;18H
. C.
1;3; 2N
. D.
2;1;0K
.
Lời giải
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u 83.(THPT Kim Ln 2017) Trong không gian với hệ tọa độ
Oxyz
, tìm tọa độ hình chiếu
B
của
điểm
5;3; 2B
trên đường thẳng
13
:
2 1 1
x y z
d


.
A.
1;3;0B
. B.
5;1;2B
. C.
3;2;1B
. D.
9;1;0B
.
Lời giải
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u 84.(THPT Chuyên Hà Tĩnh 2019) Trong không gian vi h tọa độ
Oxyz
, cho điểm
1;2;2A
và đường thng
6 1 5
:
2 1 1
x y z
d

. Tìm tọa độ đim
B
đối xng vi
A
qua
d
.
A.
3;4; 4B 
. B.
2; 1;3B
. C.
3;4; 4B
. D.
3; 4;4B
.
Lời giải
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u 85.(THPT Kim Liên 2018) Trong không gian vi h tọa độ
Oxyz
, tìm tọa độ đim
A
đối xng
với điểm
1;0;3A
qua mt phng
: 3 2 7 0P x y z
.
A.
1; 6;1A

. B.
0;3;1A
. C.
1;6; 1A
. D.
11;0; 5A
.
Lời giải
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u 86.(THPT Chuyênn La 2019) Trong không gian hệ tọa độ
,Oxyz
cho đường thẳng
:d
1 1 1
3 2 1
x y z


và điểm
5;0;1A
. Điểm đối xứng của
A
qua đường thẳng
d
có tọa độ là
A.
1;1;1
. B.
5;5;3
. C.
4; 1;0
. D.
3; 2; 1
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài toán 2. Tìm hình chiếu ca đưng thng
0
0
0
:
x x at
y y bt
z z ct


xung
:0mp P Ax By Cz D
Trường hp 1. Đưng thng
0
0
0
:
x x at
y y bt
z z ct


song song vi
:0mp P Ax By Cz D
1. Phương pháp.
Do
0 0 0
0
0
//
0
Aa Bb Cc
nu
mp P
Ax By Cz D
M

Gi
là hình chiếu vuông góc ca
lên mt phng
P
là giao tuyến ca hai mt phng
P
mp Q
Viết phương trình mt phng
Q
đi qua điểm
O
M
nhn cp véc tơ chỉ phương
,
QP
n u n


làm véc tơ
pháp tuyến.
viết dượi dng giao tuyến ca hai
,.mp Q mp P
4. Bài tp minh ha.
Bài tp 15. Trong không gian cho đường thng
112
:
2 1 1
x y z
. Tìm hình chiếu vuông
góc ca
trên mt phng
Oxy
.
Lời giải
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Q
'
n
P
M
O
u
P
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 87.(THPT Chuyên Phan Bội Châu 2019) Trong không gian với hệ tọa độ
Oxyz
, cho đường
thẳng
1 2 1
:
2 1 3
x y z
d

một mặt phẳng
: 3 0P x y z
. Đường thẳng
'd
hình
chiếu của
d
theo phương
Ox
lên
P
,
'd
nhận
; ;2019u a b
một vec chỉ phương . Xác
định tổng
ab
A.
2019
. B.
2020
. C.
2018
. D.
2019
.
Lời giải.
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u 88.(Sở GD & ĐT Quãng Bình 2019) Trong không gian hệ tọa độ
Oxyz
, cho mặt phẳng
( ):P
30x y z
đường thẳng
21
:
2 1 3
x y z
d


. Hình chiếu vuông góc của đưng thng
d
trên
()P
có phương trình là:
A.
12
.
5 8 13
x y z

B.
12
.
2 7 5
x y z

C.
12
.
4 3 7
x y z

D.
12
.
2 3 5
x y z

Lời giải
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u 89.(THPT Nguyn Công Tr 2019) Cho đường thng
d
:
21
2 3 2
x y z

mt phng
()P
:
20x y z
. Phương trình hình chiếu vuông góc ca
d
trên
()P
A.
1
12
23
xt
yt
zt


. B.
1
12
23
xt
yt
zt


.
C.
1
12
23
xt
yt
zt


. D.
1
12
23
xt
yt
zt



.
Lời giải
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u 90.(Chuyên T Trng Cn Thơ) Trong không gian vi h tọa độ
Oxyz
, cho đường thng
1 3 1
:
3 4 1
x y z
d

mt phng
:2 2 12 0P x y z
. Viết phương trình đường thng
d
là hình chiếu vuông góc của đường thng
d
trên mt phng
P
A.
1 2 3
:
2 1 2
x y z
d

B.
1 4 3
:
3 4 1
x y z
d

.C.
42
:
3 1 1
x y z
d


. D.
1 4 2
:
3 4 1
x y z
d

.
Lời giải
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u 91.(THPT Kinh Môn 2018) Trong không gian cho đường thng
112
:
2 1 1
x y z
. Tìm
hình chiếu vuông góc ca
trên mt phng
Oxy
.
A.
0
1
0
x
yt
z
. B.
12
1
0
xt
yt
z

. C.
12
1
0
xt
yt
z

. D.
12
1
0
xt
yt
z
.
Lời giải
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Trường hp 2. Đưng thng
0
0
0
:
x x at
y y bt
z z ct


ct
:0mp P Ax By Cz D
tại điểm
.A
1. Phương pháp.
Do
0Aa Bb Cc
n
u
không cùng phương
Suy ra
cắt
tại
A
.
Tọa độ giao điểm
A
là nghiệm của hệ :
0
0
0
0 (a)
(b)
Ax By Cz D
x x at
y y bt
z z ct



Gi
là hình chiếu vuông góc ca
lên mt phng
P
là giao tuyến ca hai mt phng
P
mp Q
Viết phương trình mt phng
Q
đi qua điểm
O
M
nhn cp véc tơ chỉ phương
,
QP
n u n


làm véc tơ
pháp tuyến.
Đưng thng
viết dưới dng giao tuyến ca hai
,.mp Q mp P
2. Ví d minh ha.
Bài tp 16. Lập phương trình của đường thng
, biết
1).
là hình chiếu vuông góc ca
12
:
1 2 1
x y z
d


lên
: 1 0mp x y z
2).
đi qua
2;3; 1A
và ct
d
tại điểm
B
sao cho
, 2 3dB
.
n
P
u
A
'
P
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải.
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Mức độ 3. Vận dụng
u 93.(THPT Chuyên Hà Tĩnh 2019) Trong không gian vi h tọa độ
Oxyz
, cho đường thng
6 1 5
:
2 1 1
x y z
d

và mt phng
: 2 3 4 0P x y z
. Viết phương trình đường thng
d
hình chiếu vuông góc ca
d
trên
P
.
A.
6
25
23
xt
yt
zt


. B.
6
25
23
xt
yt
zt

. C.
6
25
23
xt
yt
zt


. D.
6
25
23
xt
yt
zt


.
Lời giải
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u 94.(THPT Chuyên Thái Bình 2019) Trong không gian
Oxyz
, gọi
d
hình chiếu vuông góc
của đường thẳng
1 2 3
:
2 3 1
x y z
d

trên mặt phẳng tọa độ
Oxy
. Vecto nào dưới đây một
vecto chỉ phương của
d
?
A.
2;3;0u
. B.
2;3;1u
. C.
2;3;0u 
. D.
2; 3;0u 
.
Lời giải
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u 95.(THPT Hồng Bàng 2018) Trong không gian với hệ toạ độ
Oxyz
, cho đường thẳng
phương trình
2
: 3 2
13
xt
d y t
zt


. Viết phương trình đường thẳng
d
hình chiếu vuông góc của
d
lên mặt phẳng
Oyz
.
A.
0
: 3 2
13
x
d y t
zt

. B.
0
: 3 2
0
x
d y t
z

. C.
2
: 3 2
0
xt
d y t
z

. D.
:2
0
xt
d y t
z
.
Lời giải
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u 96.(Cụm Trần Kim Hưng 2019) Trong không gian với hệ trục tọa độ
Oxyz
cho mặt phẳng
: 1 0P x y z
đường thẳng
4 2 1
:
2 2 1
x y z
d

. Viết phương trình đường thẳng
d
là hình chiếu vuông góc của
d
trên mặt phẳng
P
.
A.
21
5 7 2
x y z

. B.
21
5 7 2
x y z

C.
21
5 7 2
x y z

. D.
21
5 7 2
x y z

.
Lời giải
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u 97.(THPT Hậu Lộc 2018) Trong không gian tọa độ
Oxyz
, cho hai điểm
1;0; 3 ,A
3; 1;0B
.
Viết phương trình tham số của đường thẳng
d
hình chiếu vuông góc của đường thẳng
AB
trên mặt phẳng
Oxy
.
A.
0
33

x
yt
zt
. B.
12
0
33

xt
y
zt
. C.
12
0


xt
yt
z
. D.
0
0
33
x
y
zt
.
Lời giải
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u 98.(THPT Hồng Bàng 2018) Trong không gian với hệ toạ độ
Oxyz
, cho đường thẳng
2
: 3 2
13
xt
d y t
zt


. Viết phương trình đường thẳng
d
là hình chiếu vuông góc của
d
lên
mp Oyz
.
A.
0
: 3 2
13
x
d y t
zt

. B.
0
: 3 2
0
x
d y t
z

. C.
2
: 3 2
0
xt
d y t
z

. D.
:2
0
xt
d y t
z
.
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Lời giải
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u 99.(THPT Thch Thành 2019) Trong không gian
Oxyz
cho mt phng
: 3 0P x y z
đưng thng
12
:
1 2 1
x y z
d


. Hình chiếu vuông góc ca
d
trên
mp P
có phương trình là
A.
1 1 1
1 4 5
x y z


. B.
1 1 1
3 2 1
x y z


. C.
1 1 1
1 4 5
x y z

. D.
1 4 5
1 1 1
x y z

.
Li gii
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u 100.Trong không gian
Oxyz
, cho đường thẳng
1 3 1
:,
2 1 2 2
x y z
d
mm


1
,2
2
m




mt
phng
: 6 0P x y z
. Gọi đường thng
là hình chiếu vuông góc ca
d
lên mặt phẳng
P
. Có bao nhiêu số thực
m
để đường thẳng
vuông góc với giá của véctơ
( 1;0;1)a 
?
A.
2
. B.
1
. C.
3
. D.
0
.
Li gii
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 101.(THPT 󰉼󰉴) Trong không gian v󰉵i h󰉪 tr󰉺c
Oxyz
, cho m󰉢t ph󰉠ng 󰉼󰉴
trình
: 5 4 0P x y z
 󰉼󰉶ng th󰉠ng
1 1 5
:
2 1 6
x y z
d

. Hình chi󰉦u vuông góc c󰉻a
󰉼󰉶ng th󰉠ng
d
trên m󰉢t ph󰉠ng
P
󰉼󰉴
A.
23
22
xt
yt
zt

. B.
2
22
xt
yt
zt

. C.
13
2
1
xt
yt
zt


. D.
3
2
1
xt
y
zt


.
L󰉶i gi󰉘i
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u 102.(THPT Chuyên Thái Nguyên 2019) 󰉵󰉪󰉭󰉳
,Oxyz
󰉼󰉶󰉠
3 1 1
:
3 1 1
x y z
d

󰉢󰉠
: 4 0P x z
󰉦󰉼󰉴󰉼󰉶󰉠hình 󰉦
󰉻󰉼󰉶󰉠
d
󰉢󰉠
P
.
A.
33
1
1
xt
yt
zt


. B.
3
1
1
xt
yt
zt


. C.
3
1
1
xt
y
zt

. D.
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
215
󰉵󰉚-󰉪 Tel: 0935.660.880
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u 103 .(THPT Chuyên Bình Long 2019) 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
2 1 2
:
1 1 2
x y z
󰉢󰉠
:0P x y z
󰉳󰉴󰉫󰉼󰉴
u
󰉻󰉼󰉶
󰉠
󰉦󰉻󰉼󰉶󰉠
󰉢󰉠
P
.
A.
1;1; 2u 
. B.
1; 1;0u 
. C.
1;0; 1u 
. D.
1; 2;1u 
.
L󰉶i gi󰉘i
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u 104.(THPT 󰉼 2019) 󰉵󰉪󰉭󰉳
Oxyz
󰉢󰉠
: 3 0P x y z
󰉼󰉶󰉠
12
:
1 2 1
x y z
d


󰉼󰉶󰉠
'd
󰉯󰉽󰉵
d
󰉢
󰉠
P
󰉼󰉴
A.
1 1 1
1 2 7
x y z

. B.
1 1 1
1 2 7
x y z

. C.
1 1 1
1 2 7
x y z

. D.
1 1 1
1 2 7
x y z

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 105.(󰉺  󰉰   󰇜    󰉵 󰉪 󰉭 󰉳
,Oxyz
 󰉢 󰉠
:3 5 2 8 0P x y z
󰉼󰉶󰉠
75
: 7
65
xt
d y t t
zt


󰉼󰉴󰉼󰉶󰉠
󰉯󰉽󰉵󰉼󰉶󰉠
d
󰉢󰉠
.P
.
A.
55
: 13
25
xt
yt
zt
. B.
17 5
: 33
66 5
xt
yt
zt

. C.
11 5
: 23
32 5
xt
yt
zt

. D.
13 5
: 17
104 5
xt
yt
zt

.
L󰉶i gi󰉘i
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D󰉗ng 4. 󰉦󰉼󰉴󰉼󰉶󰉻, c󰉻󰉼󰉶ng th󰉠ng.
Bài toán 1. 󰉦󰉼󰉴󰉼󰉶󰉻
.ABC
Xét tam giác
,ABC
󰉼󰉶
A
󰉴󰉫󰉼󰉴
11
u AB AC
AB AC

󰉼󰉹󰉗󰉼󰉶
A
󰉴󰉫
󰉼󰉴
11
u AB AC
AB AC

.
1. Ví d󰉺 minh h󰉭a.
Bài t󰉝p 17. Trong không gian v󰉵i h󰉪 to󰉗 󰉳 󰉧 các vuông góc
,Oxyz
cho
1
1 1 1
:
1 2 2
x y z
2
13
:
1 2 2
x y z
c󰉞t nhau cùng n󰉟m trong m󰉢t ph󰉠ng
P
. Vi󰉦󰉼󰉴󰉼󰉶ng phân
giác c󰉻a góc t󰉗o b󰉷󰉼󰉶ng th󰉠ng
12
,
và n󰉟m trong m󰉢t ph󰉠ng
P
.
󰉶󰉘
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Phân giác ngoài
Phân giác trong
C
B
A
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
217
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 18. Trong không gian v󰉵i h󰉪 to󰉗 󰉳 󰉧 các vuông góc
Oxyz
cho tam giác
,ABC
v󰉵󰉨m
1; 2;1 , 2;2;1 , 1; 2;2 .A B C
H󰉮i 󰉼󰉶 
A
󰉻tam giác
ABC
󰉞󰉢
󰉠
Oyz
󰉨 ?
A.
48
0; ; .
33



B.
24
0; ;
33



C.
D.
󰉶󰉘
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Bài toán 2. 󰉦󰉼󰉴󰉼󰉶󰉭󰉻󰉼󰉶󰉠
1
d
c󰉞t nhau t󰉗󰉨m
; ; .
A A A
A x y z
1 1 1 1 2 2 2 2
; ; , ; ;u a b c u a b c
󰉘󰉿󰉼󰉶󰉠
1
d
2
d
󰉞󰉗
0 0 0
;;A x y z
󰉚󰉼󰉹󰉴󰉫󰉼󰉴
1 1 1 1 2 2 2 2
; ; , ; ;u a b c u a b c
󰉼󰉶󰉻󰉼󰉶󰉠
1
d
2
d
󰉴󰉫󰉼󰉴󰉼󰉹󰉬󰉷
12
12
11
u u u
uu
1 1 1 2 2 2
2 2 2 2 2 2
1 1 1 2 2 2
11
; ; ; ;a b c a b c
a b c a b c

󰉼󰉶ng h󰉹p:
2
d
Phân giác góc tù
Phân giác góc nhọn
u
2
u
1
A
d
2
d
1
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
󰉼󰉶ng h󰉹p 1: 󰉦
1 2 1 2
12
11
0u u u u u
uu
là véc tơ chỉ phương của phân giác to bi
góc nhn giữa hai đường thng
1
d
2
d
12
12
11
u u u
uu
là véc tơ chỉ phương của phân
giác to bi góc tù giữa hai đường thng
1
d
2
d
.
󰉼󰉶ng h󰉹p 2: 󰉦
1 2 1 2
12
11
0u u u u u
uu
là véc tơ chỉ phương của phân giác to bi
góc tù giữa hai đường thng
1
d
2
d
12
12
11
u u u
uu
là véc tơ chỉ phương của phân
giác to bi góc nhn giữa hai đường thng
1
d
2
d
.
2. Ví d󰉺 minh h󰉭a.
Bài t󰉝p 19. Trong không gian
,Oxyz
cho
1
1 1 1
:
1 2 2
x y z
;
2
13
: 1 4
1
xt
yt
z


. Vi󰉦󰉼󰉴
󰉼󰉶ng phân giác
c󰉻a góc nh󰉭n t󰉗o b󰉷󰉼󰉶ng th󰉠ng
12
,
.
L󰉶i gi󰉘i
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Bài t󰉝p 20. Trong không gian v󰉵i h󰉪 to󰉗 󰉳 󰉧 các vuông góc
,Oxyz
cho
1
13
: 1 4
1
xt
yt
z

. G󰉭󰉼󰉶ng
th󰉠ng
2
󰉨m
(1;1;1)A
có véc tơ chỉ phương
2
2;1;2u 
. Vi󰉦󰉼󰉴󰉼󰉶ng phân
giác
c󰉻a góc nh󰉭n t󰉗o b󰉷󰉼󰉶ng th󰉠ng
12
,
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
219
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 21. 󰉼󰉶ng th󰉠ng
12
1 1 3 1
: , : .
2 1 1 1 2 1
x y z x y z

1). Ch󰉽ng minh r󰉟󰉼󰉶ng th󰉠ng
1
2
c󰉞t nhau và l󰉝󰉼󰉴󰉢t ph󰉠ng ch󰉽a
󰉼󰉶ng th󰉠
2). 󰉨m
M
thu󰉳c
1
có kho󰉘󰉦n
2
b󰉟ng
210
.
3
3). L󰉝󰉼󰉴󰉯 󰉼󰉶ng phân giác c󰉻a các góc t󰉗o b󰉷󰉼󰉶ng th󰉠ng.
L󰉶i gi󰉘i.
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󰉽󰉳. 󰉝󰉺
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 106.(THPT Chuyên Lê H󰉰ng Phong 2019) Trong không gian h󰉪 t󰉭󰉳
,Oxyz
󰉼󰉶ng
th󰉠ng
1
1
:2
3
xt
d y t
z


2
1
: 2 7 .
3
x
d y t
zt


󰉼󰉴󰉼󰉶ng phân giác c󰉻a góc nh󰉭n gi󰊀a
1
d
2
d
A.
1 2 3
5 12 1
x y z

. B.
1 2 3
5 12 1
x y z

. C.
1 2 3
5 12 1
x y z

. D.
1 2 3
5 12 1
x y z

.
L󰉶i gi󰉘i
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u 107.( S󰉷 Phú Th󰉭 2019) Trong không gian h󰉪 t󰉭󰉳
,Oxyz
󰉨m
1;2; 2A
. Bi󰉦t m c󰉻󰉼󰉶ng tròn n󰉳i ti󰉦p tam giác . Giá tr󰉬 c󰉻a
b󰉟ng
A. 1. B. 3. C. 2. D. 0.
L󰉶i gi󰉘i
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8 4 8
;;
333
B



;;I a b c
OAB
a b c
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
221
󰉵󰉚-󰉪 Tel: 0935.660.880
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u 108.(Chuyên 󰉗i H󰉭c Vinh 2019) Trong không gian h󰉪 t󰉭󰉳 , cho tam giác c
󰉨m 󰉼󰉶ng phân giác trong k󰉤 t󰉾 c󰉻a tam giác 
󰉨󰉨m sau?
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 109.󰇛󰉧󰉽) Trong không gian
Oxyz
󰉼󰉶󰉠
13
:3
54



xt
dy
zt
󰉭
󰉼󰉶󰉠󰉨
1; 3;5A
󰉴󰉫󰉼󰉴
1;2; 2u
󰉼󰉶󰉻
󰉭󰉗󰉷
d
󰉼󰉴
A.
12
25
6 11


xt
yt
zt
. B.
12
25
6 11

xt
yt
zt
. C.
17
35
5


xt
yt
zt
. D.
.
L󰉶i gi󰉘i
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Oxyz
ABC
1;2;3 , 3; 1;2 , 2; 1;1A B C
A
ABC
0;4;4P
2;0;1M
1;5;5N
3; 2;2Q
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 110.󰇛󰉧 Chính Th󰉽c 2018 ) Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng
1
: 2 .
3
xt
d y t
z


G󰉭i
󰉼󰉶ng th󰉠󰉨m
(1;2;3)A
và có 󰉴 ch󰉫 󰉼󰉴
(0; 7; 1).u
󰉼󰉶ng phân giác c󰉻a góc
nh󰉭n t󰉗o b󰉷i
d
󰉼󰉴
A.
16
2 11 .
38
xt
yt
zt



B.
45
10 12 .
2
xt
yt
zt

C.
45
10 12 .
2
xt
yt
zt
D.
15
2 2 .
3
xt
yt
zt



L󰉶i gi󰉘i
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u 111.(THPT H󰉘i H󰉝u-2018) Trong không gian h󰉪 t󰉭󰉳
Oxyz
󰉼󰉶ng th󰉠ng c󰉞t nhau
1
2
: 2 2
1
xt
yt
zt

,
2
1
:
2
xt
yt
zt

,tt
. Vi󰉦󰉼󰉴󰉼󰉶ng phân giác c󰉻a góc nh󰉭n t󰉗o b󰉷i
1
2
.
A.
1
2 3 3
x y z

. B.
1
1 1 1
x y z

. C.
1
2 3 3
x y z

. D.
1
1 1 1
x y z

.
L󰉶i gi󰉘i
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u 112.(󰉭󰉱󰉤) Trong không gian 󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠󰉞
nhau
1
2
: 2 2
1
xt
yt
zt

,
2
1
:
2
xt
yt
zt

,tt
󰉦󰉼󰉴󰉼󰉶phân 󰉻󰉭󰉗
󰉷
1
2
.
A.
1
2 3 3
x y z

. B.
1
1 1 1
x y z

. C.
1
2 3 3
x y z

. D. 󰉘A, B, C 󰉧
L󰉶i gi󰉘i
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
223
󰉵󰉚-󰉪 Tel: 0935.660.880
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u 113.(󰉭󰉱󰉤) Trong không gian h󰉪 t󰉭󰉳
Oxyz
󰉼󰉶ng th󰉠ng
󰉼󰉴
1
11
:,
1 1 2
x y z
d


2
13
:
2 4 2
x y z
d



. Vi󰉦󰉼󰉴trình 󰉼󰉶ng phân giác c󰉻a
nh󰊀ng góc tù t󰉗o b󰉷i
12
,dd
.
A.
13
3 5 4
x y z


. B.
13
1 1 1
x y z

. C.
11
2 1 1
x y z

. D.
13
2 1 1
x y z

.
L󰉶i gi󰉘i
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u 114.(THPT Chuyên B󰉞c Giang) Trong không gian v󰉵i h󰉪 t󰉭󰉳
Oxyz
cho tam giác
ABC
bi󰉦t
2;1;0A
,
3;0;2B
,
4;3; 4C
. Vi󰉦󰉼󰉴󰉼󰉶ng phân giác trong c󰉻a góc
A
.
A.
2
1
0
x
yt
z

. B.
2
1
x
y
zt
. C.
2
1
0
xt
y
z

. D.
2
1
xt
y
zt

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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Nh󰉝n xét:
󰉼󰉶ng phân giác trong c󰉻a góc
BAC
󰉴󰉫 󰉼󰉴
11
u AB AC
AB AC
.
.
D󰉗ng 5. M󰉳t s󰉯i toán 󰉦n góc, kho󰉘󰉼󰉴
1. 󰉼󰉴
Ta v󰉝n d󰉺ng các ki󰉦n th󰉽c sau:
N󰉦u
0
0
0
0
00
, , .: , ( )
x x at
A Ay y bt t R
z
t
z
x at y z
c
b ct
t


V󰉝n d󰉺ng kho󰉘󰉼󰉵󰇛󰇜󰉼󰉴󰉠ng hàng󰉨 thi󰉦t l󰉝p
󰉼󰉴󰉪 󰉼󰉴trình.
2. Bài t󰉝p minh h󰉭a.
Bài t󰉝p 22. Tìm
m
󰉨 󰉼󰉶ng th󰉠ng sau c󰉞t nhau và tìm t󰉭󰉳 󰉨m c󰉻a chúng.
12
6 2 3 4 3 2
: ; :
2 4 1 4 1 2
x y z x y z
dd
m

.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
225
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 23. Trong không gian
Oxyz
cho m󰉢t ph󰉠ng
:2 2 0x y z n
󰉼󰉶ng th󰉠ng
1 1 3
:
2 1 2 1
x y z
m
. Tìm
,mn
󰉨:
1). 󰉼󰉶ng th󰉠ng
n󰉟m trong
mp
2). 󰉼󰉶ng th󰉠ng
song song v󰉵i
mp
.
󰉶󰉘
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Bài t󰉝p 24. Tìm
m
󰉨
1). 󰉼󰉶ng th󰉠ng
1
6 3 1
:
2 2 1
x y z m
d
m


2
42
:
4 3 2
x y z
d


c󰉞t nhau. Tìm giao
󰉨m c󰉻a chúng.
2). 󰉼󰉶ng th󰉠ng
2
2
2
21
: 1 4 4 1
2
m
x m m t
d y m m t
z m m t
song song v󰉵i
:2 2 0P x y
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 25. Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng :
1
1 1 3
:
1 2 1
x y z
2
2 3 9
:
3 2 2
x y z
1). Ch󰉽ng minh r󰉟󰉼󰉶ng th󰉠ng
1
,
2
chéo nhau. Tính góc và kho󰉘ng cách gi󰊀a hai
󰉼󰉶ng th󰉠ng
1
2
.
2). 󰉨m
,AB
󰉱i trên
1
sao cho
3AB
󰉨m
C
󰉼󰉶ng th󰉠ng
2
sao cho
ABC
có di󰉪n tích nh󰉮 nh󰉙t.
3). Vi󰉦󰉼󰉴󰉼󰉶ng th󰉠ng
d
c󰉞󰉼󰉶ng th󰉠ng
12
,
l󰉚󰉼󰉹t t󰉗i
,MN
th󰉮a mãn
65MN
d
t󰉗o v󰉵i
1
m󰉳t góc
th󰉮a
8
cos
15
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
227
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 26. Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng
4 3 2 3 8 7
:
2 1 1 4 3
m
x m y m z m
d
m m m

v󰉵i
31
1; ;
42
m



. Ch󰉽ng minh r󰉟ng khi
m
󰉱󰉼󰉶ng th󰉠ng
m
d
luôn n󰉟m trong m󰉳t
m󰉢t ph󰉠ng c󰉯 󰉬nh. Vi󰉦󰉼󰉴h m󰉢t ph󰉠
󰉶󰉘
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Bài t󰉝p 27. Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng
12
12
: ; : ,
1 1 2
1
xt
x y z
d d y t t
zt

Xét v󰉬 󰉼󰉴󰉯i gi󰊀a
1
d
2
d
. Tìm t󰉭󰉳 󰉨m
12
,M d N d
sao cho
MN
song song v󰉵i
:0mp P x y z
󰉳 dài
2MN
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 28. Tìm t󰉭󰉳 󰉨m thu󰉳󰉼󰉶ng th󰉠ng
12
:
2 1 3
x y z
kho󰉘ng cách t󰉾 
󰉦n m󰉢t ph󰉠ng
:2 2 1 0Q x y z
b󰉟ng
1
.
󰉶󰉘
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Bài t󰉝p 29. Trong không gian v󰉵i h󰉪 t󰉭󰉳
Oxyz
󰉼󰉶ng th󰉠ng :
1
3 3 3
:
2 2 1
x y z
d

;
2
52
:
6 3 2
x y z
d


Ch󰉽ng minh r󰉟ng
1
d
2
d
c󰉞t nhau t󰉗i
I
. Tìm t󰉭󰉳 󰉨m
,AB
l󰉚󰉼󰉹t thu󰉳c
12
,dd
sao cho
tam giác
AIB
cân t󰉗i
I
và có di󰉪n tích b󰉟ng
41
42
.
󰉶󰉘
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
229
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 30. Trong không gian v󰉵i h󰉪 to󰉗 󰉳 󰉧 các vuông góc
Oxyz
cho b󰉯󰉨m
1;0;0 , 1;1;0 , 0;1;0 , 0;0;A B C D m
v󰉵i
0m
là tham s󰉯.
1). Tính kho󰉘ng cách gi󰊀󰉼󰉶ng th󰉠ng
AC
BD
khi
2m
.
2). G󰉭i
H
hình chi󰉦u vuông góc c󰉻a
O
trên
BD
. Tìm các giá tr󰉬 c󰉻a tham s󰉯
m
󰉨 di󰉪n tích
tam giác
OBH
󰉗t giá tr󰉬 l󰉵n nh󰉙t.
󰉶󰉘
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Bài t󰉝p 31. Trong không gian
Oxyz
cho
1). 󰉨m
1; 1;0M
󰉼󰉶ng th󰉠ng
:
2 1 1
2 1 1
x y z

:mp P
20x y z
.
Tìm t󰉭󰉳 󰉨m
A
thu󰉳c m󰉢t ph󰉠ng
P
bi󰉦t r󰉟󰉼󰉶ng th󰉠ng
AM
vuông góc v󰉵i
kho󰉘ng cách t󰉾
A
󰉦󰉼󰉶ng th󰉠ng
b󰉟ng
33
.
2
2). 󰉨m
2;5;3A
󰉼󰉶ng th󰉠ng
12
:
2 1 2
x y z
d


. Vi󰉦󰉼󰉴󰉢t ph󰉠ng
Q
ch󰉽󰉼󰉶ng th󰉠ng
d
sao cho kho󰉘ng cách t󰉾
A
󰉦n
Q
l󰉵n nh󰉙t.
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
󰉶󰉘
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Bài t󰉝p 32. Trong không gian h󰉪 t󰉭󰉳
,Oxyz
cho
0;1;0 , 2;2;2 , 2;3;1A B C
󰉼󰉶ng th󰉠ng
1 2 3
:
2 1 2
x y z
d

󰉨m
M
󰉼󰉶ng th󰉠ng
d
󰉨:
a). Th󰉨 tích t󰉽 di󰉪n
MABC
b󰉟ng
3
.
b). Di󰉪n tích tam giác
MAB
có di󰉪n tích nh󰉮 nh󰉙t .
󰉶󰉘
................................................................................................
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
231
󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 33. Trong không gian h󰉪 t󰉭󰉳
,Oxyz
cho
: 2 5 0P x y z
󰉼󰉶ng th󰉠ng
:d
3
1 3,
2
x
yz
󰉨m
. G󰉭i
󰉼󰉶ng th󰉠ng n󰉟m trên
P
󰉨m c󰉻a
d
P
󰉰ng th󰉶i vuông góc v󰉵i
d
.
Tìm trên
nh󰊀󰉨m
M
sao cho kho󰉘ng cách
AM
ng󰉞n nh󰉙t.
󰉶󰉘
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Bài t󰉝p 34. Trong không gian h󰉪 t󰉭󰉳
,Oxyz
󰉨m
10;2; 1A
󰉼󰉶ng th󰉠ng
d
󰉼󰉴
31
.
23
xz
y


L󰉝p 󰉼󰉴 trình m󰉢t ph󰉠ng
P

A
, song song v󰉵i
d
kho󰉘ng cách t󰉾
d
󰉦n m󰉢t
ph󰉠ng
P
là l󰉵n nh󰉙t.
󰉶󰉘
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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Bài t󰉝p 35. Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng
:d
3 2 1
2 1 1
x y z

m󰉢t ph󰉠ng
P
:
20x y z
. L󰉝󰉼󰉴󰉼󰉶ng th󰉠ng
n󰉟m trong m󰉢t ph󰉠ng
P
, c󰉞t
d
t󰉗o v󰉵i
d
góc l󰉵n nh󰉙t.
󰉶󰉘
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3. Câu h󰉮i tr󰉞c nghi󰉪m.
󰉽󰉳. 󰉝󰉦
u 115.(THPT Yên 2019) Trong không gian h󰉪 t󰉭󰉳
Oxyz
, 󰉼󰉶ng th󰉠ng 󰉼󰉴
12
:2
22
xt
d y t t
zt

󰉨m
1;2;Mm
. m g tr󰉬 tham s󰉯
m
󰉨 󰉨m
M
thu󰉳󰉼󰉶ng th󰉠ng
d
.
A.
2m
. B.
2m 
. C.
1m
. D.
0m
.
L󰉶i gi󰉘i
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
233
󰉵󰉚-󰉪 Tel: 0935.660.880
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u 116.(THPT chuyên 󰉩 trãi 2019) 󰉨󰉻󰉼󰉶󰉠
32
: 2 3
64
xt
d y t
zt

5
: 1 4
20
xt
d y t
zt



󰉭󰉳
A.
5; 1;20
. B.
3; 2;1
. C.
3;7;18
. D.
3; 2;6
.
L󰉶i gi󰉘i
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u 117.(THPT chuyên Lam 󰉴 2019) Trong không gian v󰉵i h󰉪 to󰉗 󰉳
Oxyz
󰉼󰉶ng th󰉠ng
1
1 3 3
:
1 2 3
x y z
d


v
2
3
: 1 2 ,
0
xt
d y t t
z
. M󰉪n󰉧 n󰉼󰉵ng ?
A.
1
d
c󰉞t v vuông gc v󰉵i
2
d
. B.
1
d
song song
2
d
.
C.
1
d
c󰉞t v không vuông gc v󰉵i
2
d
. D.
1
d
cho
2
d
.
L󰉶i gi󰉘i
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u 118.(THPT chuyên Biên a) 󰉵󰉪󰉺󰉗󰉳
Oxyz
󰉼󰉶󰉠
1
1 2 3
:
2 3 4
x y z
d

và
2
1
: 2 2
32
xt
d y t
zt



󰉦󰉝󰉧󰉬󰉼󰉴󰉯󰉼󰉶󰉠
A. 󰉞󰉼 B. 󰉾󰉞󰉾vuông góc.
C. 󰉞 D. 󰉼󰉞
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 119.󰇛󰉼󰉶󰉪 2019) 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
󰉼󰉴
1
:
12
x mt
d y t
zt

1
: 2 2
3
xt
d y t
zt




󰉼󰉶󰉠󰉞
A.
5m
. B.
0m
. C.
1m
. D.
1m 
.
L󰉶i gi󰉘i
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u 120.󰇛󰉷󰉰󰇜 Trong không gian
Oxyz
󰉼󰉶󰉠
d
'd
󰉼󰉴󰉚󰉼󰉹
2 4 1
:
2 3 2
x y z
d

4
' : 1 6 ;
14
xt
d y t t
zt
󰉬󰉼󰉴󰉯󰉻
󰉼󰉶󰉠
d
'd
là :
A.
d
'd
󰉵 B.
d
'd
󰉞
C.
d
'd
trùng nhau. D.
d
'd
chéo nhau.
L󰉶i gi󰉘i
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u 121.󰇛 2017) Trong không gian 󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
1
12
:
2 1 1
x y z
d
2
12
:1
3
xt
d y t
z
󰉪󰉧󰉼󰉵
A.
12
,dd
song song. B.
12
,dd
chéo nhau. C.
12
,dd
󰉞. D.
12
,dd
vuông góc.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
235
󰉵󰉚-󰉪 Tel: 0935.660.880
u 122.󰇛󰉩󰉭󰇜󰉼󰉶󰉠
1
12
: 2 3
34
xt
d y t
zt



2
34
: 5 6
78
xt
d y t
zt



󰉪󰉧󰉪󰉧
A.
1
d
trùng
2
d
. B.
12
dd
. C.
1
d
2
d
chéo nhau. D.
1
d
󰉞
2
d
.
L󰉶i gi󰉘i
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u 123.(THPT Chuyên Vinh 2017) 󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
21
:
3
2
12
x yz
d



42
:
6 2 4
x y z
d


󰉪󰉧
A.
//dd
. B.
dd
. C.
d
d
chéo nhau. D.
d
d
󰉞
L󰉶i gi󰉘i
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u 124.(THPT Ng.T.Minh Khai) 󰉼󰉶󰉠 󰉼󰉴
1
21
:
4 6 8
x y z
d



2
72
:
6 9 12
x y z
d


󰉬󰉼󰉴󰉯󰊀
1
d
2
d
:
A. Song song. B. Trùng nhau. C. 󰉞. D. Chéo nhau.
L󰉶i gi󰉘i
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u 125.󰇛󰉩󰉪-󰉦󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
1
13
:
1 2 3
x y z
d


2
2
: 1 4
26
xt
d y t
zt


󰉠󰉬󰉠󰉬?
A. 󰉼󰉶󰉠
1
d
,
2
d
󰉞. B. 󰉼󰉶󰉠
1
d
,
2
d
trùng nhau.
C. 󰉼󰉶󰉠
1
d
,
2
d
chéo nhau. D. 󰉼󰉶󰉠
1
d
,
2
d
󰉵.
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
L󰉶i gi󰉘i
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u 126.󰇛󰉺󰉼󰉶󰉴󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
, cho hai
󰉼󰉶󰉠
1
:
12
x at
d y t
zt

1
: 2 2
3
xt
d y t
zt



󰉬󰉻
a
󰉨󰉼󰉶󰉠
d
d
󰉞
A.
2a 
. B.
1a 
. C.
0a
. D.
1a
.
L󰉶i gi󰉘i
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u 127.󰇛󰉷󰉰󰇜Tìm
m
󰉨󰉼󰉶󰉠󰉞
tz
ty
mtx
d
21
1
:
'3
'22
'1
:'
tz
ty
tx
d
.
A.
2m
. B.
1m
. C.
0m
. D.
1m 
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
237
󰉵󰉚-󰉪 Tel: 0935.660.880
u 128. 󰉼󰉶󰉠
1
1
:
12
x mt
d y t
zt

2
1 2 3
:
1 2 1
x y z
d


. Tìm
m
󰉨󰉼󰉶󰉠
1
d
2
d
.
A.
0m
. B.
1m 
. C.
1m
. D.
2m
.
L󰉶i gi󰉘i
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u 129.󰇛󰉚󰇜󰉵󰉪󰉭󰉳
Oxyz
󰉬󰉼󰉴󰉯󰉻
󰉼󰉶󰉠
1
12
: 2 3
54
xt
d y t
zt


2
7 3 '
: 2 2 '
1 2 '
xt
d y t
zt


là:
A. 󰉞 B. Trùng nhau. C. Song song. D. Chéo nhau.
L󰉶i gi󰉘i
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u 130.󰇛󰉷󰉼󰉵󰇜󰉵󰉪󰉭󰉺󰉭󰉳
Oxyz
, cho hai
󰉼󰉶󰉠
1
32
:1
14
xt
yt
zt
2
4 2 4
:
3 2 1
x y z
󰉠󰉬
A.
1
2
chéo nhau và vuông góc nhau. B.
1
2
󰉵
C.
1
󰉞󰉵
2
. D.
1
󰉞󰉵
2
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 131󰇛󰉷󰉰󰇜 Trong không gian 󰉵󰉪󰉭󰉳
Oxyz
󰉦󰉼󰉶󰉠
1
11
1
1
:
12
x mt
d y t
zt

2
22
2
1
: 2 2
3
xt
d y t
zt



󰉞
m
󰉟
A.
2m
. B.
1
2
m
. C.
3m
. D.
0m
.
L󰉶i gi󰉘i
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V󰉬 󰉼󰉴󰉯i c󰉻󰉼󰉶ng th󰉠ng v󰉵i m󰉢t ph󰉠ng
󰉼󰉶ng th󰉠ng
d

0 0 0 0
;;M x y z
và có vtcp
( ; ; )
d
u a b c
:0mp Ax By Cz D
vtpt
;;n A B C
. 
󰉼󰉴:
d
c󰉞t
( ) 0Aa Bb Cc
(
n
không vuông góc v󰉵i
d
u
)
0 0 0
0
/ /( )
0
Aa Bb Cc
d
Ax By Cz D
(
n
vuông góc v󰉵i
d
u
0
()M
)
0 0 0
0
()
0
Aa Bb Cc
d
Ax By Cz D

(
n
vuông góc v󰉵i
d
u
0
()M
)
( ) / / , 0
dd
d u n u n



󰉼󰉴:
Cho m󰉢t ph󰉠ng
()
: 󰉼󰉶ng th󰉠ng
01
02
03
:
x x a t
d y y a t
z z a t



󰉼󰉴
0 1 0 2 0 3
( ) ( ) ( ) 0A x a t B y a t C z a t
(󰉛n
t
) (*)
d
//
()
(*) vô nghi󰉪m.
d
c󰉞t
()
󰇛󰇜󰉳t nghi󰉪m.
d
()
(*) có vô s󰉯 nghi󰉪m.
u 132.(THPT Kim Liên 2017) 󰉵󰉪󰉭󰉳
Oxyz,
󰉼󰉶󰉠󰉼󰉴
trình
14
:.
5 3 1
x y z
d


󰉮󰉼󰉶󰉠
d
󰉵󰉢󰉠󰉢󰉠
󰉼󰉴󰉼󰉵
A.
( ): 2 2 0x y z
. B.
( ): 2 9 0x y z
.
C.
( ):5 3 2 0x y z
. D.
( ):5x 3y z 9 0
.
L󰉶i gi󰉘i
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0Ax By Cz D
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
239
󰉵󰉚-󰉪 Tel: 0935.660.880
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u 133.(THPT chuyên Thánh Tông) 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
11
:
2 1 3
x y z
d


󰉢󰉠
: 5 4 0x y z
󰉬󰉬󰉼󰉴󰉯󰉻
d
A.
d
󰉞󰉵
. B.
d
.
C.
d
. D.
//d
.
L󰉶i gi󰉘i
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u 134.(Chuyên KHTN 2019) 󰉵󰉪󰉭󰉳
Oxyz
󰉭
󰉢󰉠󰉽
󰉼󰉶󰉠
23
( ):
1 1 2
x y z
d


󰉵󰉢󰉠
: 2z 1 0xy
󰉮
󰉦󰉻
󰉨
A.
0;1;3
. B.
2;3;3
. C.
5;6;8
D.
1; 2;0
L󰉶i gi󰉘i
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u 135.󰇛󰉩󰇜󰉼󰉶󰉠
1 1 2
:
1 2 3
x y z
d

󰉢󰉠
: 4 0.x y z
󰉠󰉬󰉠󰉬
A.
d
. B.
d
. C.
//d
. D.
d
󰉞
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 136.󰇛󰉦󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉢󰉠
:2 0xy

󰉪󰉧󰉪󰉧
A.
//Oz
. B.
Oy
. C.
Oz
. D.
// Oyz
.
L󰉶i gi󰉘i
................................................................................................
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u 137.󰇛󰇜󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
:
1 1 2
x y z
󰉵󰉢󰉠󰉢󰉠
A.
: 2 0x y z
. B.
: 2 0Q x y z
. C.
:0x y z
. D.
:0P x y z
.
L󰉶i gi󰉘i
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u 138.󰇛󰉷-󰉰󰇜󰉵󰉪󰉺󰉗󰉳
Oxyz
󰉼󰉶󰉠
d
󰉢󰉠
P
󰉼󰉴󰉽󰉼󰉴
3 1 2
2 1 1
x y z

3 5 5 0x y z
󰉭
󰉢󰉠
Q
󰉢󰉠
Oxz
󰉭󰉪󰉧󰉯󰉪󰉧
A.
dP
d
󰉞
Q
. B.
//dP
//dQ
.
C.
//dP
d
󰉞
Q
. D.
d
󰉞
P
d
󰉞
Q
.
L󰉶i gi󰉘i
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u 139.󰇛󰉩󰇜 Trong 󰉵󰉪󰉭󰉳
Oxyz
󰉢󰉠
: 2 3 1 0 x y z
 󰉼󰉶 󰉠
3
: 2 2
1

xt
d y t
z
   󰉪󰉧  󰉪 󰉧

A.
d
. B.
d
. C.
//d
. D.
d
󰉞
.
L󰉶i gi󰉘i
................................................................................................
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠
241
󰉵󰉚-󰉪 Tel: 0935.660.880
................................................................................................
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u 140.(THPT chuyên Quý 󰇜 󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
15
:
1 3 1
x y z
d



󰉢󰉠
:3 3 2 6 0P x y z
󰉪󰉧
A.
d
󰉞󰉵
P
. B.
d
󰉵
P
.
C.
d
󰉟
P
. D.
d
󰉵
P
.
L󰉶i gi󰉘i
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u 141.󰇛󰉼󰉴󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
d
󰉼󰉴
1 2 3
2 4 1
x y z

. 󰉢󰉠
: 2 7 0P x y mz
,
m
tham 󰉯󰊁. Tìm
󰉙󰉘󰉬󰉻
m
󰉨󰉼󰉶󰉠
d
󰉵󰉢󰉠
P
?
A.
1
2
m 
. B.
6m 
. C.
2m 
. D.
10m
.
L󰉶i gi󰉘i
................................................................................................
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u 142.(THPT Huy 󰉝 2019) 󰉵󰉪󰉺󰉭󰉳
,Oxyz
󰉼󰉶󰉠
󰉼󰉴
2 1 1
:.
1 1 1
x y z
d

󰉢󰉠
2
: 1 7 0,P x my m z
󰉵
m
tham
󰉯󰊁
m
󰉼󰉶󰉠
d
󰉵󰉢󰉠
.P
A.
1
2
m
m

. B.
2m
. C.
1m
. D.
1m 
.
L󰉶i gi󰉘i
................................................................................................
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 143.(󰉷  󰉼󰉵c) Tron󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
15
:
1 3 1
x y z
d



󰉢󰉠
:3 3 2 6 0P x y z
󰉪󰉧sau 
A.
d
󰉵
P
. B.
d
󰉵
P
.
C.
d
󰉞󰉵
P
. D.
d
󰉟
P
.
L󰉶i gi󰉘i
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u 144.(THPT chuyên Quý 󰇜 󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶
󰉠
15
:
1 3 1
x y z
d



󰉢󰉠
:3 3 2 6 0P x y z
󰉪󰉧
A.
d
󰉞󰉵
P
. B.
d
󰉵
P
.
C.
d
󰉟
P
. D.
d
󰉵
P
.
L󰉶i gi󰉘i
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u 145. Trong không gian
Oxyz
 󰉼󰉶 󰉠
3 2 4
:
4 1 2
x y z
 󰉢 󰉠
: 4 4 5 0x y z
󰉠󰉬󰉠󰉬
A.

. B. 󰊀
󰉟g 30
0
.
C.

. D.

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 146.(󰉧󰉭 ) 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
15
:
1 3 1
x y z
d



󰉢󰉠
:3 3 2 6 0xyP z
󰉪󰉧󰉼󰉵
A.
d
󰉟
P
. B.
d
son󰉵
P
.
C.
d
󰉞󰉵
P
. D.
d
󰉵
P
.
L󰉶i gi󰉘i
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u 147. 󰉵󰉪󰉭󰉳
Oxyz
󰉢󰉠
:3 4 2 2016 0P x y z
. Trong
󰉼󰉶󰉠󰉼󰉶󰉠󰉵󰉢󰉠
P
.
A.
4
1 1 1
:
3 4 2
x y z
d

. B.
3
1 1 1
:
354
x y z
d

.
C.
2
1 1 1
:
4 3 1
x y z
d

. D.
1
1 1 1
:
2 2 1
x y z
d

.
L󰉶i gi󰉘i
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u 148.󰇛󰉰󰇜 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶
󰉠
11
:
1 2 2
x y z
d


󰉢󰉠
:2 15 0P x y
󰉨 ?
A.
dP
. B.
||dP
. C.
dP
. D.
1; 1;0d P I
.
L󰉶i gi󰉘i
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u 149.(Chuyên 󰉗󰉭 Vinh 2019) 󰉵󰉪 󰉺 󰉭󰉳
Oxyz
󰉢󰉠
: 2 3 6 0x y z
󰉼󰉶󰉠
1 1 3
:
1 1 1
x y z

󰉪󰉧 
A.
. B.
󰉞󰉵
.
C.

. D.

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 150.(󰉩󰉭) 󰉵󰉪󰉭󰉳
Oxyz
󰉢󰉠
:3 4 2 2016 0P x y z
.   󰉼󰉶 󰉠  󰉼󰉶 󰉠    󰉵 󰉢
󰉠
()P
.
A.
1
1 1 1
:
2 2 1
x y z
d

. B.
3
1 1 1
:
354
x y z
d

.
C.
4
1 1 1
:
3 4 2
x y z
d

. D.
2
1 1 1
:
4 3 1
x y z
d

.
L󰉶i gi󰉘i
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u 151.(THPT 󰉢 Thúc 󰉽 󰇜    󰉵 󰉪 󰉺 󰉭 󰉳
,Oxyz
 󰉼󰉶
󰉠
1
:
1 1 2
x y z
 󰉢󰉠
2
: 1 0,P x my m z m
󰉯󰊁󰉙󰉘
󰉬󰉻
m
󰉨󰉢󰉠
P
󰉵󰉼󰉶󰉠
.
.
A.
1
2
m 
. B.
1m
. C.
1m
1
2
m 
. D.
0m
1
2
m
.
L󰉶i gi󰉘i
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u 152. 󰉵󰉪󰉭󰉳
Oxyz
󰉢󰉠
: 2 1 0P mx my z
󰉼󰉶
󰉠
1
11
x y z
nm

󰉵
0, 1mm
. Khi
Pd
󰉱
mn
󰉟
A.
2mn
. B. 󰉦󰉘 C.
1
2
mn
. D.
2
3
mn
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 153.(TTLT  󰉪 󰉧 2019) 󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶󰉠
1 2 1
:
2 1 1
x y z
d

󰉵󰉢󰉠
:0P x y z m
󰉬󰉻
m
.
A.
m
. B.
2m
. C.
0m
. D.
0m
.
L󰉶i gi󰉘i
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u 154.(THPT Tiên Lãng 2019)    󰉵 󰉪 󰉺 󰉭 󰉳
Oxyz
  󰉢 󰉠
: 3 2 5 0P x y z
󰉼󰉶󰉠
1 2 3
:
2 1 2
x y z
d
mm

󰉨󰉼󰉶󰉠
d
󰉵
P
thì:
A.
2m 
. B.
1m
. C. .
0m
. D.
1m 
.
L󰉶i gi󰉘i
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u 155.(THPT 󰉩) 󰉼󰉶󰉠
23
: 5 7
43
xt
d y t
z m t


󰉢 󰉠
:3 7 13 91 0P x y z
󰉬󰉻󰉯
m
󰉨
d
󰉵
P
.
A.
10
. B.
13
. C.
10
. D.
13
.
L󰉶i gi󰉘i
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u 156.(THPT Chuyên Võ Nguyên giáp 2019) 󰉵󰉪󰉭󰉳
,Oxyz
󰉢
󰉠
,PQ
R
󰉚󰉼󰉹󰉼󰉴
: 2 0P x my z
;
: 1 0Q mx y z
:3 2 5 0R x y z
󰉭
m
d
󰉦󰉻󰉢󰉠
P
Q
. Tìm
m
󰉨
󰉼󰉶󰉠
m
d
󰉵󰉢󰉠
R
.
A. 󰉬
m
. B.
1
1
3
m
m

. C.
1
3
m 
. D.
1m
.
L󰉶i gi󰉘i
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 157.(󰉼-󰉦2019) Tron󰉵󰉪󰉺󰉭󰉳
Oxyz
, cho 󰉼󰉶
󰉠
d
󰉼󰉴
21
35
xt
yt
zt


󰉟
: 4 0P mx y nz n
. 
2mn
󰉟
A.
0
. B.
4
. C.
2
. D.
3
.
L󰉶i gi󰉘i
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u 158.(THPT Chuyên Thái Nguyên 2017)  󰉵 󰉪󰉭󰉳
,Oxyz
󰉼󰉶
󰉠
2
:2
x
d y m t
z n t

󰉢󰉠
:2 0P mx y mz n
󰉦󰉼󰉶󰉠
d
󰉟󰉢
󰉠
.P

mn
.
A.
12
. B.
8
. C.
12
. D.
8
.
L󰉶i gi󰉘i
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u 159. 󰉵󰉬󰉻
, mn
󰉼󰉶󰉠
34
: 1 4
3
xt
D y t t
zt


󰉟󰉢󰉠
: 1 2 4 9 0P m x y z n
?
A.
4; 14mn
. B.
4; 10mn
. C.
3; 11mn
. D.
4; 14mn
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 160.󰇛󰉰󰇜 Trong không gian 󰉵󰉪󰉭󰉳
,Oxyz
󰉼󰉶
󰉠󰉼󰉴
4 1 2
:.
211
x y z
d
󰉢󰉠
: 3 2 4 0,P x y mz
󰉵
m
󰉯󰊁
m
󰉼󰉶󰉠
d
󰉵󰉢󰉠
.P
A.
1
3
m
. B.
2m
. C.
1
2
m
. D.
1m
.
L󰉶i gi󰉘i
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u 161.(THPT chuyên Thánh Tông 2019)    󰉵 󰉪 󰉭 󰉳
Oxyz
 󰉢
󰉠
2
: 1 2 1 0m x y mz m
󰉬
m
󰉦
󰉵
Ox
.
A.
1m
. B.
1m 
. C.
0m
. D.
1m 
.
L󰉶i gi󰉘i
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u 162.(THPT 󰉢 Thúc 󰉽 󰇜    󰉵 󰉪 󰉺 󰉭 󰉳
,Oxyz
 󰉼󰉶
󰉠
1
:
1 1 2
x y z
 󰉢󰉠
2
: 1 0,P x my m z m
󰉯󰊁󰉙󰉘
󰉬󰉻
m
󰉨󰉢󰉠
P
󰉵󰉼󰉶󰉠
.
A.
1
2
m 
. B.
1m
. C.
1m
1
2
m 
. D.
0m
1
2
m
.
L󰉶i gi󰉘i
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u 163.󰇛 󰉩 󰇜  󰉨
1;2;1A
4;5; 2B
󰉢󰉠
P
󰉼󰉴
3 4 5 6 0x y z
󰉼󰉶󰉠
AB
󰉞
P
󰉗󰉨
M
󰊃󰉯
MB
MA
.
A.
2
. B.
1
4
. C.
4
. D.
3
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 164.󰇛󰉷-󰉰󰇜 󰉵󰉪󰉺󰉭󰉳
Oxy
󰉢󰉠
P
󰉼󰉶󰉠
󰉼󰉴󰉽󰉼󰉴
3 1 0x y z
22
21
x y z
m


󰉵
m
󰉯󰊁
0
. Tìm
m
󰉨󰉼󰉶󰉠
󰉵󰉢󰉠
P

󰉘
d
󰊀󰉼󰉶󰉠
󰉢󰉠
P
.
A.
1m 
3
11
d
. B.
1m
3
11
d
. C.
2m
3
11
d
. D.
1m
4
11
d
.
L󰉶i gi󰉘i
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󰉨m gi󰊀a 󰉼󰉶ng th󰉠ng và m󰉢t ph󰉠ng
Cho m󰉢t ph󰉠ng
( ): 0Ax By Cz D
󰉼󰉶ng th󰉠ng
01
02
03
:
x x a t
d y y a t
z z a t



.
x󰉼󰉴
0 1 0 2 0 3
( ) ( ) ( ) 0A x a t B y a t C z a t
(󰉛n
t
) (*)
d
//
()
(*) vô nghi󰉪m
d
c󰉞t
()
󰇛󰇜󰉳t nghi󰉪m󰉨m c󰉻a
mp
󰉼󰉶ng th󰉠ng
.d
d
()
(*) có vô s󰉯 nghi󰉪m
u 165.(THPT  󰉬󰇜 󰉨󰉻
31
:
1 1 2
x y z
d


( ):2 7 0P x y z
.
A.
0;2; 4M
. B.
1;4; 2M
. C.
6; 4;3M
. D.
3; 1;0M
.
L󰉶i gi󰉘i
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u 166. 󰉭󰉳󰉨󰉻󰉼󰉶󰉠
1
: 2 3
3
xt
d y t
zt



󰉢󰉠
Oyz
.
A.
1;2;2
. B.
0;5;2
. C.
0; 1;4
. D.
0;2;3
.
L󰉶i gi󰉘i
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 167. 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
3 1 3
:
2 1 1
x y z
d

󰉢
󰉠
P
󰉼󰉴
2 5 0x y z
. 󰉭󰉳󰉨󰉻
d
P
.
A.
–1;0; 4M
. B.
1;0;4M
. C.
–5;0; 2M
. D.
–5; 2;2M
.
L󰉶i gi󰉘i
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u 168.Trong không gian
Oxyz
   󰉨
3;1;1 , 4;8; 3 , 2;9; 7M N P
 󰉢 󰉠
: 2 6 0 Q x y z
󰉼󰉶󰉠
d

G
󰉵
Q
󰉨
A
󰉻󰉢
󰉠
Q
󰉼󰉶󰉠g
d
󰉦
G
󰉭
MNP
.
A.
1; 2; 1A 
. B.
1;2; 1A
. C.
1;2;1A
. D.
1; 2; 1A
.
L󰉶i gi󰉘i
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u 169.󰇛󰉼󰉗󰉳󰇜 󰉵󰉪󰉭󰉳
Oxyz
󰉺
󰉽
1 1 1
.ABC A B C
 󰉵
0; 3;0A
,
4;0;0B
,
0;3;0C
,
1
4;0;4B
. 󰉭
M
  󰉨 󰉻
11
AB
󰉢󰉠
P
qua
A
,
M
󰉵
1
BC
󰉞
11
AC
󰉗
N
󰉳󰉗󰉠
MN
.
A.
3
. B.
4
. C.
17
2
. D.
23
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 170.󰇛󰉼󰉗󰉳󰇜 󰉵󰉪󰉭󰉳
Oxyz
, cho hai
󰉼󰉶󰉠
1
1 2 1
:
3 1 2
x y z
d

2
33
:5
2
xt
d y t
zt

.
󰉢󰉠
Oxz
󰉞󰉼󰉶󰉠
1
d
,
2
d
󰉚󰉼󰉹󰉗󰉨
A
,
B
󰉪
OAB
là.
A.
5
. B.
15
. C.
10
. D.
55
.
L󰉶i gi󰉘i
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u 171.󰇛󰉺-TPHCM 2017) Trong không gian 󰉪󰉺󰉗󰉳
Oxzy
, cho
1;2;3A
,
1;0; 5B
,
:2 3 4 0P x y z
. Tìm
MP
sao cho
A
,
B
,
M
󰉠
A.
1;2;0M
. B.
3;4;11M
. C.
0;1; 1M
. D.
2;3;7M
.
L󰉶i gi󰉘i
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u 172.(THPT Chuyên Quang Trung 2019) Trong không gian 󰉪󰉗󰉳
Oxyz
, cho
2;3;1M
,
5;6; 2N
󰉼󰉶󰉠
M
,
N
󰉞󰉢󰉠
xOz
󰉗
A
󰉨
A
chia 󰉗
MN
󰊃󰉯
A.
1
4
. B.
1
4
. C.
1
2
. D.
2
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 173. 󰉵󰉪󰉭󰉳
,Oxyz
󰉨
4;5; 2A
2; 1;7 .B
󰉼󰉶󰉠
AB
󰉞󰉢󰉠
Oyz
󰉗󰉨
M
󰉫󰉯
.
MA
MB
A.
2
MA
MB
. B.
1
2
MA
MB
. C.
1
3
MA
MB
. D.
3
MA
MB
.
L󰉶i gi󰉘i
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u 174.(THPT chuyên  Phúc)Trong 󰉵󰉪󰉭󰉳
Oxyz
󰉨
1;2; 2A
2; 1;0 .B
󰉼󰉶󰉠
AB
󰉞󰉢󰉠
: 1 0P x y z
󰉗󰉨
I
.󰉫󰉯
IA
IB
󰉟
A.
4
. B.
2
. C.
6
. D.
3
.
L󰉶i gi󰉘i
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c gi󰊀󰉼󰉶ng th󰉠ng
󰉼󰉶ng th󰉠ng
d
có vtcp
( ; ; )u a b c
󰉼󰉶ng th󰉠ng
'd
có vtcp
' ( '; '; ')u a b c
.
G󰉭i
là góc gi󰊀󰉼󰉶ng th󰉠
0
2 2 2 2 2 2
.'
. ' ' '
cos (0 90 )
. ' ' '
.'
uu
a a bb cc
a b c a b c
uu




c gi󰊀󰉼󰉶ng th󰉠ng v󰉵i m󰉢t ph󰉠ng
󰉼󰉶ng th󰉠ng
d
có vtcp
( ; ; )u a b c
và m󰉢t ph󰉠ng
()
có vtpt
;;n A B C
.
G󰉭i
là góc h󰉹p b󰉷󰉼󰉶ng th󰉠ng
d
và m󰉢t ph󰉠ng
()
ta có:
2 2 2 2 2 2
.
sin
.
.
un
Aa Bb Cc
A B C a b c
un



󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
u 175.󰇛󰉼󰇜 󰉭
󰊀󰉼󰉶󰉠
5 2 2
:
2 1 1
xyz
d

và m󰉢
󰉠:
3 4 5 0x y z

A.
90

. B.
45

. C.
60

. D.
30

.
L󰉶i gi󰉘i
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u 176.(THPT Chuyên B󰉞c Giang 2019) 󰉵󰉪󰉭󰉳
,Oxyz
󰉼󰉶󰉠
:d
1
22
3
xt
yt
zt



󰉢󰉠
:P
30xy
󰉯󰊀󰉼󰉶󰉠
d
mp P
.
A.
60
. B.
30
. C.
120
. D.
45
.
L󰉶i gi󰉘i
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u 177.(󰉗i H󰉭c Vinh) Trong không gian t󰉭󰉳
,Oxyz
󰉼󰉶ng th󰉠ng 󰉼󰉴
trình . Góc gi󰊀󰉼󰉶ng th󰉠ng b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 178.(󰉗i H󰉭c Vinh 2019) Trong không gian h󰉪 t󰉭󰉳
,Oxyz
󰉼󰉶ng th󰉠ng
󰉼󰉴  m󰉢t ph󰉠ng . Góc gi󰊀 󰉼󰉶ng
th󰉠ng v󰉵i m󰉢t ph󰉠ng b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
..........................................................................................................................................................................................................
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1
1 2 3
:
2 1 2
x y z
2
3 1 2
:
1 1 4
x y z
12
,
0
30
0
45
0
60
0
135
1 2 3
:
2 1 2
x y z
( ): 4 2019 0P x y z
P
0
30
0
45
0
60
0
135
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 179.(Chuyên 󰉗i H󰉭c Vinh 2019) Trong không gian h󰉪 t󰉭󰉳
Oxyz
󰉼󰉶ng th󰉠ng
1
1 2 3
:
2 1 2
x y z
2
3 1 2
:
1 1 4
x y z
. Góc gi󰊀a 󰉼󰉶ng th󰉠ng
12
,
b󰉟ng
A.
0
30
. B.
0
45
. C.
0
60
. D.
0
135
.
L󰉶i gi󰉘i
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u 180.(THPT Kim Liên 2018) Trong không gian v󰉵i h󰉪 tr󰉺c t󰉭 󰉳
yOx z
 󰉼󰉶ng th󰉠ng
32
:
2 1 1
x y z
và m󰉢t ph󰉠ng
:3 4 5 8 0x y z
. Góc gi󰊀󰉼󰉶ng th󰉠ng
và m󰉢t ph󰉠ng
có s󰉯 
A.
45
. B.
90
. C.
30
. D.
60
.
L󰉶i gi󰉘i
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u 181.(Chuyên 󰉗i H󰉭c Vinh) Trong không gian 󰉼󰉶ng th󰉠ng m󰉢t
ph󰉠ng . Góc gi󰊀󰉼󰉶ng th󰉠ng
và m󰉢t ph󰉠ng b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 182.(Chuyên 󰉗i H󰉭c Vinh 2019) Trong không gian v󰉵i h󰉪 tr󰉺c t󰉭󰉳 󰉼󰉶ng th󰉠ng
m󰉢t ph󰉠ng . G󰉭i góc gi󰊀󰉼󰉶ng th󰉠ng
m󰉢t ph󰉠ng  b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
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Oxyz
:
1 2 1
x y z
: 2 0x y z
30
60
150
120
Oxyz
2 1 1
:
1 2 3
x y z
d

: 2 3 0x y z
d
0
0
0
45
0
90
0
60
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 183.(THPT 󰉰ng Phong 2019) 󰉵󰉪󰉺󰉭󰉳
,Oxyz
󰉭
d
giao
󰉦󰉻󰉢󰉠󰉼󰉴󰉚󰉼󰉹
2 2017 0x y z
5 0.x y z
Tính
󰉯󰉳󰊀󰉼󰉶󰉠
d
󰉺
.Oz
A.
O
45
. B.
O
0
. C.
O
30
. D.
O
60
.
L󰉶i gi󰉘i
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u 184.󰇛󰉩󰇜󰉵󰉪󰉭󰉳
,Oxyz
󰊀󰉼󰉶
󰉠
1
11
:
1 1 2
x y z
d


1
13
:.
1 1 1
x y z
d


A.
30 .
B.
60 .
C.
45
. D.
90 .
L󰉶i gi󰉘i
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u 185.󰇛󰉪󰉻-󰉘󰇜󰉭󰉳
,Oxyz
󰉨
3; 1; 0A
,
0; 7; 3B
,
2; 1; 1C 
,
3;2;6D
󰊀󰉼󰉶󰉠
, AB CD
là:
A.
30
. B.
90
. C.
45
. D.
60
.
L󰉶i gi󰉘i
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u 186.(󰉩󰉦n 2019) 󰉵󰉪󰉭󰉳
,Oxyz
󰉢󰉠
:3 4 5 8 0P x y z
󰉼󰉶󰉠
d
󰉦󰉻󰉢󰉠
: 2 1 0xy
: 2 3 0.xz
󰉭
󰊀󰉼󰉶󰉠
d
󰉢󰉠
.P
Tính
.
A.
90 .

B.
60 .

C.
30 .

D.
45 .

L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 187.(THPT Chuyên Nguy󰉩n Du 2019) Trong không gian 󰉭󰉳
Oxyz
, cho 󰉼󰉶󰉠
1
2 1 3
:
11
2
x y z
d

2
5 3 5
:
1
2
x y z
d
m

󰉗󰉵
60
󰉬󰉻󰉯
m
A.
1m 
. B.
3
2
m
. C.
1
2
m
. D.
1m
.
L󰉶i gi󰉘i
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u 188.(THPT Yên-Knh 2019) Trong không gian
Oxyz
󰉼󰉶ng th󰉠ng
d
giao tuy󰉦n c󰉻a
hai m󰉢t ph󰉠ng
( ): .sin cos 0;( ): .cos sin 0; 0;
2
P x z Q y z



. Góc gi󰊀a
()d
tr󰉺c
Oz
là:
A.
30
. B.
45
. C.
60
. D.
90
.
L󰉶i gi󰉘i
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u 189.󰇛󰉷󰇜Trong không gian
Oxyz
󰉭
d
󰉼󰉶 󰉠  󰉨
1; 1;2A
   󰉵 󰉢 󰉠
:2 3 0P x y z
 󰉰 󰉶 󰉗 󰉵 󰉼󰉶 󰉠
11
:
1 2 2
x y z
󰉳󰉵󰉙󰉼󰉴󰉼󰉶󰉠
d
A.
112
4 5 3
x y z

. B.
112
4 5 3
x y z

. C.
112
4 5 3
x y z

. D.
112
4 5 3
x y z

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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Kho󰉘ng cách t󰉾 󰉨m
1 1 1 1
;;M x y z
󰉦󰉼󰉶ng th󰉠ng
vtcp
u
.
S󰉿 d󰉺ng công th󰉽c:
10
1
,
,
M M u
dM
u


(v󰉵i
0
M 
)
u 190.(THPT H󰉘i H󰉝u 2019) Trong không gian
Oxyz
󰉨m
;;P a b c
. Kho󰉘ng cách t󰉾
P
󰉦n tr󰉺c t󰉭󰉳
Oy
b󰉟ng:
A.
22
ac
. B.
b
. C.
b
. D.
22
ac
.
L󰉶i gi󰉘i
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u 191.󰇛󰉚󰇜Trong không gian 󰉪
Oxyz
󰉨
2;1;1A
󰉼󰉶󰉠
1 2 3
:
1 2 2
x y z
d

󰉘󰉾
A
󰉦󰉼󰉶󰉠
d
là.
A.
5
. B.
35
2
. C.
25
. D.
35
.
L󰉶i gi󰉘i
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u 192.󰇛󰉷󰉰󰇜 󰉵󰉪󰉭󰉳
Oxyz
, cho tam giác
ABC
󰉵
1; 2; 1A
,
0; 3; 4B
,
2; 1; 1C
󰉳󰉼󰉶󰉾
A
󰉦
BC
󰉟
A.
50
33
. B.
33
50
. C.
6
. D.
53
.
L󰉶i gi󰉘i
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u 193.(THPT Chuyên SPHN 2019) 󰉵󰉪󰉺󰉭󰉳
Oxyz
, cho
4;4;0A
,
2;0;4B
,
1; 2;1C
󰉘󰉾
C
󰉦󰉼󰉶󰉠
AB
là:
A.
32
. B.
13
. C.
23
. D.
3
.
L󰉶i gi󰉘i
u
M
o
x
o
;
y
o
;
z
o
M
1
x
1
;
y
1
;
z
1
Δ
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 194.󰇛󰉚󰇜Trong không gian 󰉪󰉺󰉭󰉳
Oxyz
󰉨
2;1; 2A
,
1; 3;1B
,
3; 5;2C
󰉳󰉼󰉶
AH
󰉻
ABC
là.
A.
17
. B.
32
. C.
17
2
. D.
2 17
.
L󰉶i gi󰉘i
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u 195.󰇛󰉪󰉻-󰉘󰇜 Trong không gian 󰉭󰉳
Oxyz
󰉘󰉾󰉨
4; 3;2M
󰉦󰉼󰉶󰉠
22
:
3 2 1
x y z
.
A.
; 3 3dM
. B.
;3dM
. C.
;3dM
. D.
; 3 2dM
.
L󰉶i gi󰉘i
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u 196.(THPT TH Cao Nguyên 2019) 󰉵󰉪󰉺󰉭󰉳
,Oxyz
󰉯󰉨
3;0;0 , 0;2;0 , 0;0;6 , 1;1;1A B C D
 󰉭
 󰉼󰉶 󰉠  
D
 󰉮  󰉱
󰉘󰉾󰉨
, , A B C
󰉦
󰉵󰉙󰉮
󰉨󰉨󰉼󰉵

A.
3;4;3M
. B.
1; 2;1M 
. C.
3; 5; 1M
. D.
7;13;5M
.
L󰉶i gi󰉘i
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Kho󰉘ng cách c󰉻a hai 󰉼󰉶ng th󰉠ng co nhau
1
2
.
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
󰉼󰉶ng th󰉠ng chéo nhau:
󰉼󰉶ng th󰉠ng
1

1 1 1 1
;;M x y z
, có vtcp
1
u
.
󰉼󰉶ng th󰉠ng
2

2 2 2 2
;;M x y z
, có vtcp
2
.u
󰉿 d󰉺ng công th󰉽c
1 2 1 2
12
12
.
,.
( , )
,
u u M M
d
uu



u 197.󰇛󰉩󰇜Trong không gian 󰉪󰉭󰉳
Oxyz
󰉘󰊀i
󰉼󰉶󰉠
1
7 5 9
:
3 1 4
x y z
d

2
4 18
:
3 1 4
x y z
d


󰉟
A.
30
. B.
20
. C.
25
. D.
15
.
L󰉶i gi󰉘i
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u 198.(THPT Kim Liên 2019) Trong không gian
Oxyz
, ch󰉼󰉶ng th󰉠ng
1
12
:
2 1 1
x y z
d


2
14
: 1 2 ,
22
xt
d y t t
zt


. Kho󰉘ng cách gi󰊀󰉼󰉶ng th󰉠󰉟ng
A.
87
6
. B.
174
6
C.
174
3
D.
87
3
L󰉶i gi󰉘i
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u
2
u
1
2
1
M
1
M
2
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 199. 󰉵󰉪󰉭󰉳
Oxyz
󰉼󰉶󰉠
1 1 1
:
2 3 2
x y z
d

1 2 3
:
2 1 1
x y z
d

󰉘
h
󰊀󰉼󰉶󰉠
d
d
.
A.
22 21
21
h
. B.
4 21
21
h
. C.
10 21
21
h
. D.
8 21
21
h
.
L󰉶i gi󰉘i
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u 200.(THPT 󰉩󰇜Trong không gian 󰉪󰉭󰉳
,Oxyz
󰉼󰉶󰉠
1
3 2 1
:
4 1 1
x y z
d

2
12
:
6 1 2
x y z
d


. Kho󰉘ng cách gi󰊀a chúng b󰉟ng
A.
5
. B.
4
. C.
2
. D.
3
.
L󰉶i gi󰉘i
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Kho󰉘ng cách gi󰊀󰉼󰉶ng th󰉠ng
song song
.mp P
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
Cho 󰉼󰉶ng th󰉠ng
song song
.mp P
󰉼󰉶ng th󰉠ng

;;
o o o
M x y z
, có vtcp
u
.
M󰉢t ph󰉠ng
P
󰉼󰉴
0Ax By Cz D
󰉿 d󰉺ng công th󰉽c
2 2 2
( ,( )) ( ,( )) .
o o o
Ax By Cz D
d P d M P
A B C

u 201.󰇛󰉼󰉴󰇜Trong không gian h󰉪 t󰉭󰉳
Oxyz
, kho󰉘ng cách gi󰊀󰉼󰉶ng
th󰉠ng
11
:
1 4 1
x y z
d


và m󰉢t ph󰉠ng
( ):2 2 9 0P x y z
b󰉟ng:
A.
10
3
. B.
4
. C.
2
. D.
4
3
.
L󰉶i gi󰉘i
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u 202.(THPT ISCHOOL Nha Trang 2019) Trong không gian
Oxyz
, kho󰉘ng cách gi󰊀 󰉼󰉶ng
th󰉠ng
1 2 3
:
2 2 3
x y z
d

và m󰉢t ph󰉠ng
: 2 2 5 0P x y z
b󰉟ng
A.
16
3
. B.
2
. C.
5
3
. D.
3
.
L󰉶i gi󰉘i
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u 203.(Toán H󰉭c Tu󰉱i Tr󰉤 2019) Trong không gian v󰉵i h󰉪 t󰉭 󰉳
Oxyz
, kho󰉘ng cách gi󰊀a
󰉼󰉶ng th󰉠ng
1 2 3
:
2 3 1
x y z
d

và m󰉢t ph󰉠ng
: 1 0P x y z
b󰉟ng:
A.
3
14
. B.
3
C.
1
3
. D.
0
.
L󰉶i gi󰉘i
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M
o
x
o
;
y
o
;
z
o
P
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 204.󰇛󰉴󰇜󰉵󰉪󰉭󰉳
Oxyz
󰉘󰊀
󰉼󰉶󰉠
1 3 2
:
2 2 1
x y z
󰉢󰉠
( ): 2 2 4 0P x y z
A.
0
. B.
1
. C.
3
. D.
2
.
L󰉶i gi󰉘i
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u 205.(THPT Chuyên Quang Trung 2019) Trong không gian v󰉵i h󰉪 t󰉭󰉳
Oxyz
, cho m󰉢t ph󰉠ng
: 2 2 1 0P x y z
󰉼󰉶ng th󰉠ng
1 2 1
:
2 2 1
x y z
. Kho󰉘ng cách gi󰊀a
P
b󰉟ng
A.
8
3
. B.
7
3
. C.
6
3
. D.
8
3
.
L󰉶i gi󰉘i
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u 206.󰇛󰉴󰇜Trong không gian h󰉪
Oxyz
, kho󰉘ng cách gi󰊀󰉼󰉶ng th󰉠ng
1
:
1 1 2
x y z
d

và m󰉢t ph󰉠ng
: 2 0P x y z
b󰉟ng:
A.
2 3.
B.
3
.
3
C.
23
.
3
D.
3.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 207.(THPT Nình Bình 2019) Trong không gian 󰉭󰉳
,Oxyz
mp P
󰉨
2;1;0A
3;0;1B
󰉵
11
:
1 1 2
x y z
󰉘󰊀
󰉢󰉠
P
.
A.
3
2
. B.
3
2
. C.
2
2
. D.
3
2
.
L󰉶i gi󰉘i
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u 208.(THPT ) Tính kho󰉘ng cách gi󰊀󰉼󰉶ng th󰉠ng
13
:
2 1 2
x y z
d


m󰉢t ph󰉠ng
( ) : 2 2 1 0P x y z
.
A.
7
.
3
B.
8
.
3
C.
5
.
3
D .
1
.
3
L󰉶i gi󰉘i
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u 209.󰇛󰉷󰉰󰇜󰉵󰉪󰉭󰉳
Oxyz
󰉘󰊀󰉢
󰉠ng
󰉵
: 5 0x y z
:2 2 2 3 0x y z
󰉟
A.
22
. B.
17
6
. C.
73
6
. D.
7
6
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 210. Trong không gian h󰉪
,Oxyz
󰉼󰉶ng th󰉠ng và m󰉢t ph󰉠ng
. Bi󰉦t
c󰉞t m󰉢t ph󰉠ng
P
t󰉗i
,AM
thu󰉳c sao cho .
Tính kho󰉘ng cách t󰉾
M
t󰉵i m󰉢t ph󰉠ng
.P
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 211.(Chuyên  2019) Trong không gian 󰉼󰉶ng th󰉠ng
󰉨m . G󰉭i 󰉨m thu󰉳󰉼󰉶ng th󰉠ng sao cho di󰉪n tích
tam giác b󰉟ng . Giá tr󰉬 c󰉻a t󰉱ng b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
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3 1 2
:
1 1 4
x y z
( ): 2 6 0P x y z
23AM
2
2
3
3
Oxyz
12
:
2 1 1
x y z
d


( 1;3;1)A
0;2; 1B
;;C m n p
d
ABC
22
m n p
1
2
3
5
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 212.(Chuyên 2019) Trong không gian 󰉼󰉶ng th󰉠ng
󰉨m . G󰉭i 󰉨m thu󰉳󰉼󰉶ng th󰉠ng sao cho tam
giác vuông t󰉗i A. Giá tr󰉬 c󰉻a t󰉱ng b󰉟ng
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 213.󰇛󰇜Trong không gian 󰉼󰉶ng th󰉠ng
và m󰉢t ph󰉠ng . G󰉭i 󰉨m thu󰉳󰉼󰉶ng th󰉠ng sao cho kho󰉘ng cách
t󰉾 󰉦n m󰉢t ph󰉠ng b󰉟ng . N󰉦u 󰉳 󰉳 c󰉻a b󰉟ng.
A. . B. . C. . D. .
L󰉶i gi󰉘i
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u 214.󰇛󰇜Trong không gian h󰉪 t , cho tam giác vuông t󰉗i ,
󰉼󰉶ng th󰉠ng  󰉼󰉴ng trình  󰉼󰉶ng th󰉠ng
n󰉟m trong m󰉢t ph󰉠ng 󰉨m 󰉳 󰉳 󰉨m .
A. . B. . C. . D. .
L󰉶i gi󰉘i
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Oxyz
12
:
2 1 1
x y z
d


( 1;3;1)A
0;2; 1B
;;C m n p
d
ABC
2m n p
0
2
3
5
Oxyz
12
:
1 2 3
x y z
d


: 2 2 3 0P x y z
M
d
M
P
2
M
M
3
21
3
1
Oxyz
ABC
A
30 , 2ABC BC
BC
4 5 7
1 1 4
x y z

AB
: 3 0a x z
C
A
3
2
3
9
2
5
2
󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 215.󰇛󰉷󰉗󰉴󰇜Trong không gian h󰉪 tr󰉺c
Oxyz
, cho hình thoi
ABCD
󰉵
1 2 1 2 3 2A ; ; ,B ; ;
. Tâm
I
󰉻󰉳󰉼󰉶󰉠
12
1 1 1
x y z
d:



󰉫
󰉫
D
󰉻
A.
0 1 2D ; ;
. B.
2 1 0D ; ;
. C.
0 1 2D ; ;
. D.
2 1 0D ; ;
.
L󰉶i gi󰉘i
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u 216.(THPT 󰉚 󰉗 2019)    󰉵󰉪 󰉺 󰉭 󰉳
Oxyz
 󰉼󰉶
󰉠
1
:
2 1 1
x y z
d

󰉢󰉠
:2 2 2 0.P x y z
󰉨
M
󰉳
d
sao
cho
M
󰉧󰉯󰉭󰉳
O
󰉢󰉠g
P
?
A. 4. B. 0. C. 2. D. 1.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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󰉽󰉳 󰉝󰉺
u 217.󰇛󰉼󰉴󰉦 Vinh 2019) Trong không gian 󰉪󰉺󰉭󰉳
Oxyz
, cho hai 󰉨
1;4;2 , 1;2;4AB
󰉼󰉶󰉠
54
: 2 2
4
xt
d y t
zt



󰉨
M
󰉳
d
󰉬󰉮󰉙󰉻
󰉪am giác
AMB
A.
23
. B.
22
. C.
32
. D.
62
.
L󰉶i gi󰉘i
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u 218. 󰇛󰉳󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
, cho hai
󰉨
1;2;3A
,
5; 4; 1B 
󰉢󰉠
P
qua
Ox
sao cho
; 2 ;d B P d A P
,
P
󰉞
AB
󰉗
;;I a b c
󰉟󰊀
AB
. Tính
a b c
.
A.
12
. B.
6
. C.
4
. D.
8
.
L󰉶i gi󰉘i.
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 219.(THPT Chuyên 󰉚󰉗󰇜󰉵󰉪󰉺󰉭󰉳
Oxyz
󰉼󰉶
󰉠
1
:
2 1 1
x y z
d

󰉢󰉠
: 2 2 5 0x y z
󰉨m
A
trên
d
󰉳
󰉼󰉴󰉘󰉾
A
󰉦
󰉟
3
.
A.
4; 2;1A
. B.
2; 1; 2A 
. C.
2; 1; 0A
. D.
0; 0; 1A
.
L󰉶i gi󰉘i
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u 220.(THPT Kim Liên 2018) Trong không gian t󰉭󰉳
Oxyz
󰉨m
0;0;1A
,
1; 2;0B 
,
2;0; 1C
. T󰉝p h󰉹  󰉨m
M
󰉧  󰉨m
,,A B C
 󰉼󰉶ng th󰉠ng
. Vi󰉦󰉼󰉴
󰉼󰉶ng th󰉠ng
.
A.
1
3
2
3
xt
yt
zt

. B.
1
3
2
3
xt
yt
zt

. C.
1
3
2
xt
yt
zt

. D.
1
2
1
1
2
xt
yt
zt

.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶 󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 221.󰇛󰉢ng Thành Nam) Trong không gian v󰉵i h󰉪 tr󰉺c t󰉭󰉳
Oxyz
󰉨m
6;0;0A
,
0; 4;0B
,
0;0;6C
. T󰉝p h󰉹p t󰉙t c󰉘 󰉨m
M
󰉧󰉨m
A
,
B
,
C
m󰉳󰉼󰉶ng th󰉠󰉼󰉴
A.
3 2 3
2 3 2
x y z

. B.
3 2 3
2 3 2
x y z

.C.
3 2 3
2 3 2
x y z

. D.
3 2 3
2 3 2
x y z

.
L󰉶i gi󰉘i
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u 222. (THPT Nguy󰉩n Trãi 2019) Trong không gian
Oxyz
, cho m󰉢t c󰉚u
2 2 2
9x y z
󰉨m
0 0 0
; ; M x y z
thu󰉳󰉼󰉶ng th󰉠ng
1
: 1 2 .
23
xt
d y t
zt



󰉨m
,A
,B
C
phân bi󰉪t cùng thu󰉳c m󰉢t c󰉚u
sao cho
,MA
,MB
MC
ti󰉦p tuy󰉦n c󰉻a m󰉢t c󰉚u. Bi󰉦t r󰉟ng m󰉢t ph󰉠ng
ABC

1; 1; 2D
.
T󰉱ng
2 2 2
0 0 0
T x y z
b󰉟ng
A.
30
. B.
26
. C.
20
. D.
21
.
L󰉶i gi󰉘i
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󰉪󰉗󰉭 󰉼󰉴-Bài 3. 󰉼󰉴󰉼󰉶󰉠

󰉵󰉚-󰉪 Tel: 0935.660.880
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u 223. 󰇛󰇜Trong không gian h󰉪
,Oxyz
󰉨m
1;2;3 , 1;2;0AB
1;3;4M
. G󰉭i
d
󰉼󰉶ng th󰉠ng qua
B
vuông góc v󰉵i
AB
󰉰ng th󰉶i cách
M
m󰉳t kho󰉘ng nh󰉮
nh󰉙t. M󰉳󰉴󰉫 󰉼󰉴󰉻a
d
có d󰉗ng
2; ;u a b
. Tính t󰉱ng
ab
.
A.
1
. B.
2
. C.
1.
D.
2.
L󰉶i gi󰉘i
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 3. Phương Trình Đường Thẳng
270
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 224.(THPT Lê Qúy Đôn 2019) Trong không gian h
,Oxyz
cho ba điểm
1;2;3 , 1;2;0AB
1;3;4M
. Gi
d
đường thng qua
B
vuông góc vi
AB
đồng thi cách
M
mt khong nh
nht. Một véc tơ chỉ phương của
d
có dng
2; ;u a b
. Tính tng
ab
.
A.
1
. B.
2
. C.
1.
D.
2.
Li gii
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u 225.(Sở GD & ĐT Vĩnh Phúc) Trong không gian
Oxyz
, cho hai điểm
2 2 1M ; ;
,
1 2 3A ; ;
đường thẳng
15
2 2 1
x y z
d:


. Tìm vectơ chỉ phương
u
của đường thẳng
đi qua
M
,
vuông góc với đường thẳng
d
đồng thời cách điểm
A
một khoảng nhỏ nhất.
A.
2 2 1u ; ;
. B.
3 4 4u ; ;
. C.
2 1 6u ; ;
. D.
1 0 2u ; ;
.
Li gii
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 226.(Sở GD & ĐT Vĩnh Phúc 2019) Trong không gian
Oxyz
, cho điểm
(10;2;1)A
đường
thẳng
11
:
2 1 3
x y z
d


. Gọi
()P
là mặt phẳng đi qua điểm
A
, song song với đường thẳng
d
sao
cho khoảng cách giữa
d
()P
lớn nhất. Khoảng cách từ điểm
( 1;2;3)M
đến mặt phẳng
()P
bằng
A.
533
2765
. B.
97 3
15
. C
2 13
13
. D.
76 790
790
.
Li gii
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u 227.(S GD & ĐT Điện Biên 2019) Trong không gian
Oxyz
, cho
: 2 2 1 0 P x y z
đường thẳng
11
:
1 2 1


x y z
d
. Biết điểm
;;A a b c
0c
điểm nằm trên đường thẳng
d
và cách
P
một khoảng bằng 1. Tính tổng
S a b c
A.
2S
. B.
2
5
S
. C.
4S
. D.
12
5
S
.
Li gii
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272
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 228.(Chuyên ĐH Vinh 2020)
Cho đường thng
(1 ; 1 ; 0), (3 ;-1 ; 4)AB
. Tìm tọa độ đim thuc
sao cho đạt giá tr nh nht.
A. B. C. D.
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u 229.(Chuyên ĐH Vinh 2020)
Cho
( ): 1 0mp x y z
hai điểm . Gi điểm
thuc mt phng sao cho đạt giá tr nh nhất. Khi đó giá trị ca là:
A. . B. . C. . D. .
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1 1 2
:
1 1 2
x y z
M
MA MB
( 1;1; 2).M 
11
; ;1 .
22
M



33
; ; 3 .
22
M




(1; 1;2).M
: 1 0x y z
1;1;0 , 3; 1;4AB
M
P MA MB
P
5P
6P
7P
8P
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 230.(Chuyên ĐH Vinh 2020) Cho hai điểm .
Biết thuc mt phng sao cho đạt giá tr nh nhất. Khi đó, giá trị ca
biu thc bng bao nhiêu?
A. . B. . C. . D. .
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u 231.(Chuyên ĐH Vinh 2020) Cho đường thng
hai điểm
Biết điểm
thuc
sao cho biu thc
đạt giá tr ln nht. Khi
đó tổng
bng:
A. . B. . C. . D. .
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: 3 5 0x y z
1; 1;2 , 5; 1;0AB
;;M a b c
MA MB
23T a b c
5T
3T 
7T 
9T 
1 1 2
:
1 1 2
x y z
(1;1;0),A
( 1;0;1).B
( ; ; )M a b c
T MA MB
a b c
8
8 33
33
8
3
4 33
8
3
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274
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
u 232.(Chuyên ĐH Vinh Ln 2020)
Cho đường thng
hai điểm Biết điểm
thuc sao
cho biu thc
đạt giá tr ln nht là Khi đó, bng bao nhiêu?
A. . B. . C. . D. .
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u 233.(THPT Kim Liên 2019 ) Trong không gian
Oxyz
, cho hai đim
( 2; 2;1)M 
,
(1;2; 3)A
đưng thng
16
:
2 2 1
x y z
d


. Gi
đưng thng qua
M
, vuông góc vi đưng thng
d
, đồng thi cách
A
mt khong nht. Khong cách nht đó
A.
29
. B.
6
. C.
5
. D.
34
9
.
Li gii
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u 234.(THPT Thanh Chương 2019) Trong không gian h tọa độ
Oxyz
, cho đường thng
1 1 2
:
2 1 1
x y z
d


. Gi
mt phng chứa đường thng
d
to vi mt phng
Oxy
mt góc nh nht. Khong cách t
0;3; 4M
đến mt phng
bng
A.
30
. B.
26
. C.
20
. D.
35
.
1
:
1 1 1
x y z
(0;1; 3),A
( 1;0;2).B
M
T MA MB
max
.T
max
T
max
3T
max
23T
max
33T
max
2T
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Li gii
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u 235.(THPT Yên Knh Ninh 2019)
Trong không gian
Oxyz
cho hai điểm
(1;2; 1)A
,
(7; 2;3)B
đường thng
d
phương trình
1 2 2
3 2 2
x y z

. Điểm
I
thuc
d
sao cho
AI BI
nh nhất. Hoành độ của điểm
I
A.
2
. B.
0
. C.
4
. D.
1
.
Li gii
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u 236.(Sở GD & ĐT Quảng Nam 2020) Trong không gian
Oxyz
, cho đường thẳng
43
: 3 4
0
xt
d y t
z


Gọi
A
là hình chiếu vuông góc của
O
trên
d
. Điểm
M
di động trên tia
Oz
, điểm
N
di động trên
đường thẳng
d
sao cho
MN OM AN
. Gọi
I
là trung điểm đoạn thẳng
OA
. Trong trường hợp
diện tích tam giác
IMN
đạt giá trị nhỏ nhất, một véctơ pháp tuyến của mặt phẳng
,Md
tọa
độ là
A.
4;3;5 2
. B.
4;3;10 2
. C.
4;3;5 10
. D.
4;3;10 10
.
Li gii
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u 237.(Chuyên KHTN Hà Nội 2020)
Trong không gian
Oxyz
, cho các điểm
2;2;2 , 2;4; 6 , 0;2; 8A B C
:0mp P x y z
.
Xét các điểm
M
thuộc mặt phẳng
P
sao cho
90AMB
, đoạn thẳng
CM
độ dài lớn nhất
bằng
A.
2 15
. B.
2 17
. C. 8. D. 9.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 238.(Chuyên Đại hc Vinh 2020)
Trong không gian h tọa độ
Oxyz
, cho đường thng
3 4 2
:
2 1 1
x y z
d

và 2 điểm
6;3; 2A
,
1;0; 1B
. Gi
đường thẳng đi qua
B
, vuông góc vi
d
tha mãn khong cách t
A
đến
là nh nht. Một vectơ chỉ phương của
có tọa độ
A.
1;1; 3
. B.
1; 1; 1
. C.
1;2; 4
. D.
2; 1; 3
.
Li gii
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u 239.(THPT Hu Lc 2020) Trong không gian tọa độ
Oxyz
, cho ba điểm
;0;0Aa
,
0, ,0Bb
,
0,0,Cc
vi
a
,
b
,
c
nhng s dương thay đổi tha mãn
2 2 2
4 16 49a b c
. Tính tng
2 2 2
S a b c
khi khong cách t
O
đến mt phng
ABC
đạt giá tr ln nht.
A.
51
5
S
. B.
49
4
S
. C.
49
5
S
. D.
51
4
S
.
Li gii
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u 240.(THPT Hàm Rồng 2020)
Trong không gian
Oxyz
, cho điểm
1;4;3A
mặt phẳng
: 2 0P y z
. Biết điểm
B
thuộc mặt
phẳng
P
, điểm
C
thuộc
Oxy
sao cho chu vi tam giác
ABC
nhỏ nhất. Hỏi giá trị nhỏ nhất đó
A.
45
. B.
65
. C.
25
. D.
5
.
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u 241.(Chuyên Đại Hc Vinh 2020) Trong không gian
Oxyz
, cho điểm
2; ;3;4A
, đường thng
12
:
2 1 2
x y z
d


mt cu
2 2 2
: 3 2 1 20S x y z
. Mt phng
P
chứa đường
thng
d
tha mãn khong cách t đim
A
đến
P
ln nht. Mt cu
S
ct
P
theo đường
tròn có bán kính bng
A.
5
. B.
1
. C.
4
. D.
2
.
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u 242.(Tp Chí Toán Hc 2020) Trong không gian
Oxyz
, cho hai điểm
0; 1;2A
,
1;1;2B
đưng thng
11
:
1 1 1
x y z
d


. Biết
;;M a b c
thuộc đường thng
d
sao cho tam giác
MAB
din tích nh nhất. Khi đó, giá trị
23T a b c
bng:
A.
5
. B.
3
. C.
4
. D.
10
.
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Li gii
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u 243.(Tp Chí Toán Hc 2020) Trong không gian
Oxyz
, cho hai điểm
0; 1;2A
,
1;1;2B
đưng thng
11
:
1 1 1
x y z
d


. bao nhiêu điểm
M
thuộc đường thng
d
sao cho tam giác
MAB
có din tích bng
1
.
A.
0
. B.
1
. C.
2
. D. Vô s.
Li gii
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u 244.(THPT Ngô Quyền Hà Nội 2020) Trong không gian trục tọa độ
Oxyz
, cho điểm
2;5;3A
,
đường thẳng
12
:
2 1 2


x y z
d
. Biết rằng phương trình mặt phẳng
P
chứa
d
sao cho khoảng
cách từ
A
đến mặt phẳng
P
lớn nhất, dạng
30 ax by cz
(với
,,abc
các số nguyên).
Khi đó tổng
T a b c
bằng
A.
3
. B.
3
. C.
2
. D.
5
.
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Li gii
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A . L THUY󰈸T
I. ĐỊNH NGHĨA PHƯƠNG TRÌNH MẶT CẦU.
1. Định nghĩa: Trong không gian tọa độ
Oxyz
, cho mt cu
;S I R
tâm
;;I a b c
và bán kính
R
.
Điểm
;;M x y z
thuộc mặt cầu khi và chỉ khi
22
IM R IM R
.
Khi đó phương trình mặt cầu có dạng:
2 2 2
2
x a y b z c R
1
.
Ngoài ra để lập phương trình mặt cầu ta có thể tìm các hệ số
,,ab
,cd
trong phương trình:
2 2 2
2 2 2 0x y z ax by cz d
2
.
Với tâm
;;I a b c
, bán kính
2 2 2 2
0R a b c d
.
Nhận xét: Phương trình
2
có các trường hợp sau.
Khi
2 2 2 2
0R a b c d
thì
2
là phương trình mặt cu.
Khi
2 2 2 2
0R a b c d
thì
0IM
và phương trình
2
xác định điểm
I
duy nht.
Khi
2 2 2 2
0R a b c d
thì phương trình
2
không phi mt cu.
d 1. Cho phương trình
2 2 2
2 6 8 1 0.x y z x y z
Hỏi phương trình y phải
mt cu ? Nếu là phương trình mặt cu, hãy tìm tâm và bán kính ca nó ?
Lời giải
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Ví d 2. Lập phương trình mặt cu
S
biết mt cu
S
có tâm
1;2;3I
bán kính
5R
.
Lời giải
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Ví d 3. Trong không gian
Oxyz
, cho hai điểm
(2; 1;2)A
,
(0;1;0)B
. Viết phương trình mặt cu
đưng kính
AB
.
Lời giải
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M
R
y
z
x
I (a;b;c)
j
i
k
O
M
§BI 4. PHƯƠNG TRÌNH MT CU
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d 4. Lập phương trình mt cu
S
biết mt cu
S
đi qua
2; 4;3C
các nh chiếu
ca
C
lên ba trc tọa độ.
Lời giải
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II. VỊ TRÍ TƯƠNG ĐỐI CỦA MẶT CẦU.
1. V trí tương đối gia một điểm vi mt mt cu
Cho mt cu
S
có tâm
I
, bán kính
R
và điểm
A
.
Đim
A
thuc mt cu
IA R
.
Đim
A
nm trong mt cu
IA R
.
Đim
A
nm ngoài mt cu
IA R
.
2. Vị trí tương đối giữa mặt cầu và mt phẳng:
Cho mt cu
2 2 2
2
( ) :S x a y b z c R
và mt phng
:0Ax By Cz D
.
Tính:
2 2 2
;
Aa Bb Cc D
d d I
A B C


dR
: mt cu
S
và mt phng
()
không có điểm chung.
dR
: mt phng
()
tiếp xúc mt cu
S
ti
H
.
Đim
H
đưc gi là tiếp điểm hay
H
là hình chiếu của
I
lên mặt phẳng ().
Mt phng
()
đưc gi là tiếp din.
dR
: mt phng
()
ct mt cu
S
theo giao tuyến là đường tròn.
()
mặt cầu
S
không giao nhau
()
mặt cầu
S
tiếp xúc nhau tại
H
()
và mặt cầu
S
cắt nhau
theo giao tuyến là đường tròn
tâm
H
, bánnh
22
r R h
dR
dR
dR
R
α
I
H
α
R
I
H
C
( )
r
R
α
I
H
M
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Ví d 5.(THPT Nguyn Hu 2020) Lp phương trình mt cu
S
tâm
3; 2;4I
tiếp xúc
vi
mp P
:
2 2 4 0x y z
.
Lời giải.
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3. Vị trí tương đối giữa mặt cầu và đường thẳng:
Cho đường thng
:
01
02
03
x x a t
y y a t
z z a t



1
và mt cu
2 2 2
2
( ) : 2 .S x a y b z c R
Khi đó mặt cầu
S
có tâm
I
, bán kính
R
.
,h d I
là khoảng cách từ tâm
I
lên đường thẳng
.
H
là hình chiếu của
I
lên đường thẳng
.
Đưng thng
mt cu
S
không điểm chung
Đưng thng
tiếp c vi
mt cu
S
Đưng thng
ct mt cu
S
tại hai điểm phân bit
,d I R
vô nghim
,d I R
có đúng một nghim.
H
gi tiếp điểm hay hình
chiếu của điểm
I
xung
,d I R
hai nghim phân bit
,AB
H
trung điểm của
AB
Do đó:
2
22
4
AB
Rh
Nhận xét:
Tìm giao điểm của đường thẳng và mặt cầu ta xét hệ phương trình:
01
02
03
2 2 2
2
x x a t
y y a t
z z a t
x a y b z c R



Thay phương trình tham số
1
vào phương trình mặt cầu
2
, giải tìm
t
.
Thay
t
vào (1) được tọa độ giao điểm.
Ví d 6. Lập phương trình mặt cu
,S I R
tâm
1;3;5I
ct
23
:
1 1 1
x y z
ti hai
đim
,AB
sao cho
12AB
Lời giải.
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3. V trí tương đi ca hai mt cu:
1
S
không cắt
2
S
và ở
ngoài nhau.
1
S
tiếp xúc
2
S
1
S
cắt
2
S
tại
,.AB
1 2 1 2
I I R R
1 2 1 2
I I R R
' ' 'R R II R R
1
C
không cắt
2
C
và lồng
vào nhau.
1
C
tiếp xúc trong với
2
C
1 2 1 2
I I R R
1 2 1 2
I I R R
B.PHÂN DẠNG VÀ VÍ DỤ MINH HỌA.
Dng 1. Xác đnh tâm và bán kính mt cầu cho trước.
1. Phương pháp .
Cho mt cu
2 2 2
2 2 2 0.x y z ax by cz d
Khi đó để tìm tâm bán kính mt cu ta tiến
hành hai cách sau:
ch 1. Nhóm hng t và thêm bt hng t để xut hin hằng đẳng thc có dng
2 2 2
2
x a y b z c R
Khi đó, mặt cu
S
có tâm là
,,I a b c
, bán kính
2 2 2
R a b c d
.
ch 2. Thc hiện phương pháp đồng nht thc hai vế:
Gi phương trình mặt cu
2 2 2 2 2 2
2 :2 2 0 0x y z Ax By Cz D B CĐK DA
thì
Đồng nht hai vế ta được :
22
22
22
a A a A
b B b B
c C c C





Suy ra tâm
,,I A B C
, bán kính
2 2 2
R A B C D
u ý :
a
2
;
b
2
R
2
R
1
a
1
;
b
1
I
1
I
2
a
1
;
b
1
R
1
R
2
a
2
;
b
2
I
1
I
2
a
2
;
b
2
R
2
R
1
a
1
;
b
1
B
A
I
1
I
2
a
1
;
b
1
R
1
R
2
a
2
;
b
2
I
1
I
2
a
2
;
b
2
R
2
R
1
a
1
;
b
1
I
1
I
2
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Trong trc nghim ta tìm tâm
,,I a b c
bng cách ly h s ca
,,x y z
chia cho
2
.
Bán kính
2 2 2
R a b c d
.
Với phương trình mặt cu
S
:
2 2 2
2 2 2 0x y z ax by cz d
vi
2 2 2
0a b c d
thì
S
có tâm
; ; I a b c
và bán kính
2 2 2
R a b c d
.
2. Bài tp minh ha.
Bài tp 1. Trong các phương trình nào sau đây, phương trình nào phương trình của mt mt
cu ? Nếu là phương trình mặt cu, hãy tìm tâm và bán kính ca nó ?
a).
2 2 2
10 4 2 30 0.x y z x y z
b).
2 2 2
0.x y z y
c).
2 2 2
2 2 2 2 3 5 2 0.x y z x y z
d).
2 2 2
3 4 8 25 0.x y z x y z
e).
2 2 2
3 3 3 6 8 15 3 0.x y z x y z
Lời giải.
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Bài tp 2. Cho phương trình
2 2 2 2
4 4 2 4 0.x y z mx y mz m m
Xác định
m
để nó là phương trình mặt cầu. Khi đó, tìm
m
để bán bán kính ca nó là nh nht ?
Lời giải.
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Bài tp 3. Cho phương trình
2 2 2 2
2 cos 2 sin 4 (4 sin ) 0.x y z x y z
Xác định
để nó là phương trình mặt cầu. Khi đó, tìm
để bán bán kính ca nó là nh nht , ln
nht ?
Lời giải.
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3. Câu hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
u 1.(THPT Chuyên Hạ Long Quảng Ninh 2018) Trong không gian với hệ tọa đ
Oxyz
, cho mặt
cầu có phương trình
22
2
1 3 9x y z
. Tìm tọa độ tâm
I
và bán kính
R
của mặt cầu đó.
A.
1;3;0I
;
3R
. B.
1; 3;0I
;
9R
. C.
1; 3;0I
;
3R
. D.
1;3;0I
;
9R
Lời giải
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u 2.(THPT Lục Ngạn 2018)
Tâm
I
và bán kính
R
của mặt cầu
2 2 2
: 1 2 3 9S x y z
A.
1;2;3 ; 3IR
. B.
1;2; 3 ; 3IR
. C.
1; 2;3 ; 3IR
. D.
1;2; 3 ; 3IR
.
Lời giải
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u 3. (Sở GD & ĐT Cần T 2018)
Trong không gian
Oxyz
, mặt cầu
2 2 2
1 2 3 4x y z
có tâm và bán kính lần lượt là
A.
1; 2;3I 
;
2R
. B.
1;2; 3I
;
2R
. C.
1;2; 3I
;
4R
. D.
1; 2;3I 
;
4R
Lời giải
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u 4.(Sở GD&ĐT Đồng Tháp 2018)
Trong không gian với hệ toạ độ
Oxyz
, cho mặt cầu
22
2
: 2 1 4S x y z
. Tâm
I
của
mặt cầu
S
A.
2;1; 1I
. B.
2;0; 1I
. C.
2;0;1I
. D.
2;1;1I
.
Lời giải
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u 5.(THPT Đức Th 2018) Phương trình mặt cầu có tâm
1; 2;3I
, bán kính
2R
là:
A.
2 2 2
1 2 3 4.x y z
B.
2 2 2
1 2 3 4.x y z
C.
2 2 2
1 2 3 2.x y z
D.
2 2 2
1 2 3 2.x y z
Lời giải
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u 6.(THPT Lương Văn Chánh 2018) Trong không gian
Oxy
, phương trình nào dưới đây
phương trình mặt cầu tâm
1;0; 2I
, bán kính
4r
?
A.
22
2
1 2 16x y z
. B.
22
2
1 2 16x y z
.
C.
22
2
1 2 4x y z
. D.
22
2
1 2 4x y z
.
Lời giải
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u 7.(THPT Trần Nhân Tông 2018) Trong không gian với hệ trục
Oxyz
, cho mặt cầu có phương
trình
2 2 2
: 2 4 4 5 0S x y z x y z
. Tọa độ tâm và bán kính của
S
A.
2; 4; 4I
2R
. B.
1; 2; 2I
2R
.
C.
1; 2; 2I 
2R
. D.
1; 2; 2I 
14R
.
Lời giải
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u 8.(Sở GD&ĐT Bình Phước) Trong không gian
Oxyz
, cho mặt cầu
S
phương trình
2 2 2
2 4 6 2 0 x y z x y z
. Tìm tọa độ tâm
I
và tính bán kính
R
của
S
.
A. Tâm
1;2; 3I
và bán kính
4R
. B. Tâm
1; 2;3I
và bán kính
4R
.
C. Tâm
1;2;3I
và bán kính
4R
. D. Tâm
1; 2;3I
và bán kính
16R
.
Lời giải
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u 9.(THPT Thanh Miện 2018) Trong không gian
Oxyz
cho mặt cầu
S
phương trình:
2 2 2
2 4 4 7 0x y z x y z
. Xác định tọa độ tâm
I
và bán kính
R
của mặt cầu
S
:
A.
1; 2;2I 
;
3R
. B.
1;2; 2I
;
2R
.
C.
1; 2;2I 
;
4R
. D.
1;2; 2I
;
4R
.
Lời giải
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u 10.(THPT Bình Xuyên 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu có phương
trình
2 2 2
: 2 1 0S x y z x y
. Tâm
I
và bán kính
R
của
S
A.
1
;1;0
2
I



1
4
R
. B.
1
;1;0
2
I



1
2
R
.
C.
1
; 1;0
2
I



1
2
R
. D.
1
; 1;0
2
I



1
2
R
.
Li gii
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u 11.(Sở GD & ĐT Đồng Tháp 2018) Trong không gian với hệ trục tọa độ
Oxyz
, phương trình
nào dưới đây là phương trình của một mặt cầu?
A.
2 2 2
2 4 3 8 0x y z x y z
. B.
2 2 2
2 4 3 7 0x y z x y z
.
C.
22
2 4 1 0x y x y
. D.
22
2 6 2 0x z x z
.
Li gii
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u 12.(THPT Đức Thọ Tĩnh 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
S
:
2 2 2
6 4 8 4 0x y z x y z
. Tìm tọa độ tâm
I
và tính bán kính
R
của mặt cầu
S
.
A.
3; 2;4I
,
25R
. B.
3;2; 4I 
,
5R
. C.
3; 2;4I
,
5R
. D.
3;2; 4I 
,
25R
Li gii
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u 13.(THPT Chuyên Quý Đôn Đà Nẵng 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho
mặt cầu
S
có phương trình
2 2 2
: 2 4 6 5 0S x y z x y z
. Tính diện tích mặt cầu
S
.
A.
42
. B.
36
. C.
9
. D.
12
.
Li gii
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u 14.(THPT Hồng Quang Hải Dương 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 2 2 4 2 0S x y z x y z
. Tính bán kính
r
của mặt cầu.
A.
22r
. B.
26r
. C.
4r
. D.
2r
.
Li gii
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u 15.(THPT Trần Quốc Tuấn 2018) Trong không gian với hệ trục tọa độ
Oxyz
, tìm tọa độ tâm
I
và tính bán kính
R
của mặt cầu
S
:
2 2 2
4 2 4 0x y z x z
.
A.
2;0; 1I
,
3R
. B.
4;0; 2I
,
3R
. C.
2;0;1I
,
1R
. D.
2;0; 1I
,
1R
.
Li gii
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u 16.(THPT Tứ Kỳ 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu có phương trình
2 2 2
: 4 2 2 3 0S x y z x y z
. Tìm tọa độ tâm
I
và bán kính
R
của
S
.
A.
2; 1;1I
3R
. B.
2;1; 1I 
3R
.
C.
2; 1;1I
9R
. D.
2;1; 1I 
và
9R
.
Li gii
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u 17.(THPT Trần P 2018)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 4 2 6 5 0S x y z x y z
. Mặt cầu
S
bán
kính là
A.
3
. B.
5
. C.
2
. D.
7
.
Li gii
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u 18.(Sở GD&ĐT Bắc Ninh 2018) Trong không gian với hệ tọa độ
Oxyz
, tính bán kính
R
của
mặt cầu
S
:
2 2 2
2 4 0x y z x y
.
A.
5
. B.
5
. C.
2
. D.
6
.
Li gii
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u 19.(THPT Tây Thụy Anh 2018) Trong các phương trình sau, phương trình nào không phải
phương trình mặt cầu?
A.
2 2 2
2 4 4 21 0x y z x y z
. B.
2 2 2
2 2 2 4 4 8 11 0x y z x y z
.
C.
2 2 2
1x y z
. D.
2 2 2
2 2 4 11 0x y z x y z
.
Li gii
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Mức độ 3. Vận dụng
u 20.(THPT Chuyên Quý Đôn 2020) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt cầu
S
có phương trình
2 2 2
: 2 4 6 5 0S x y z x y z
. Tính diện tích mặt cầu
S
.
A.
42
. B.
36
. C.
9
. D.
12
.
Li gii
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u 21.(THPT Chuyên Thoại Ngọc Hầu 2020) Trong không gian
Oxyz
, tìm tất cả các giá trị của
m
để phương trình
2 2 2
4 2 2 0x y z x y z m
là phương trình của một mặt cầu.
A.
6m
. B.
6m
. C.
6m
. D.
6m
.
Li gii
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u 22.(THPT Trần Nhân Tông 2020) Trong không gian với hệ tọa độ
Oxyz
, tìm tất cả các giá trị
m
để phương trình
2 2 2
2 2 4 0x y z x y z m
là phương trình của một mặt cầu.
A.
6m
. B.
6m
. C.
6m
. D.
6m
.
Li gii
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u 23.(TT Diệu Hiền Cần T 2018)
Trong không gian vi h trc tọa độ
Oxy
, có tt c bao nhiêu s t nhiên ca tham s
m
để
phương trình
2 2 2 2
2 2 2 3 3 7 0x y z m y m z m
là phương trình ca mt mt cu.
A.
2
. B.
3
. C.
4
. D.
5
.
Li gii
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u 24.(THTT Số 4-487-2018) Trong không gian với hệ toạ độ
Oxyz
cho phương trình mặt cầu
2 2 2 2
2 2 4 2 5 9 0x y z m x my mz m
.Tìm
m
để phương trình đó phương trình của
một mặt cầu.
A.
55m
. B.
5m 
hoặc
1m
. C.
5m 
. D.
1m
.
Li gii
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u 25.(THPT Chuyên Hùng Vương 2020) Trong không gian với hệ tọa độ
,Oxyz
cho mặt cầu
S
có phương trình
2 2 2
2 4 4 0x y z x y z m
có bán kính
5.R
Tìm giá trị của
m
.
A.
4m
. B.
4m 
. C.
16m
. D.
16m 
.
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u 26.(THPT Can Lộc 2018) Cho mặt cầu
2 2 2
: 2 4 1 0S x y z x y mz
. Khẳng định o
sau đây luôn đúng với mọi số thực
m
?
A.
S
luôn tiếp xúc với trục
Oy
. B.
S
luôn tiếp xúc với trục
Ox
.
C.
S
luôn đi qua gốc tọa độ
O
. D.
S
luôn tiếp xúc với trục
Oz
.
Li gii
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u 27. Trong hệ tọa độ
Oxyz
,
cho mặt cầu
.Mặt cầu tiếp xúc với mặt phẳng với giá trị của là:
A.
3; 1.mm
. B.
1
1;
2
mm
. C.
1; 2mm
. D.
1; 3.mm
Li gii
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u 28.(THPT Nguyễn Huệ 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt cầu
S
phương trình
2
2
22
2 5 3 5x y z m m
. Các giá trị
m
để mặt cầu
S
cắt trục
Oz
tại
hai điểm phân biệt là
A.
m
. B.
4m 
.
C.
4m 
hoặc
1m
. D.
41m
.
Li gii
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2 2 2 2
: 2 4 2 2 2 0S x y z x y mz m m
Oxy
m
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u 29. Trong không gian vi h tọa độ
Oxyz
, tìm tt c các giá tr ca
m
để phương trình
2 2 2
2 2 4 0x y z x y z m
là phương trình của mt mt cu
A.
6m
. B.
6m
. C.
6m
. D.
6m
.
Li gii
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u 30. Cho phương trình
2 2 2 2
2 2 4 2 5 9 0x y z m x my mz m
.
Tìm
m
để phương trình đó là phương trình của một mặt cầu.
A.
51m
. B.
5m 
hoc
1m
.
C.
5m 
hoc
1m
. D.
1m
.
Li gii
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u 31. Cho phương trình:
2 2 2 2
2 1 4 1 2 1 5 10 14 0x y z m x m y m z m m
.
Tìm
m
để phương trình đó là phương trình một mặt cầu.
A.
42m
. B.
42mm
. C.
42mm
. D.
42m
.
Li gii
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Dng 2. Viết phương trình mt cu thỏa mãn điều kiện cho trước.
1. Phương pháp chung.
Cho phương trình mặt cu
2 2 2 2 2 2
2 :2 2 0 0x y z ax by cz d a bĐK cd
thì
Tâm
,,I a b c
: Tính
,,abc
bng cách ly h s ca chia cho
2
.
Bán kính
2 2 2
R a b c d
.
Chú ý:
Với phương trình mặt cu
S
:
2 2 2 2 2 2
2 :2 2 0 0x y z ax by cz d a bĐK cd
thì
S
tâm
; ; I a b c
và bán kính
2 2 2
R a b c d
.
2. Bài toán tng quát và minh ha.
Bài toán 1. Phương trình mặt cu tâm
I
đi qua điểm
A
.
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tp 4. Lập phương trình mặt cu
S
biết mt cu
S
có tâm nm trên
Ox
và đi qua
1;2;1 , 3;1; 2AB
Lời giải.
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2. Câu hi trc nghim.
Mc độ 1,2. Nhận biết-Thông hiểu
u 32.(THPT Can Lộc 2018)
Mặt cầu
S
có tâm
1; 3;2I
và đi qua
5; 1;4A
có phương trình:
A.
2 2 2
13 242x y z
. B.
2 2 2
13 242x y z
.
C.
2 2 2
13 242x y z
. D.
2 2 2
13 242x y z
.
Lời giải
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B
R
A
I(a;b;c)
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u 33.(THPT Chun Thái nh 2018) Trong không gian
Oxyz
, cho hai điểm
1; 0; 1I
2; 2; 3A
. Mặt cầu
S
tâm
I
và đi qua điểm
A
có phương trình là
A.
22
2
1 1 3x y z
. B.
22
2
1 1 3x y z
.
C.
22
2
1 1 9x y z
. D.
22
2
1 1 9x y z
.
Lời giải
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u 34.(Sở GD&ĐT Đồng Tháp 2018) Mặt cầu
S
tâm
3; 3;1I
đi qua điểm
5; 2;1A
phương trình là
A.
2 2 2
5 2 1 5x y z
. B.
2 2 2
3 3 1 25x y z
.
C.
2 2 2
3 3 1 5x y z
. D.
2 2 2
5 2 1 5x y z
.
Lời giải
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u 35.(THPT Can Lộc 2018) Mặt cầu
S
m
1; 3;2I
đi qua
5; 1;4A
phương
trình:
A.
2 2 2
13 242x y z
. B.
2 2 2
13 242x y z
.
C.
2 2 2
13 242x y z
. D.
2 2 2
13 242x y z
.
Lời giải
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u 36.(THPT Trần Quốc Tuấn 2018) Trong không gian với hệ tọa độ
Oxyz
cho tam giác
ABC
(2;2;0)A
,
(1;0;2)B
,
(0;4;4)C
. Viết phương trình mặt cầu tâm
A
đi qua trọng tâm
G
của
tam giác
ABC
.
A.
2 2 2
( 2) ( 2) 4x y z
. B.
2 2 2
( 2) ( 2) 5x y z
.
C.
2 2 2
( 2) ( 2) 5x y z
. D.
2 2 2
( 2) ( 2) 5x y z
.
Lời giải
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u 37. Trong không gian
Oxyz
, cho hai điểm
(1;2;3)A
,
(1; 2;5)B
. Phương trình của mặt cầu đi
qua 2 điểm
A
,
B
và có tâm thuộc trục
Oy
A.
2 2 2
4 22 0x y z y
. B.
2 2 2
4 26 0x y z y
.
C.
2 2 2
4 22 0x y z y
. D.
2 2 2
4 26 0x y z y
.
Lời giải
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u 38. Trong không gian
Oxyz
, cho 3 điểm:
1;3;0 , 1;1;2 , 1; 1;2A B C
. Mặt cầu
S
tâm
I
là trung điểm đoạn thẳng
AB
S
đi qua điểm
C
. Phương trình mặt cầu
S
là:
A.
2 2 2
1 1 1 5x y z
. B.
22
2
2 1 11x y z
.
C.
22
2
2 1 11x y z
D.
22
2
2 1 11x y z
.
Lời giải.
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u 39. Trong không gian
Oxyz
, cho hai điểm
0;2;3A
0;4; 1B
. Mặt cầu có tâm thuộc trục
Oy
đồng thời đi qua hai điểm
A
B
có bán kính bằng
A.
1
. B.
5
. C.
10
. D.
7
.
Lời giải
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u 40. Trong không gian
Oxyz
, cho hai điểm
1; 1; 0B 
3;1; 1C
.
Tọa độ điểm
M
thuộc trục
Oy
M
cách đều
B
,
C
A.
9
0; ;0
4
M



. B.
9
0; ;0
2
M



. C.
9
0; ;0
4
M



. D.
9
0; ;0
2
M



.
Lời giải
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u 41. Mặt cầu đi qua hai điểm
1;2;3A
,
2;1;0B
và tâm thuộc trục
Ox
có đường kính là
A.
173
. B.
173
4
.
C.
173
2
. D.
173
2
.
Lời giải
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Bài toán 2. Phương trình mt cu đưng kính
AB
Tâm
I
là trung điểm
AB
:
2
2
2
AB
I
AB
I
AB
I
xx
x
yy
y
zz
z
Bán kính
2
AB
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
3. Câu hi trc nghim.
B
R
A
I(a;b;c)
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Mức độ 1,2. Nhận biết-Thông hiểu
u 42.(THPT Hồng Bàng 2018) Trong không gian với hệ toạ độ
Oxyz
, cho hai điểm
2;1;1A
,
0;3; 1B
. Mặt cầu
S
đường kính
AB
có phương trình là
A.
2
22
23x y z
. B.
22
2
1 2 3x y z
.
C.
2 2 2
1 2 1 9x y z
. D.
22
2
1 2 9x y z
.
Lời giải
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u 43.(Cụm 5 Đồng Bằng Sông Cửu long 2018) Trong không gian với hệ tọa độ
Oxyz
, viết
phương trình chính tắc của mặt cầu có đường kính
AB
với
2;1;0A
,
0;1;2B
.
A.
2 2 2
1 1 1 4x y z
. B.
2 2 2
1 1 1 2x y z
.
C.
2 2 2
1 1 1 4x y z
. D.
2 2 2
1 1 1 2x y z
.
Lời giải
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u 44.(Sở GD&ĐT Đồng Tháp 2018) Trong không gian
Oxyz
, cho hai điểm
3; 2;0A
,
1;0; 4B
.
Mặt cầu nhận
AB
làm đường kính có phương trình là
A.
2 2 2
4 2 4 15 0x y z x y z
. B.
2 2 2
4 2 4 15 0x y z x y z
.
C.
2 2 2
4 2 4 3 0x y z x y z
. D.
2 2 2
4 2 4 3 0x y z x y z
.
Lời giải
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u 45.(Sở GD & ĐT Quãng Tr2018) Trong không gian hệ tọa độ
Oxyz
, cho hai điểm
2;1;0A
,
2; 1;2B
. Phương trình của mặt cầu có đường kính
AB
là:
A.
2
22
1 24x y z
. B.
2
22
16x y z
.
C.
2
22
16x y z
. D.
2
22
1 24x y z
.
Lời giải
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u 46.(S GD&ĐT Cần Thơ 2018) Trong không gian với hệ tọa độ
Oxyz
, cho hai điểm
1;2;3M
1;2; 1N 
. Mặt cầu đường kính
MN
có phương trình là
A.
22
2
2 1 20x y z
. B.
22
2
2 1 5x y z
.
C.
22
2
2 1 5x y z
. D.
22
2
2 1 20x y z
.
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Lời giải
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u 47.(THPT Hậu Lộc 2018) Trong không gian
Oxyz
, cho hai điểm
6; 2; 5A
,
4; 0; 7B
. Viết
phương trình mặt cầu đường kính
AB
.
A.
2 2 2
5 1 6 62 x y z
. B.
2 2 2
5 1 6 62 x y z
.
C.
2 2 2
1 1 1 62 x y z
. D.
2 2 2
1 1 1 62 x y z
.
Lời giải.
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u 48.(THPT Hoàng Hoa Thám 2018) Trong không gianhệ tọa độ
Oxyz
, cho điểm
2;1; 2A
4;3;2B
. Viết phương trình mặt cầu
S
đường kính
AB
.
A.
22
2
: 3 2 24S x y z
. B.
22
2
: 3 2 6S x y z
.
C.
22
2
: 3 2 24S x y z
. D.
22
2
: 3 2 6S x y z
.
Lời giải
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u 49.(THPT Hải Hậu 2018) Trong không gian với hệ tọa độ
Oxy
, cho hai điểm
1;2;1A
,
0;2;3B
. Viết phương trình mặt cầu có đường kính
AB
A.
2
22
15
22
24
x y z



. B.
2
22
15
22
24
x y z



C.
2
22
15
22
24
x y z



D.
2
22
15
22
24
x y z



Lời giải
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u 50.(THPT Chuyên ĐH Vinh 2018) Trong không gian
Oxyz
, mặt phẳng
:2 6 3 0P x y z
cắt trục
Oz
đường thẳng
56
:
1 2 1
x y z
d


lần lượt tại
A
,
B
. Phương trình mặt cầu đường
kính
AB
A.
2 2 2
2 1 5 36x y z
. B.
2 2 2
2 1 5 9x y z
.
C.
2 2 2
2 1 5 9x y z
. D.
2 2 2
2 1 5 36x y z
.
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Lời giải
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Bài toán 3. Mt cu tâm
;;I a b c
tiếp xúc mt phng
:0Ax By Cz D
Tâm
;;I a b c
.
Bán kính
2 2 2
;
Aa Bb Cc D
R d I
A B C


Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tp 5. Lập phương trình mặt cu
S
biết mt cu
S
có tâm
3; 2;4I
và tiếp xúc vi
: 2 2 4 0mp P x y z
.
Lời giải.
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Bài tp 6. Lập phương trình mặt cầu
S
có tâm
1;1;2I
và tiếp xúc với
: 2 2 1 0P x y z
Lời giải.
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Bài tp 7. Lập phương trình mặt cầu
S
có bán kính
3R
và tiếp xúc với mặt phẳng
: 2 2 3 0P x y z
tại điểm
1;1; 3A
;
Lời giải.
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n
p
P
R
H
I(a;b;c)
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u hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
u 51.(THPT Chuyên Lam n 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 3 0P x y z
và điểm
1;1;0I
. Phương trình mặt cầu tâm
I
và tiếp xúc với
P
A.
22
2
5
11
6
x y z
. B.
22
2
25
11
6
x y z
.
C.
22
2
5
11
6
x y z
. D.
22
2
25
11
6
x y z
.
Lời giải
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u 52.(THPT Chuyên Tiền Giang 2018) Trong hệ tọa độ
Oxyz
, cho điểm
2;1;1A
mặt phẳng
:2 2 1 0xyP z
. Phương trình của mặt cầu tâm
A
và tiếp xúc với mặt phẳng
P
A.
2 2 2
2 1 1 9x y z
. B.
2 2 2
2 1 1 2x y z
.
C.
2 2 2
2 1 1 4x y z
. D.
2 2 2
2 1 1 36x y z
.
Lời giải
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u 53.(Sở GD&ĐT Nam Định 2018) Trong không gian với hệ tọa độ
Oxyz
, viết phương trình mặt
cầu
S
có tâm
0;1; 1I
và tiếp xúc với mặt phẳng
:2 2 3 0P x y z
A.
22
2
1 1 4x y z
. B.
22
2
1 1 4x y z
.
C.
22
2
1 1 4x y z
. D.
22
2
1 1 2x y z
.
Lời giải
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u 54.(THPT Chuyên Thái Bình 2018) Mt cu
S
tâm
1;2;1I
tiếp xúc vi mt phng
P
:
2 2 2 0x y z
có phương trình là:
A.
S
:
2 2 2
1 2 1 3x y z
. B.
S
:
2 2 2
1 2 1 3x y z
.
C.
S
:
2 2 2
1 2 1 9x y z
. D.
S
:
2 2 2
1 2 1 9x y z
.
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Lời giải
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u 55.(Sở GDT Bình Phước) Trong không gian với hệ tọa độ
Oxyz
, mặt cầu
S
m
2;1; 1I
, tiếp xúc với mặt phẳng tọa độ
Oyz
. Phương trình của mặt cầu
S
A.
2 2 2
2 1 1 4x y z
. B.
2 2 2
2 1 1 1x y z
.
C.
2 2 2
2 1 1 4x y z
. D.
2 2 2
2 1 1 2x y z
.
Lời giải
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u 56.(THPT Chuyên Hùng ơng 2018) Trong không gian với hệ tọa độ
Oxyz
, cho điểm
1;2; 5I
mặt phẳng
: 2 2 8 0P x y z
. Viết phương trình mặt cầu tâm
I
tiếp xúc
với mặt phẳng
P
.
A.
2 2 2
1 2 5 25x y z
. B.
2 2 2
1 2 5 25x y z
.
C.
2 2 2
1 2 5 5x y z
. D.
2 2 2
1 2 5 36x y z
.
Lời giải
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u 57.(THPT Chuyên Lam Sơn 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 3 0P x y z
và điểm
1;1;0I
. Phương trình mặt cầu tâm
I
và tiếp xúc với
P
là:
A.
22
2
5
11
6
x y z
. B.
22
2
25
11
6
x y z
.
C.
22
2
5
11
6
x y z
. D.
22
2
25
11
6
x y z
.
Lời giải
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u 58.(THPT Chuyên Thoại Ngọc Hầu 2018) Trong không gian
Oxyz
, phương trình nào dưới đây
là phương trình mặt cầu có tâm
1;2; 1I
và tiếp xúc với mặt phẳng
: 2 2 8 0P x y z
?
A.
2 2 2
1 2 1 9x y z
. B.
2 2 2
1 2 1 9x y z
.
C.
2 2 2
1 2 1 3x y z
. D.
2 2 2
1 2 1 3x y z
.
Lời giải
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u 59.(Sở GD&ĐT Nam Định 2018) Trong không gian với hệ tọa độ
Oxyz
, viết phương trình mặt
cầu
S
có tâm
0;1; 1I
và tiếp xúc với mặt phẳng
:2 2 3 0P x y z
A.
22
2
1 1 4x y z
. B.
22
2
1 1 4x y z
.
C.
22
2
1 1 4x y z
. D.
22
2
1 1 2x y z
.
Lời giải
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u 60.(THPT Chuyên Tiền Giang 2018) Trong hệ tọa độ
Oxyz
, cho điểm
2;1;1A
mặt phẳng
:2 2 1 0xyP z
. Phương trình của mặt cầu tâm
A
và tiếp xúc với mặt phẳng
P
A.
2 2 2
2 1 1 9x y z
. B.
2 2 2
2 1 1 2x y z
.
C.
2 2 2
2 1 1 4x y z
. D.
2 2 2
2 1 1 36x y z
.
Lời giải
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u 61.Cho mặt cầu
2 2 2
: 2 6 8 1 0S x y z x y z
. Xác định bán kính
R
của mặt cầu
S
và viết phương trình mặt phẳng
P
tiếp xúc với mặt cầu tại
1;1;1M
?
A. Bán kính của mặt cầu
5R
, phương trình mặt phẳng
:4 3 1 0P y z
.
B. Bán kính của mặt cầu
5R
, phương trình mặt phẳng
:4 3 1 0P x z
.
C. Bán kính của mặt cầu
5R
, phương trình mặt phẳng
:4 3 1 0P y z
.
D. Bán kính của mặt cầu
3R
, phương trình mặt phẳng
:4 3 7 0P y y
Lời giải
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u 62. Trong không gian
Oxyz
, cho điểm
2;4; 3I
. Phương trình mặt cầu có tâm
I
tiếp xúc
với mặt phẳng
Oxz
A.
2 2 2
2 4 3 4x y z
. B.
2 2 2
2 4 3 29x y z
.
C.
2 2 2
2 4 3 9x y z
. D.
2 2 2
2 4 3 16x y z
.
Lời giải
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u 63. Trong không gian
Oxyz
, cho điểm
1;2;3I
. Mặt cầu
S
tâm
I
tiếp xúc với mặt
phẳng
Oxz
có phương trình là
A.
2 2 2
1 2 3 9x y z
. B.
2 2 2
1 2 3 1x y z
.
C.
2 2 2
1 2 3 14x y z
. D.
2 2 2
1 2 3 4x y z
.
Lời giải
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u 64. Trong không gian
Oxyz
, cho
1;2;3I
. Phương trình mặt cầu
S
tâm
I
, tiếp xúc với
Oxy
là.
A.
2 2 2
1 2 3 5.x y z
B.
2 2 2
1 2 3 9.x y z
C.
2 2 2
1 2 3 9.x y z
D.
2 2 2
1 2 3 14.x y z
Lời giải
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u 65. Cho điểm
(1; 2;3)M
. Gọi
I
là hình chiếu vuông góc của
M
lên trục
Ox
.
Phương trình nào dưới đây là phương trình của mặt cầu tâm
I
, bán kính
IM
?
A.
2 2 2
( 1) 13x y z
B.
2 2 2
( 1) 13x y z
C.
2 2 2
( 1) 13x y z
D.
2 2 2
( 1) 17x y z
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Lời giải
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u 66. Trong không gian với hệ tọa độ
Oxyz
, mặt cầu tâm
0;0;3I
và tiếp xúc với mặt phẳng
Oxy
có phương trình là
A.
2 2 2
( 3) 3x y z
. B.
2 2 2
( 3) 9x y z
.
C.
2 2 2
( 3) 3x y z
. D.
2 2 2
( 3) 9x y z
.
Lời giải
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u 67. Trong không gian
Oxyz
, cho điểm
1;2; 3I 
. Phương trình mặt cầu tâm
I
tiếp xúc với
trục
Oy
A.
2 2 2
1 2 3 10x y z
. B.
2 2 2
1 2 3 100x y z
.
C.
2 2 2
1 2 3 10x y z
. D.
2 2 2
1 2 3 100x y z
.
Lời giải
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u 68. Trong không gian
Oxyz
, cho mặt cầu
S
tâm
1;2; 3A
tiếp xúc với trục
Ox
.
Phương trình của
S
A.
2 2 2
1 2 3 13x y z
. B.
2 2 2
1 2 3 13x y z
.
C.
2 2 2
1 2 3 13x y z
. D.
2 2 2
1 2 3 13x y z
.
Lời giải
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u 69. Trong bốn phương trình mặt cầu sau, tìm phương trình của mặt cầu tiếp xúc với trục
Oz
.
A.
2 2 2
2 1 3 5x y z
. B.
2 2 2
2 1 3 12x y z
.
C.
2 2 2
2 1 3 10x y z
. D.
2 2 2
2 1 3 13x y z
.
Lời giải
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u 70. Trong không gian
Oxyz
, cho mặt cầu
S
tâm
2; 4;3I
và tiếp xúc với trục
Ox
.
Phương trình của mặt cầu
S
là:
A.
2 2 2
2 4 3 25x y z
. B.
2 2 2
2 4 3 4x y z
.
C.
2 2 2
2 4 3 4x y z
. D.
2 2 2
2 4 3 25x y z
.
Lời giải
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Mức độ 3. Vận dụng
u 71. Trong không gian với hệ tọa độ
,Oxyz
mặt cầu
S
bán kính bằng
2,
tiếp xúc với mặt
phẳng
Oyz
và có tâm nằm trên tia
.Ox
Phương trình của mặt cầu
S
A.
2
22
: 2 4S x y z
. B.
2
22
: 2 4S x y z
.
C.
2
22
: 2 4S x y z
. D.
2
22
: 2 4S x y z
.
Lời giải
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Suy ra mt cu
1;1;0A
có tâm
1; 1;0A
và bán kính
2R
nên
2
22
: 2 4S x y z
u 72. Cho mặt cầu
2 2 2
: 2 4 6 0S x y z x y z m
. Tìm
m
để
S
tiếp xúc với mặt phẳng
: 2 2 1 0P x y z
.
A.
2m 
. B.
2m
. C.
3m 
D.
3m
.
Lời giải
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Bài toán 4. Mt cu ngoi tiếp t din
ABCD
(đi qua 4 điểm
, , ,A B C D
)
Gi s mt cu
S
dng:
2 2 2
2 2 2 0x y z ax by cz d
2
Thế tọa độ của điểm
, , ,A B C D
vào phương trình
2
ta
đưc 4 phương trình.
Gii h phương trình tìm
, , ,a b c d
ri viết phương trình
mt cu.
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tp 8. Lập phương trình mặt cu
S
biết mt cu
S
đi qua
2; 4;3C
các nh chiếu
ca
C
lên ba trc tọa độ.
Lời giải.
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Bài tp 9. Lập phương trình mặt cầu
S
đi qua bốn điểm
0;1;0 , 2;3;1 , 2;2;2A B C
1; 1;2D
;
Lời giải.
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B
R
A
I(a;b;c)
D
C
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 73.(THPT Chuyên Lương Thế Vinh 2018) Trong không gian
Oxyz
, cho bốn điểm
2;1;0A
;
1; 1;3B
;
3; 2;2C
1;2;2D
. Hỏi có bao nhiêu mặt cầu tiếp xúc với tất cả bốn mặt phẳng
ABC
,
BCD
,
CDA
,
DAB
.
A.
7
. B.
8
. C. vô số. D.
6
.
Lời giải
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u 74.(THPT Chuyên ĐHSP- Nội 2018) Trong không gian với hệ tọa độ
Oxyz
, cho
1;0;0A
,
0;0;2B
,
0; 3;0C
. Bán kính mặt cầu ngoại tiếp tứ diện
OABC
A.
14
3
. B.
14
4
. C.
14
2
. D.
14
.
Lời giải
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u 75. Trong không gian
Oxyz
cho ba điểm
2;0;0A
,
0;3;0B
,
2;3;6C
.
Thể tích khối cầu ngoại tiếp tứ diện
OABC
A.
1372
3
. B.
343
6
. C.
49
. D.
341
6
.
Lời giải
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u 76.(THPT Chuyên ĐHSP 2018) Trong không gian tọa độ
Oxyz
, mặt cầu
S
đi qua điểm
O
và cắt các tia
Ox
,
Oy
,
Oz
lần lượt tại các điểm
A
,
B
,
C
khác
O
thỏa mãn
ABC
có trọng tâm là
điểm
2;4;8G
.
Tọa độ tâm của mặt cầu
S
A.
1;2;3
. B.
4 8 16
;;
3 3 3



. C.
2 4 8
;;
3 3 3



. D.
3;6;12
.
Lời giải
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u 77.(THPT Trần Phú 2018) Trong không gian với hệ tọa độ
Oxyz
, cho tứ diện
ABCD
tọa
độ đỉnh
2; 0; 0A
,
0; 4; 0B
,
0; 0; 6C
,
2; 4; 6A
. Gọi
S
mặt cầu ngoại tiếp tứ diện
ABCD
. Viết phương trình mặt cầu
S
có tâm trùng với tâm của mặt cầu
S
và có bán kính gấp
2
lần bán kính của mặt cầu
S
.
A.
2 2 2
1 2 3 56x y z
. B.
2 2 2
2 4 6 0x y z x y z
.
C.
2 2 2
1 2 3 14x y z
. D.
2 2 2
2 4 6 12 0x y z x y z
.
Lời giải
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u 78. Cho 4 điểm
1; 1;0A
,
1;3;2B
,
4;3;2C
,
4; 1;2D
. Viết phương trình mặt cầu đi
qua 4 điểm
, , ,A B C D
.
A.
2 2 2
: 5 2 2 1 0S x y z x y z
. B.
2 2 2
: 5 4 4 0S x y z x y z
.
C.
2 2 2
: 11 10 26 3 0S x y z x y z
. D.
2 2 2
: 5 8 10 5 0S x y z x y z
.
Lời giải
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u 79. Cho bốn điểm
1;1;0A
,
3;1;2B
,
3;4;2C
,
1;4;2D
. Viết phương trình mặt cầu đi
qua
4
điểm
A
,
B
,
C
,
D
.
A.
2 2 2
: 2 5 2 1 0S x y z x y z
. B.
2 2 2
: 5 4 4 0S x y z x y z
.
C.
2 2 2
: 10 11 26 3 0S x y z x y z
. D.
2 2 2
: 8 5 10 5 0S x y z x y z
.
Lời giải
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u 80. Trong không gian
Oxyz
, mặt cầu qua bốn điểm
4;4;4 , 2;5;3 , 0;0;2 , 1; 1;0A B C D
,
có tâm là
;;I a b c
. Giá trị
bằng
A.
5
. B.
6
. C.
7
. D.
4
.
Lời giải
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u 81. Phương trình mặt cầu
()S
đi qua các điểm
, (4;0;0) (0; 2;0) (0;0;2)O A B C
A.
2 2 2
( 2) ( 1) ( 1) 6x y z
. B.
2 2 2
( 2) ( 1) ( 1) 24x y z
.
C.
2 2 2
( 4) ( 2) ( 2) 24x y z
. D.
2 2 2
( 2) ( 1) ( 1) 6x y z
.
Lời giải
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u 82. Cho
2;0;0 , 0;2;0 , 0;0;2 , 2;2;2M N E F
. Tính bán kính mặt cầu ngoại tiếp tứ diện
MNEF
.
A.
3
2
R
. B.
22R
. C.
23R
. D.
3R
.
Lời giải
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u 83. Trong không gian
Oxyz
, cho
1; 2;0A 
,
5; 3;1B 
,
2; 3;4C 
. Trong các mặt cầu đi
qua ba điểm
,,A B C
mặt cầu có diện tích nhỏ nhất có bán kính
R
bằng
A.
6R
. B.
36
2
R
. C.
3R
. D.
52
2
R
.
Lời giải
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Bài toán 5. Mt cu đi qua
,,A B C
tâm
:0I Ax By Cz D
:
Gi s mt cu
S
có dng:
2 2 2
2 2 2 0x y z ax by cz d
2
Thế tọa độ của điểm
,,A B C
vào phương trình
2
ta được 3
phương trình.
; ; 0I a b c Aa Bb Cc D
Gii h 4 phương trình tìm
, , ,a b c d
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tp 10. Lập phương trình mặt cu
S
biết mt cu
S
tâm nm trên
mp Oxy
đi
qua
1;0;2 , 2;1;1 ,MN
1; 1;1P 
.
Lời giải.
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Bài tp 11. Lập phương trình mặt cầu
S
có tâm thuộc mp
: 2 0P x y z
đi qua ba
điểm
2;0;1 , 1;0;0AB
,
1;1;1C
;
Lời giải.
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B
R
A
I(a;b;c)
C
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u hi trc nghim.
Mức độ 3. Vận dụng
u 84.(Tạp Chí Toán Học Tuổi Trẻ 2020) Trong không gian với hệ trục tọa độ
Oxyz
, cho ba điểm
1;2; 4A
,
1; 3;1B
,
2;2;3C
. Tính đường kính
l
của mặt cầu
S
đi qua ba điểm trên
tâm nằm trên mặt phẳng
Oxy
.
A.
2 13l
. B.
2 41l
. C.
2 26l
. D.
2 11l
.
Lời giải
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u 85.(THPT Chuyên Thái Bình 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho
1;2;3A
;
4;2;3B
;
4;5;3C
. Diện tích mặt cầu nhận đường tròn ngoi tiếp tam giác
ABC
làm đường
tròn lớn là
A.
9
. B.
36
. C.
18
. D.
72
.
Lời giải
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u 86.(THPT Chuyên Thái Nguyên 2018) Trong h tọa độ
Oxyz
, cho mt cu
S
tâm thuc
mp
Oxy
đi qua ba điểm
1 ; 3 ; 3A
,
2 ; 1 ; 0B
1 ; 1 ; 1C
. Mt cu
S
bán kính
R
bng bao nhiêu?
A.
4R
. B.
26R
. C.
5R
. D.
21R
.
Lời giải
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Bài toán 6. Mt cu
S
đi qua hai điểm
,AB
tâm thuc đường thng
d
Tâm
01
0 2 0 1 0 2 0 3
03
, , , tI
x x a t
y y a t t x a t y a t z a
z z a t
I


Ta có
, ( )A B S
22
IA IB R IA IB
.
Gii phương trình tìm ra
t
tọa độ
I
, tính được
R
.
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tập 12.(THPT Chuyên Qúy Đôn 2020) Lập phương trình mặt cầu
S
có tâm nằm trên
đường thẳng
2 1 1
:
3 2 2
x y z
d

tiếp xúc với hai mặt phẳng
: 2 2 2 0P x y z
: 2 2 4 0Q x y z
;
Lời giải.
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Bài tp 13. Lập phương trình mặt cầu
,S I R
a). Mặt cầu
S
có tâm thuộc đường thẳng
2 1 1
:
1 2 2
x y z
và tiếp xúc với mặt phẳng
1
: 3 2 6 0x y z
và mặt phẳng
2
: 2 3 0x y z
b). Mặt cầu
S
có tâm thuộc đường thẳng
23
:,
1 1 2
x y z
d


đi qua
1;1;4M
và tiếp xúc
với
2 2 4
:
1 1 4
x y z
d

.
Lời giải.
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B
d
R
A
I(a;b;c)
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Bài tp 14. Lập phương trình mặt cầu
S
biết
a). Có tâm
6;3; 4I
và tiếp xúc với
Oy
b). Có m nằm trên đường thẳng
2
:
0
x
d
y

tiếp xúc vi hai mặt phẳng
: 2 8 0P x z
: 2 5 0Q x z
.
Lời giải.
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 15. Lập phương trình mặt cầu
S
tâm thuộc đường thẳng
2
: 1 2
1
xt
d y t
zt



đồng
thời tiếp xúc với
2
mặt phẳng
: 2 2 5 0P x y z
: 2 2 13 0Q x y z
.
Lời giải.
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 87.(THPT Bình Xun 2018) Trong không gian
Oxyz
, cho đường thẳng
1
:
2 1 2
x y z
d

hai điểm
2;1;0A
,
2;3;2B
. Phương trình mặt cầu
S
đi qua hai điểm
A
,
B
tâm thuộc
đường thẳng
:d
A.
2 2 2
1 1 2 17x y z
. B.
2 2 2
1 1 2 9x y z
.
C.
2 2 2
1 1 2 5x y z
. D.
2 2 2
1 1 2 16x y z
.
Lời giải
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u 88.(THPT Lục Ngạn 2018) Trong không gian với hệ tọa độ
Oxyz
, phương trình mặt cầu đi
qua hai điểm
3; 1;2A
,
1;1; 2B
và có tâm thuộc trục
Oz
A.
2 2 2
2 10 0x y z z
. B.
2
22
1 11x y z
.
C.
2
22
1 11x y z
. D.
2 2 2
2 11 0x y z y
.
Li gii
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài toán 7. Mt cu
S
tâm
I
cắt đường thng
d
tại hai điểm
,AB
phân bit.
Tính độ dài
IH
chính là khong cách t tâm
I
đến đường
thng
d
01
02
03
,
x x a t
y y a t t
z z a t


Tính chất đường kính dây cung
2
AB
HA HB
.
Áp dụng định lý Py ta go tính
22
R IH HA
Phương trình
2 2 2
2
;:S I R x a y b z c R
Bài tp 16. Lập phương trình mặt cầu
,S I R
tâm
1;3;5I
cắt
23
:
1 1 1
x y z
tại
hai điểm
,AB
sao cho
12AB
Lời giải.
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Bài tp 17. Trong không gian
,Oxyz
cho đường thẳng
d
là giao tuyến của hai mặt phẳng
vi
:2 2 1 0,x y z
: 2 2 4 0x y z
và mặt cầu
S
có phương trình
2 2 2
4 6 0x y z x y m
. Tìm
m
để đường thẳng
d
cắt mặt cầu
S
tại hai điểm phân biệt
,AB
sao cho
8AB
.
Lời giải.
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R
H
B
I(a;b;c)
A
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
318
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 18. Tìm tham số thực
m
để đường thẳng
: 2 1 1d x y z
cắt mặt cầu
:S
2 2 2
4 6 0x y z x y m
tại
2
điểm phân biệt
,MN
sao cho độ dài day cung
8MN
Lời giải.
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u hi trc nghim.
Mức độ 3. Vận dụng
u 89.(THPT Chuyên ĐHSP 2018) Trong không gian tọa độ
Oxyz
, cho điểm
1; 2;3 .A
Gọi
S
là mặt cầu chứa
A
có tâm
I
thuộc tia
Ox
và bán kính bằng
7
. Phương trình mặt cầu
S
A.
2
22
5 49x y z
. B.
2
22
7 49x y z
.
C.
2
22
3 49x y z
. D.
2
22
7 49x y z
.
Lời giải
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u 90. (THPT Trần Phú 2018) Trong không gian hệ tọa độ
Oxyz
, cho điểm
1;0; 1A
mặt
phẳng
: 3 0P x y z
. Gọi
S
là mặt cầu có tâm
I
nằm trên mặt phẳng
P
, đi qua điểm
A
và gốc tọa độ
O
sao cho diện tích tam giác
OIA
bằng
17
2
. Tính bán kính
R
của mặt cầu
S
.
A.
3R
. B.
9R
. C.
1R
. D.
5R
.
Lời giải
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 91. Trong không gian
Oxyz
,cho điểm
1;0;3I
đường thẳng
1 1 1
:
2 1 2
x y z
d

. Viết
phương trình mặt cầu
S
tâm
I
và cắt
d
tại hai điểm
,AB
sao cho tam giác
IAB
vuông tại
I
.
A.
22
2
40
13
9
x y z
. B.
22
2
40
13
9
x y z
.
C.
22
2
20
13
3
x y z
. D.
22
2
40
13
3
x y z
.
Lời giải
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u 92.(THPT Chuyên Ngữ Nội 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
:2 2 0P x y z
và đường thẳng
1
:
1 2 1
x y z
d

. Gọi
là một đường thẳng chứa trong
P
,
cắt và vuông góc với
d
. Vectơ
;1;u a b
là một vectơ chỉ phương của
. Tính tổng
S a b
.
A.
1S
. B.
0S
. C.
2S
. D.
4S
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
320
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 93.(THPT Hậu Lộc 2018)
Trong không gian tọa độ
Oxyz
cho mặt cầu
2 2 2
: 4 6 0S x y z x y m
và đường thẳng
giao tuyến của hai mặt phẳng
: 2 2 4 0x y z
:2 2 1 0x y z
. Đường thẳng
cắt mặt cầu
S
tại hai điểm phân biệt
,AB
thỏa mãn
8AB
khi:
A.
12m
. B.
12m 
. C.
10m 
. D.
5m
.
Lời giải
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u 94. Trong không gian
Oxyz
, cho điểm
3;4;0I
đường thẳng
1 2 1
:
1 1 4
x y z
.
Phương trình mặt cầu
S
tâm
I
cắt
tại hai điểm
A
,
B
sao cho diện tích tam giác
IAB
bằng
12
A.
22
2
3 4 25x y z
. B.
22
2
3 4 5x y z
.
C.
22
2
3 4 5x y z
. D.
22
2
3 4 25x y z
.
Lời giải
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u 95. (S GD&ĐT Phú Th 2018) Trong không gian vi h tọa độ
Oxyz
, mt cu m
(2;5;3)I
cắt đường thng
12
:
2 1 2
x y z
d


tại hai điểm phân bit
A
,
B
vi chu vi tam giác
IAB
bng
14 2 31
có phương trình
A.
2 2 2
2 3 5 49x y z
. B.
2 2 2
2 3 5 196x y z
.
C.
2 2 2
2 3 5 31x y z
. D.
2 2 2
2 3 5 124x y z
.
Lời giải
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u 95. (Sở GD&ĐT Gia Lai 2018)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 1 2 3 25S x y z
hai điểm
3; 2;6A
,
0;1;0B
. Mặt phẳng
: 2 0P ax by cz
chứa đường thẳng
AB
cắt
S
theo giao tuyến là
đường tròn có bán kính nhỏ nhất. Tính giá trị của biểu thức
2M a b c
.
A.
2M
. B.
3M
. C.
1M
. D.
4M
.
Lời giải
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u 96. (THPT Chuyên Lamn 2018)
Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 1 2 3 9S x y z
tâm
I
mặt phẳng
:2 2 24 0P x y z
. Gọi
H
hình chiếu vuông góc của
I
trên
P
. Điểm
M
thuộc
S
sao cho đoạn
MH
có độ dài lớn nhất. Tìm tọa độ điểm
M
.
A.
1;0;4M
. B.
0;1;2M
. C.
3;4;2M
. D.
4;1;2M
.
Lời giải
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u 97.(Đề Minh Họa 2019)
Trong không gian
Oxyz
, cho điểm
2;1;3E
, mặt phẳng
:2 2 3 0P x y z
mặt cầu
2 2 2
: 3 2 5 36S x y z
. Gọi
đường thẳng đi qua
E
, nằm trong
P
cắt
S
tại hai điểm khoảng cách nhỏ nhất. Biết
một vec-chỉ phương
00
2018; ;u y z
. Tính
00
.T z y
A.
0T
. B.
. C.
2018T
. D.
1009T
.
Lời giải
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u 98.(Phát triển đề minh hoạ 2019) Trong không gian
Oxyz
, cho điểm
13
; ;0
22
M




và mặt cầu
2 2 2
: 8.S x y z
Đường thẳng
d
thay đổi, đi qua điểm
,M
cắt mặt cầu
S
tại hai điểm phân
biệt
,.AB
Tính diện tích lớn nhất
S
của tam giác
.OAB
A.
7S
. B.
4S
. C.
27S
. D.
22S
.
Lời giải
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u 99. (THPT Chuyên Lào Cai 2020)
Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 2 4 6 3 0S x y z x y z m
. Tìm
m
để
1
:1
2
xt
d y t
z


cắt
S
tại hai điểm phân biệt
A.
31
2
m
. B.
31
2
m
. C.
31
2
m
. D.
31
2
m
.
Lời giải
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u 100.(TH&TT) Trong không gian với hệ tọa độ
Oxyz
, cho đường thẳng
d
mặt phẳng
P
lần lượt phương trình
12
2 1 1
x y z

2 8 0x y z
, điểm
2; 1;3A
. Phương trình
đường thẳng
cắt
d
P
lần lượt tại
M
N
sao cho
A
là trung điểm của đoạn thẳng
MN
A.
1 5 5
3 4 2
x y z

. B.
2 1 3
6 1 2
x y z

.
C.
5 3 5
6 1 2
x y z

. D.
5 3 5
3 4 2
x y z

.
Lời giải
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u 101. Cho mt cu
S
:
2 2 2
9x y z
, điểm
1;1;2M
mt phng
P
:
40x y z
.
Gi
đường thẳng đi qua
M
, thuc
P
ct
S
tại 2 điểm
,AB
sao cho
AB
độ dài nh
nht. Biết
có một véc tơ chỉ phương là
1; ;u a b
.Tính giá tr
T a b
A.
2T 
. B.
1T
. C.
1T 
. D.
0T
.
Lời giải
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Bài toán 8. Mt cu
S
tâm
I
tiếpc vi mt cu
T
cho trước:
1. Phương pháp
Xác định tâm
J
và bán kính
'R
ca mt cu
T
S dụng điều kin tiếp xúc ca hai mt cầu để tính bán kính
R
ca mt cu
.S
(Xét hai trường hp tiếp xúc trong và tiếp xúc ngoài)
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
2. Bài tp minh ha
Bài tp 19. Trong không gian
,Oxyz
cho mặt cầu
2 2 2
1
: 6 12 12 72 0S x y z x y z
mặt cầu
2 2 2
2
: 9 0.S x y z
Lập phương trình mặt cầu
S
có tâm nằm trên đường nối tâm
của hai mặt cầu
1
S
2
,S
tiếp xúc với hai mặt cầu đó và có bán kính lớn nhất.
Lời giải.
R'
R
J
I
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3. Câu hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
u 102.(THPT Thanh Chương 2019) Trong không gian
Oxyz
, cho điểm
3; 1;4I
mt cu
22
2
1
: 1 2 1S x y z
. Phương trình của mt cu
S
tâm
I
tiếp xúc ngoài vi mt
cu
1
S
A.
2 2 2
3 1 4 4x y z
. B.
2 2 2
3 1 4 16x y z
.
C.
2 2 2
3 1 4 4x y z
. D.
2 2 2
3 1 4 2x y z
.
Lời giải
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u 103.(S GD&ĐT Thanh Hóa 2020) Trong không gian vi h tọa độ
Oxyz
cho các mt cu
1
S
,
2
S
,
3
S
bán kính
1r
lần lượt tâm các đim
,
2;1; 1B 
,
4; 1; 1C 
.
Gi
S
là mt cu tiếp xúc vi c ba mt cu trên. Mt cu
S
có bán kính nh nht là
A.
2 2 1R 
. B.
10R
. C.
22R
. D.
10 1R 
.
Lời giải
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Bài toán 9. Mt cu
'S
đối xng Mt cu
S
qua mt phng
P
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1. Phương pháp
Tìm điểm
I
đối xng vi tâm
I
qua mp
P
(xem cách làm phn mt phng)
Viết phương trình mặt cầu (S’) tâm
I
có bán kính
RR
.
Phương trình
2 2 2
2
;:S I R x a y b z c R
2. Câu hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
u 104. Cho
()S
tâm
(1;2; 1)I
và bán kính
3R
. Phương trình mặt cầu
( ')S
đối xứng với
()S
qua gốc tọa độ là
A.
2 2 2
( 1) ( 2) ( 1) 9x y z
. B.
2 2 2
( 1) ( 2) ( 1) 9x y z
C.
2 2 2
2 4 2 3 0x y z x y z
. D.
2 2 2
9x y z
.
Lời giải
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Bài toán 10. Mt cu
'S
đối xng mt cu
S
qua đưng thng
d
1. Phương pháp
Tìm điểm
I
đối xng vi tâm
I
qua đường thng
d
(xem cách làm phần đường thng)
Viết phương trình mặt cầu (S’) tâm
I
có bán kính
RR
.
Phương trình
2 2 2
2
;:S I R x a y b z c R
2. Câu hi trc nghim.
Mức độ 1,2. Nhận biết-Thông hiểu
u 105. Mặt cầu đối xứng với mặt cầu
2 2 2
: 2 3 1 9S x y z
qua trục
Ox
phương trình là
A.
2 2 2
2 3 1 9.x y z
B.
2 2 2
2 3 1 9.x y z
C.
2 2 2
2 3 1 9.x y z
D.
2 2 2
2 3 1 9.x y z
Lời giải
n
p
P
R
R
I
J
H
R
R
u
J
I
H
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Bài toán 11. Tìm tiếp điểm
H
là hình chiếu cam
I
trên mt phng
()
:
1. Phương pháp
Viết phương trình đường thng
d
qua
I
vuông
góc mp
()
: ta có
d
un
.
Ta độ
H
là giao điểm ca
d
()
.
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
2. Bài tp minh ha
Bài tp 20. Trong không gian với hệ trục tọa độ
Oxyz
cho
2
:2 2 3 0P x y z m m
mặt cầu
2 2 2
: 1 1 1 9S x y z
. Tìm
m
để mặt phẳng
P
tiếp xúc với mặt cầu
S
.
Với
m
vừa tìm được hãy xác định tọa độ tiếp điểm.
Lời giải.
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Bài tập 21. Trong không gian vi h tọa độ , cho hai điểm mt
phng .Viết phương trình mt cu qua tiếp xúc vi mp ti
đim .
Lời giải
n
p
P
R
H
I(a;b;c)
Oxyz
0;3;2 ,A
1; 1;1B
: 2 2 1 0P x y z
S
A
P
B
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3. Câu hi trc nghim.
Mức độ 4. Vận dụng và Vận dụng cao
u 106.(THPT Chuyên Hùng Vương 2018) Trong không gian với h tọa độ
Oxyz
, cho mặt
phẳng
: 2 6 0P x y z
mặt phẳng
: 2 2 0P x y z
. Xác định tập hợp tâm các mặt
cầu tiếp xúc với
P
và tiếp xúc với
P
.
A. Tập hợp là hai mặt phẳng có phương trình
2 8 0x y z
.
B. Tập hợp là mặt phẳng có phương trình
: 2 8 0P x y z
.
C. Tập hợp là mặt phẳng có phương trình
2 8 0x y z
.
D. Tập hợp là mặt phẳng có phương trình
2 4 0x y z
.
Lời giải
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u 107.(Toán Học Tuổi Trẻ)
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Trong không gian với hệ trục tọa độ
,Oxyz
cho mặt cầu
2 2 2
:0S x y z ax by cz d
bán kính
19,R
đường thẳng
5
: 2 4
14
xt
d y t
zt

và mặt phẳng
:3 3 1 0.P x y z
Trong các số
; ; ;a b c d
theo thứ tự dưới đây, số nào thỏa mãn
43,a b c d
đồng thời tâm
I
của
S
thuộc đường thẳng
d
S
tiếp xúc với mặt phẳng
?P
A.
6; 12; 14;75 .
B.
6;10;20;7 .
C.
10;4;2;47 .
D.
3;5;6;29 .
Lời giải
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u 108.(Sở GD&ĐT Đồng Tháp 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 4 2 4 0S x y z x y
một điểm
1;1;0A
thuộc
S
. Mặt phẳng tiếp xúc với
S
tại
A
có phương trình là
A.
10xy
. B.
10x
. C.
20xy
. D.
10x 
.
Lời giải
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u 109.(THPT Hậu Lộc 2 2018) Trong không gian tọa độ
Oxyz
, cho mặt cầu
S
đường kính
AB
, với
6;2; 5A
,
4;0;7B
. Viết phương trình mặt phẳng
P
tiếp xúc với mặt cầu
S
tại
A
A.
: 5 6 62 0 P x y z
. B.
: 5 6 62 0 P x y z
.
C.
: 5 6 62 0 P x y z
. D.
:5 6 62 0 P x y z
.
Lời giải
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u 110. (THPT Chuyên Vinh 2020)
Trong không gian
Oxyz
, cho mt cu
2 2 2
: 1 2 1 6S x y z
tiếp xúc vi hai mt
phng
: 2 5 0P x y z
,
:2 5 0Q x y z
lần lượt tại các điểm
A
,
B
. Độ dài đoạn
AB
A.
32
. B.
3
. C.
26
. D.
23
.
Lời giải
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u 111.(THPT Hồng Lĩnh 2018) Trong không gian với hệ tọa đ
Oxyz
, cho hai điểm
1;0;0A
,
0;0;2B
mặt cầu
2 2 2
: 2 2 1 0S x y z x y
. Số mặt phẳng chứa hai điểm
A
,
B
tiếp
xúc với mặt cầu
S
A.
1
mặt phẳng. B.
2
mặt phẳng. C.
0
mặt phẳng. D. Vô số mặt phẳng.
Lời giải
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u 112.(THPT Chuyên Nguyn Du 2020)
Trong không gian
Oxyz
, cho các mt phng
:2 4 7 0P x y z
,
:4 5 14 0Q x y z
,
: 2 2 2 0R x y z
: 2 2 4 0S x y z
.
Biết mt cu
2 2 2
x a y b z c D
tâm nm trên
P
Q
, cùng tiếp xúc vi
R
S
. Giá tr
abc
bng
A.
2
. B.
3
. C.
5
. D.
4
.
Lời giải
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331
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 113.(THPT Ngô Sỹ Liên 2019)
Cho hai mặt cầu
2 2 2
1
:6S x y z
2 2 2
2
: 1 1 1 6S x y z
. Biết rằng mặt
phẳng
: 6 0 0P ax by cz a
vuông góc với mặt phẳng
:3 2 1 0Q x y z
đồng thời
tiếp xúc với cả hai mặt cầu đã cho. Tích
abc
bằng
A.
2
. B.
2
. C.
0
. D.
1
.
Lời giải
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u 114. (THPT Chuyên Lam n 2018) Trong không gian với hệ trục tọa độ
Oxyz
cho mặt cầu
2 2 2
: 2 6 4 2 0S x y z x y z
, mặt phẳng
: 4 11 0x y z
. Gọi
P
mặt phẳng
vuông góc với
, P
song song với giá của vecto
1;6;2v
P
tiếp xúc với
S
. Lập
phương trình mặt phẳng
P
.
A.
2 2 2 0x y z
2 21 0x y z
. B.
2 2 3 0x y z
2 21 0x y z
.
C.
2 2 3 0x y z
2 2 21 0x y z
. D.
2 2 5 0x y z
2 2 2 0x y z
.
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 115. Trong không gian với hệ trục tọa độ
Oxyz
, cho ba điểm
;0;0Aa
,
0; ;0Bb
,
0;0;Cc
với
, , 0a b c
. Biết rằng
ABC
đi qua điểm
1 2 3
;;
777
M



và tiếp xúc với mặt cầu
2 2 2
72
: 1 2 3
7
S x y z
. Tính
2 2 2
1 1 1
abc

.
A.
14
. B.
1
7
. C.
7
. D.
7
2
.
Lời giải
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u 116. Trong không gian vi h trc tọa độ
Oxyz
, cho đường thng
2
:
2 1 4
x y z
d

mt
cu
2 2 2
: 1 2 1 2S x y z
. Hai mt phng
P
Q
cha
d
tiếp xúc vi
S
.
Gi
M
,
N
là tiếp điểm. Tính độ dài đoạn thng
MN
.
A.
22
. B.
4
3
. C.
6
. D.
4
.
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 117.(THPT n Tây Nội 2019) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 2 2 1 0S x y z x z
đường thẳng
2
:
1 1 1
x y z
d

. Hai mặt phẳng
P
Q
chứa
d
và tiếp xúc với mặt cầu
S
tại
A
B
. Gọi
;;H a b c
là trung điểm
AB
. Giá trị
abc
A.
1
6
. B.
1
3
. C.
2
3
. D.
5
6
.
Lời giải
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u 118. (THPT Chuyên Lamn 2020)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 2 4 6 2 0S x y z x y z
mặt phẳng
: 4 3 12 10 0x y z
. Lập phương trình mặt phẳng
thỏa mãn đồng thời các điều kiện:
tiếp xúc với
S
; song song với
và cắt trục
Oz
ở điểm có cao độ dương.
A.
4 3 12 78 0x y z
. B.
4 3 12 26 0x y z
.
C.
4 3 12 78 0x y z
. D.
4 3 12 26 0x y z
.
Lời giải
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u 119.(Đề Chính Thức 2018)
Trong không gian
Oxyz,
cho mặt cầu
2 2 2
: 2 3 1 16S x y z
điểm
1; 1; 1 .A 
Xét các điểm
M
thuộc
S
sao cho đường thẳng
AM
tiếp xúc với
.S
M
luôn thuộc một mặt
phẳng cố định có phương trình là
A.
3 4 2 0xy
. B.
3 4 2 0xy
. C.
6 8 11 0.xy
D.
6 8 11 0xy
.
Lời giải
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u 120. (Đề thi THQG 2018)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 1 2 3 1S x y z
điểm
2;3;4A
. Xét
các điểm
M
thuộc
S
sao cho đường thẳng
AM
tiếp xúc với
S
,
M
thuộc mặt phẳng
phương trình là?
A.
70x y z
. B.
2 2 2 15 0x y z
. C.
70x y z
. D.
2 2 2 15 0x y z
.
Lời giải
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Bài toán 12. Tìm bán kính
r
tâm
H
đưng tròn giao tuyến ca mt phng và mt cu:
1. Phương pháp
Viết phương trình đường thng
d
qua
I
vuông
góc mp
()
: ta có
d
un
.
Tọa độ
H
là giao điểm ca
d
()
.
Bán kính
22
r R d
vi
;d IH d I

.
Bán kính
R IA
Phương trình
2 2 2
2
;:S I R x a y b z c R
2. Bài tp minh ha
Bài tp 22. Cho mặt cầu
2 2 2
: 1 1 1 25S x y z
mặt phẳng
phương
trình
2 2 7 0x y z
.
a). Chứng minh rằng mặt phẳng
cắt mặt cầu
S
theo một đường tròn. Xác định tâm và
tìm bán kính của đường tròn đó.
b). Lập phương trình mặt phẳng
P
đi qua hai điểm
1; 1;2 , 3;5; 2AB
P
cắt mặt cầu
S
theo một đường tròn có bán kính nhỏ nhất.
Lời giải.
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M
n
α
α
r
R
H
I (a;b;c)
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Bài tp 23. Lập phương trình mặt cầu
S
đi qua điểm
1; 5;2M
qua đường tròn
C
giao của mặt cầu
2 2 2
' : 2 4 4 40 0S x y z x y z
và mp
: 2 2 9 0x y z
.
Lời giải.
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Bài tp 24. Trong không gian với hệ toạ độ
Oxyz
cho đường thẳn
:2
62
xt
d y t
zt
và mặt cầu
2 2 2
: 2 2 2 1 0S x y z x y z
. Viết phương trình mặt phẳng
P
chứa
d
sao cho giao
tuyến của mặt phẳng
P
và mặt cầu
S
là đường tròn có bán kính
1r
.
Lời giải.
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Bài tp 25. Cho mặt cầu
2 2 2
: 2 4 6 0.S x y z x y z m
Tìm
m
sao cho
a). Mặt cầu tiếp xúc với mặt phẳng
: 2 2 1 0.P x y z
b). Mặt cầu cắt mặt phẳng
:2 2 1 0Q x y z
theo giao tuyến là một đường tròn có diện
tích bằng
4
.
c). Mặt cầu cắt đường thẳng
12
:
1 2 2
x y z

tại hai điểm phân biệt
,AB
sao cho tam giác
IAB
vuông (
I
là tâm mặt cầu).
Lời giải.
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Bài tp 26. Cho đường tròn
C
là giao tuyến của
: 2 2 1 0x y z
và mặt cầu
2 2 2
: 4 6 6 17 0S x y z x y z
a). Xác định tâm và bán kính của đường tròn
C
.
b). Viết phương trình mặt cầu
'S
chứa đường tròn
C
và có tâm nằm trên mặt phẳng
: 3 0P x y z
.
Lời giải.
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Bài tp 27. Trong không gian với hệ tọa độ
Oxyz
, cho
:2 2 14 0P x y z
mặt cầu
S
2 2 2
2 4 2 3 0.x y z x y z
a). Viết phương trình mặt phẳng
Q
chứa trục
Ox
và cắt
S
theo một đường tròn có bán
kính bằng
3
.
b). Tìm tọa độ điểm
M
thuộc mặt cầu
S
sao cho khoảng cách từ
M
đến mặt phẳng
P
lớn
nhất.
Lời giải.
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Bài tp 28. Trong không gian với hệ tọa độ
Oxyz
cho
1;2; 2I
và mp
:2 2 5 0P x y z
a). Lập phương trình mặt cầu
S
tâm
I
sao cho giao của
S
với
mp P
là đường tròn
C
có chu vi bằng
8
.
b). Chứng minh rằng mặt cầu
S
ở câu a tiếp xúc với đường thẳng
:2 2 3x y z
,
c). Lập phương trình mặt phẳng
Q
chứa đường thẳng
và tiếp xúc với
S
.
Lời giải.
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Bài tp 29. Trong không gian
,Oxyz
cho
1; 1;2 ,A
1;3;2 ,B
4;3;2 ,C
4; 1;2D
mặt
phẳng
:P
20x y z
. Gọi
'A
hình chiếu của
A
lên
.Oxy
Gọi
S
mặt cầu đi qua
4
điểm
', , ,A B C D
. Xác định tọa độ tâm và bán kính đường tròn là giao của
P
S
.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài tp 30. Cho
;0;0 , 0; ;0 , 0;0;A a B b C c
với
, , 0abc
1 1 1
2
abc
a).Tìm tâm và bán kính
R
mặt cầu ngoại tiếp tứ diện
OABC
.
Tìm giá trị nhỏ nhất của bán kính
R
.
b). Gọi
r
là bán kính mặt cầu ngoại tiếp tứ diện
OABC
. Chứng minh rằng:
13
4
2 3 1
r
.
Lời giải.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
3. Câu hi trc nghim
u 121. Gọi
S
mặt cầu tâm
1;2;1I
cắt mặt phẳng
: 2 2 2 0P x y z
theo một
đường tròn có bán kính
4r
. Viết phương trình của
S
.
A.
2 2 2
1 2 1 13x y z
. B.
2 2 2
1 2 1 16x y z
.
C.
2 2 2
1 2 1 25x y z
. D.
2 2 2
1 2 1 9x y z
.
Lời giải
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u 122.Đường tròn giao tuyến của mặt cầu
2 2 2
: 3 2 3 25S x y z
khi cắt bởi mặt
phẳng
Oxy
có chu vi bằng
A.
8
. B.
4
. C.
2
. D.
10
.
Lời giải
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u 123.(THPT Gia Bình I Bắc Ninh 2018)
Trong không gian tọa độ
,Oxyz
cho mặt cầu
2 2 2
: 1 2 3 25.S x y z
Mặt phẳng
Oxy
cắt mặt cầu
S
theo một thiết diện là đường tròn
.C
Diện tích của đường tròn
C
A.
8
B.
12
C.
16
D.
4
Lời giải
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u 124. Trong hệ toạ độ
Oxyz
cho
1;1;1I
mặt phẳng
:2 2 4 0P x y z
. Mặt cầu
S
tâm
I
cắt
P
theo một đường tròn bán kính
4r
. Phương trình của
S
A.
2 2 2
1 1 1 16x y z
. B.
2 2 2
1 1 1 5x y z
.
C.
2 2 2
1 1 1 9x y z
. D.
2 2 2
1 1 1 25x y z
.
Lời giải
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u 125.(THPT Đức Th 2018)
Trong hệ tọa độ
Oxyz
cho
1;1;1I
mặt phẳng
P
:
2 2 4 0x y z
. Mặt cầu
S
m
I
cắt
P
theo một đường tròn bán kính
4r
. Phương trình của
S
A.
2 2 2
1 1 1 16x y z
. B.
2 2 2
1 1 1 9x y z
.
C.
2 2 2
1 1 1 5x y z
. D.
2 2 2
1 1 1 25x y z
.
Lời giải
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u 126.(THPT Trần Quốc Tuấn 2018) Trong không gian với hệ trục tọa độ
Oxyz
viết phương
trình mặt cầu
S
tâm
( 2;3;4)I
biết mặt cầu
S
cắt mặt phẳng tọa độ
Oxz
theo một hình
tròn giao tuyến có diện tích bằng
16
.
A.
2 2 2
2 3 4 25x y z
. B.
2 2 2
2 3 4 5 x y z
.
C.
2 2 2
2 3 4 16 x y z
. D.
2 2 2
( 2) ( 3) ( 4) 9 x y z
.
Lời giải
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u 127.(THPT Can Lộc 2018) Cho mặt cầu
2 2 2
: 2 4 1 0S x y z x y mz
. Khẳng định o
sau đây luôn đúng với mọi số thực
m
?
A.
S
luôn tiếp xúc với trục
Oy
. B.
S
luôn tiếp xúc với trục
Ox
.
C.
S
luôn đi qua gốc tọa độ
O
. D.
S
luôn tiếp xúc với trục
Oz
.
Lời giải
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u 128.(Sở GD&ĐT Nội 2018) Trong không gian
Oxyz
, mặt cầu tâm
1;2; 1I
cắt mặt
phẳng
:2 2 1 0P x y z
theo một đường tròn có bán kính bằng
8
có phương trình
A.
2 2 2
1 2 1 9x y z
. B.
2 2 2
1 2 1 9x y z
.
C.
2 2 2
1 2 1 3x y z
. D.
2 2 2
1 2 1 3x y z
.
Lời giải
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u 129.(THPT Chuyên Nguyễn Quang Diệu 2018) Trong không gian với hệ trục toạ độ
Oxyz
, cho
điểm
2;1;3I
mặt phẳng
P
:
2 2 10 0x y z
. Tính bán kính
r
của mặt cầu
S
, biết
rằng
S
có tâm
I
và nó cắt
P
theo một đường tròn
T
có chu vi bằng
10
.
A.
5r
. B.
34r
. C.
5r
. D.
34r
.
Lời giải
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u 130.(THPT Trần Nhân Tông 2018) Trong không gian với hệ trục
Oxyz
, cho mặt cầu
S
tâm
0; 2;1I
mặt phẳng
: 2 2 3 0P x y z
. Biết mặt phẳng
P
cắt mặt cầu
S
theo
giao tuyến là một đường tròn có diện tích là
2
.Viết phương trình mặt cầu
S
.
A.
22
2
: 2 1 3S x y z
. B.
22
2
: 2 1 1S x y z
.
C.
22
2
: 2 1 3S x y z
. D.
22
2
: 2 1 2S x y z
Lời giải.
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u 131.(THPT Phan Đình Phùng 2018) Trong không gian với hệ tọa độ
Oxyz
cho mặt cầu
S
tâm
1;4;2I
và có thể tích bằng
256
3
. Khi đó phương trình mặt cầu
S
A.
2 2 2
1 4 2 16x y z
. B.
2 2 2
1 4 2 4x y z
.
C.
2 2 2
1 4 2 4x y z
. D.
2 2 2
1 4 2 4x y z
.
Lời giải
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u 132.(SGD&ĐT Nội 2018) Trong không gian
Oxyz
, mặt cầu tâm
1;2; 1I
cắt mặt
phẳng
:2 2 1 0P x y z
theo một đường tròn có bán kính bằng
8
có phương trình là
A.
2 2 2
1 2 1 9x y z
. B.
2 2 2
1 2 1 9x y z
.
C.
2 2 2
1 2 1 3x y z
. D.
2 2 2
1 2 1 3x y z
.
Lời giải
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u 133.(Phát triển đề minh họa 2019) Trên hệ toạ độ
Oxyz
cho mặt phẳng
P
phương trình
2x y z
và mặt cầu
S
phương trình
2 2 2
2x y z
. Gọi điểm
;;M a b c
thuộc giao
tuyến giữa
P
S
. Khẳng định nào sau đây là khẳng định đúng?
A.
min 1;1c
. B.
min 1;2b
. C.
max minab
. D.
max 2;2c


.
Lời giải
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u 134. Trong không gian Oxyz cho mặt cầu
S
tâm
1;2;3I
bán kính
3R
hai điểm
2;0;0M
,
0;1;0N
.
:0X x by cz d
mặt phẳng qua
MN
cắt
S
theo giao tuyến
đường tròn có bán kính r lớn nhất. Tính
T b c d
.
A.
1
. B.
4
. C.
2
. D.
3
.
Lời giải
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u 135. Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 6 4 2 5 0S x y z x y z
. Phương trình
mặt phẳng
Q
chứa trục
Ox
và cắt
S
theo giao tuyến là một đường tròn bán kính bằng
2
A.
:2 0Q y z
. B.
:2 0Q x z
. C.
: 2 0Q y z
. D.
:2 0Q y z
.
Lời giải
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u 136.(THPT Chuyên Thái Bình 2018) Trong không gian với hệ trục tọa độ
Oxyz
,
cho mặt cầu
2 2 2
( ): 2 4 6 3 0S x y z x y z m
. Tìm số thực
m
để
:2 2 8 0x y z
cắt
S
theo
một đường tròn có chu vi bằng
8
.
A.
4m 
. B.
2m 
. C.
3m 
. D.
1m 
Lời giải
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u 137.(Sở GD & ĐT Hà Nam 2020) Trong không gian với hệ tọa độ
,Oxyz
cho mặt phẳng
: 2 7 0P x y z
mặt cầu
2 2 2
: 2 4 10 0S x y z x z
. Gọi
Q
mặt phẳng song
song với mặt phẳng
P
và cắt mặt cầu
S
theo giao tuyến là đường tròn có chu vi bằng
6
. Hỏi
Q
đi qua điểm nào trong số các điểm sau?
A.
6;0;1M
. B.
3;1;4N
. C.
2; 1;5J 
. D.
4; 1; 2K 
.
Lời giải
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u 138.(S GD&ĐT Nam Định 2019) Trong không gian
Oxyz
, mt cu tâm
1; 2; 1I
ct mt
phng
:2 2 1 0P x y z
theo một đường tròn có bán kính bng
8
có phương trình là
A.
2
22
1 2 1 9x y z
. B.
2
22
1 2 1 9x y z
.
C.
2
22
1 2 1 3x y z
. D.
2
22
1 2 1 3x y z
.
Lời giải
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u 139.(THPT Thun Thành 2020) Trong không gian vi h ta độ
Oxyz
, cho mt cu
()S
tâm
(2;1;1)I
mt phng
( ):2 2 2 0P x y z
. Biết mt phng
()P
ct mt cu
()S
theo giao
tuyến là một đường tròn có bán kính bng
1
. Viết phương trình của mt cu
()S
.
A.
2 2 2
( ):( 2) ( 1) ( 1) 8S x y z
. B.
2 2 2
( ):( 2) ( 1) ( 1) 10S x y z
.
C.
2 2 2
( ):( 2) ( 1) ( 1) 8S x y z
. D.
2 2 2
( ):( 2) ( 1) ( 1) 10S x y z
.
Lời giải
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u 140.(THPT Trần Đại Nghĩa 2020) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt phẳng
P
:
2 2 7 0x y z
mặt cầu
S
:
2 2 2
2 4 6 11 0x y z x y z
. Mặt phẳng
Q
song
song với
P
và cắt
S
theo một đường tròn có chu vi bằng
6
có phương trình là
A.
:2 2 17 0Q x y z
. B.
:2 2 7 0Q x y z
.
C.
:2 2 19 0Q x y z
. D.
:2 2 17 0Q x y z
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 141.(THPT Toàn Thắng 2020)
Trong không gian với htọa độ
Oxyz
, cho mặt cầu
2 2 2
: 2 4 6 3 0S x y z x y z m
. Tìm
số thực
m
để
:2 2 8 0x y z
cắt
S
theo một đường tròn có chu vi bằng
8
.
A.
3m 
. B.
4m 
. C.
1m 
. D.
2m 
.
Lời giải
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u 142.(Tp Chi Toán Hc 2020)
Trong không gian
Oxyz
, cho mt cu
S
:
2 2 2
2 1 2 4x y z
và mt phng
P
:
4 3 0x y m
. Tìm tt c các giá tr thc ca tham s
m
để mt phng
P
mt cu
S
đúng
1
đim chung.
A.
1m
. B.
1m 
hoc
21m 
.
C.
1m
hoc
21m
. D.
9m 
hoc
31m
.
Lời giải
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u 143.(Tp Chí Tn Hc 2020)
Trong không gian
Oxyz
, cho mt cu
S
:
2 2 2
2 4 1 4xyz
và mt phng
P
:
3 1 0x my z m
. Tìm tt c các giá tr thc ca tham s
m
để mt phng
P
ct mt cu
S
theo giao tuyến là đường tròn có đường kính bng
2
.
A.
1m
. B.
1m 
hoc
2m 
.
C.
1m
hoc
2m
. D.
1m 
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải
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u 144.(THPT Chuyên Hunh Mẫn Đạt 2019)
Trong không gian vi h trc tọa độ
Oxyz
, cho mt phng
: 2 2 1 0P x y z
; hai điểm
1;0;0A
,
1;2;0B
mt cu
22
2
: 1 2 25S x y z
. Viết phương trình mặt phng
vuông góc vi mt phng
P
, song song với đưng thng
AB
, đồng thi ct mt cu
S
theo đường tròn có bán kính bng
22r
.
A.
2 2 3 11 0;2 2 3 23 0x y z x y z
.
B.
2 2 3 11 0;2 2 3 23 0x y z x y z
.
C.
2 2 3 11 0;2 2 3 23 0x y z x y z
.
D
2 2 3 11 0;2 2 3 23 0x y z x y z
.
Lời giải
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u 145.(THPT Chun Lê Hồng Phong 2018) Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 2 2 0P x y z
điểm
1;2; 1I 
. Viết phương trình mặt cầu
S
tâm
I
cắt mặt
phẳng
P
theo giao tuyến là đường tròn có bán kính bằng
5
.
A.
2 2 2
: 1 2 1 25.S x y z
B.
2 2 2
: 1 2 1 16.S x y z
C.
2 2 2
: 1 2 1 34.S x y z
D.
2 2 2
: 1 2 1 34.S x y z
Lời giải.
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u 146.(THPT Nguyễn Trãi-Đà Nẵng 2018) Trong không gian với htọa độ
Oxyz
, cho mặt
phẳng
: 4 4 0P x y z
mặt cầu
2 2 2
: 4 10 4 0S x y z x z
. Mặt phẳng
P
cắt
mặt cầu
S
theo giao tuyến là đường tròn có bán kính bằng
A.
2r
. B.
3r
. C.
7
. D.
5r
.
Lời giải
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u 147.(THPT Xoay 2020)
Trong không gian với hệ trục tọa độ
,Oxyz
mặt phẳng
: 2 3 0P x y z
cắt mặt cầu
2 2 2
:5S x y z
theo giao tuyến là một đường tròn có diện tích là
A.
11
4
. B.
9
4
. C.
15
4
. D.
7
4
.
Lời giải
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u 148.(THPT Chuyên Vĩnh Phúc 2018)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 2 2 4 1 0S x y z x y z
mặt phẳng
:0P x y z m
. Tìm tất cả
m
để
P
cắt
S
theo giao tuyến là một đường tròn có bán kính
lớn nhất.
A.
4m 
. B.
0m
. C.
4m
. D.
7m
.
Lời giải
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u 149.(THPT Lê Hồng Phong 2020)
Trong không gian
Oxyz
cho mặt cầu
2 2 2
: 2 2 4 3 0S x y z x y z
mặt phẳng
:2 2 0P x y z
. Mặt phẳng
P
cắt khối cầu
S
theo thiết diện một hình tròn. Tính diện
của hình tròn đó.
A.
5
. B.
25
. C.
25
. D.
10
.
Lời giải
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u 150.(THPT Chuyên Vĩnh Phúc 2018)
Trong không gian
Oxyz
, cho mặt cầu
2 2 2
: 2 2 4 1 0S x y z x y z
mặt phẳng
:0P x y z m
. Tìm tất cả
m
để
P
cắt
S
theo giao tuyến là một đường tròn có bán kính
lớn nhất.
A.
4m 
. B.
0m
. C.
4m
. D.
7m
.
Lời giải
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u 151.(THPT Chuyên Hoàng Văn Thụ 2018) Trong không gian
Oxyz
, cho điểm
1;0; 1A
, mặt
phẳng
: 3 0P x y z
. Mặt cầu
S
tâm
I
nằm trên mặt phẳng
P
, đi qua điểm
A
gốc tọa độ
O
sao cho chu vi tam giác
OIA
bằng
62
. Phương trình mặt cầu
S
A.
2 2 2
2 2 1 9x y z
2 2 2
1 2 2 9x y z
.
B.
2 2 2
3 3 3 9x y z
2 2 2
1 1 1 9x y z
.
C.
2 2 2
2 2 1 9x y z
2
22
39x y z
.
D.
2 2 2
1 2 2 9x y z
2 2 2
2 2 1 9x y z
.
Lời giải
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u 152.(THPT Hậu Lộc 2 2018)
Trong không gian tọa độ
Oxyz
, cho mặt cầu
2 2 2
: 2 4 4 16 0S x y z x y z
mặt phẳng
: 2 2 2 0P x y z
. Mặt phẳng
P
cắt mặt cầu
S
theo giao tuyến là một đường tròn có bán
kính là:
A.
6r
. B.
22r
. C.
4r
. D.
23r
.
Lời giải
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u 153. (THTT số 6-489 tháng 3 năm 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt
phẳng
:2 2 0P x y z m
mặt cầu
2 2 2
: 2 4 6 2 0S x y z x y z
. bao nhiêu giá
trị nguyên của
m
để mặt phẳng
P
cắt mặt cầu
S
theo giao tuyến là đường tròn
T
có chu vi
bằng
43
.
A.
3
. B.
4
. C.
2
. D.
1
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải
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u 154.(THPT Chuyên Lam Sơn 2018) Trong không gian với hệ tọa đ
Oxyz
, cho mặt cầu
2 2 2
: 1 2 3 16S x y z
các điểm
1;0;2A
,
1;2;2B
. Gọi
P
mặt phẳng đi
qua hai điểm
A
,
B
sao cho thiết diện của
P
với mặt cầu
S
diện tích nhỏ nhất. Khi viết
phương trình
P
dưới dạng
: 3 0P ax by cz
. Tính
T a b c
.
A.
3
. B.
3
. C.
0
. D.
2
.
Lời giải
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u 155.(Tạp Chí Toán Học 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho mặt phẳng
:2 2 0P x y z m
mặt cầu
2 2 2
: 2 4 6 2 0S x y z x y z
. bao nhiêu giá trị
nguyên của
m
để mặt phẳng
P
cắt mặt cầu
S
theo giao tuyến đường tròn
T
chu vi
bằng
43
.
A.
3
. B.
4
. C.
2
. D.
1
.
Lời giải
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Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
353
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 156.(THPT Kinh Môn 2018) Trong không gian
Oxyz
cho các mặt phẳng
: 2 1 0P x y z
,
:2 1 0Q x y z
. Gọi
S
mặt cầu tâm thuộc trục hoành, đồng thời
S
cắt mặt phẳng
P
theo giao tuyến một đường tròn bán kính bằng
2
S
cắt mặt phẳng
Q
theo giao
tuyến một đường tròn n kính bằng
r
. Xác định
r
sao cho chỉ đúng một mặt cầu
S
thỏa yêu cầu.
A.
3r
. B.
3
2
r
. C.
2r
. D.
32
2
r
.
Lời giải
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u 157.(Tạp Chí Tn Học 2020) Trong không gian với hệ tọa độ
,Oxyz
cho mặt phẳng
: 2 5 0Q x y z
mặt cầu
22
2
: 1 2 15.S x y z
Mặt phẳng
P
song song với
mặt phẳng
Q
cắt mặt cầu
S
theo giao tuyến đường tròn chu vi
6
đi qua điểm nào
sau đây?
A.
0; 1; 5A 
B.
1; 2; 0B
C.
2; 2; 1C
D.
2; 2; 1D 
Lời giải
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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Bài toán 13. Tp hợp điểm và bài toán tiếp tuyến
1. Bài tp minh ha
Bài tp 31. Cho các điểm
2;3;1 , 5; 2;7 , 1;8; 1A B C
. Tìm tập hợp các điểm
M
trong
không gian thỏa mãn
a).
2 2 2
MA MB MC
b).
AM AB BM CM
Lời giải.
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Bài tp 32. Trong không gian với hệ toạ độ Đề-các vuông góc
Oxyz
cho hai mặt phẳng song
song các phương trình tương ứng là:
1
:2 2 1 0P x y z
;
2
:2 2 5 0P x y z
điểm
1;1;1A
nằm trong khoảng giữa hai mặt phẳng đó. Gọi
S
là mặt cầu bất kỳ qua
A
và tiếp
xúc với cả hai mặt phẳng
12
,.PP
a). Chứng tỏ rằng bán kính của hình cầu
S
là một hằng số và tính bán kính đó.
b). Gọi
I
là tâm của hình cầu
.S
Chứng tỏ rằng
I
thuộc một đường tròn cố định.
Xác định toạ độ của tâm và tính bán kính của đường tròn đó.
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Lời giải.
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Bài tập 33. Cho 4 điểm
1;2;1 ; 2;0; 1 ; 1;3; 4 ; 0; 2;2A B C D
. Chứng minh rằng tập hợp
các điểm
M
sao cho
2 2 2 2
4MA MB MC MD
là một mặt cầu. Viết phương trình mặt cầu đó.
Lời giải
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2. Câu hi trc nghim
u 158.(THPT Chuyên Quý Đôn 2018) Trong không gian với hệ trục tọa độ
Oxyz
, cho ba điểm
1;0;0A
,
0;0;3C
,
0;2;0B
. Tập hợp các điểm
M
thỏa mãn
2 2 2
MA MB MC
mặt cầu
bán kính là:
A.
2R
. B.
3R
. C.
3R
. D.
2R
.
Lời giải
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356
Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 159. Trong không gian
Oxyz
, cho mặt cầu
22
2
: 1 4 9S x y z
. Từ điểm
4;0;1A
nằm ngoài mặt cầu, kẻ một tiếp tuyến bất kỳ đến
S
với tiếp điểm
M
. Tập hợp
M
đường
tròn có bán kính bằng:
A.
3
2
. B.
32
2
. C.
33
2
. D.
5
2
.
Lời giải
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u 160.Trong Trong không gian với hệ tọa độ
Oxyz
cho điểm
1;0;1 , 2; 1;0 , 0; 3; 1A B C
.
Tìm tập hợp các điểm
M
thỏa mãn
2 2 2
AM BM CM
.
A. Mặt cầu
2 2 2
2 8 4 13 0x y z x y z
. B. Mặt cầu
2 2 2
2 4 8 13 0.x y z x y z
C. Mặt cầu
2 2 2
2 8 4 13 0x y z x y z
. D. Mặt phẳng
2 8 4 13 0x y z
.
Lời giải
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u 161. Trong không gian to độ
Oxyz
, cho điểm
1; 1;3A
và hai điểm
,MB
tho mãn
4 . . 0MA MA MB MB
. Gi s đim
M
thay đổi trên mt cu
2 2 2
1 1 3 4x y z
. Khi
đó điểm
B
thay đổi trên mt mt cầu có phương trình là:
A.
2 2 2
1
: 1 1 3 4S x y z
. B.
2 2 2
2
: 1 1 3 8S x y z
.
C.
2 2 2
3
: 2 4 6 4S x y z
. D.
2 2 2
4
: 2 4 6 8S x y z
.
Lời giải
Trung Tâm Luyện Thi Đại Học Amsterdam Chương III-Bài 4. Phương Trình Mặt Cầu
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
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u 162. Trong không gian
,Oxyz
cho mt cu
2
22
: 2 3.S x y z
bao nhiêu điểm
;;A a b c
(
,,abc
các s nguyên) thuc mt phng
Oxy
sao cho có ít nht hai tiếp tuyến ca
S
đi qua
A
và hai tiếp tuyến đó vuông góc với nhau.
A.
12.
B.
8.
C.
16.
D.
4.
Lời giải
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u 163. Bán kính mt cu tâm
1;3;5I
tiếp xúc với đường thng
:1
2
xt
d y t
zt

là:
A.
14
. B.7. C.14. D.
7
.
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Lời giải
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u 164.(Chuyên Phan Bi Châu 2019) Trong không gian với hệ tọa độ
Oxyz
, cho mặt cầu
S
:
22
22
3 2 4x y z m
. Tập các giá trị của
m
để mặt cầu
S
tiếp xúc với mặt phẳng
Oyz
là:
A.
5
. B.
5
. C.
0
. D. .
Lời giải
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Câu 165.(Chuyên KHTN 2019) Trong không gian
Oxyz
, cho hai điểm
3;1; 3A
,
0; 2;3B
mặt cầu
22
2
: 1 3 1S x y z
. Xét điểm
M
thay đổi thuộc mặt cầu
S
, gtrị lớn nhất
của
22
2MA MB
bằng
A.
102
. B.
78
. C.
84
. D.
52
.
Lời giải
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u 166.(THPT Gia Bình 2018) Cho mặt cầu
22
2
: 1 4 8S x y z
các điểm
3;0;0A
,
4;2;1B
. Gọi
M
một điểm bất kỳ thuộc mặt cầu
S
. Tìm giá trị nhỏ nhất của biểu thức
2M A MB
?
A.
22
. B.
42
. C.
32
. D.
62
.
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Lớp Toán Thầy-Diệp Tuân Tel: 0935.660.880
Lời giải
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u 167. (THPT Gia Bình 2018)
Trong không gian với hệ toạ độ
Oxyz
, cho mặt cầu
2 2 2
: 2 2 2 0S x y z x y z
điểm
2;2;0A
. Viết phương trình mặt phẳng
OAB
, biết rằng điểm
B
thuộc mặt cầu
S
, hoành
độ dương và tam giác
OAB
đều.
A.
0x y z
. B.
0x y z
. C.
20x y z
. D.
20x y z
.
Lời giải
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u 168. Trong không gian với hệ tọa độ
Oxyz
, cho mặt phẳng
: 2 2 3 0P x y z
và mặt cầu
S
m
5; 3;5I
, bán kính
25R
. Từ một điểm
A
thuộc mặt phẳng
P
kẻ một đường
thẳng tiếp xúc với mặt cầu
S
tại
B
. Tính
OA
biết
4AB
.
A.
11OA
. B.
5OA
. C.
3OA
. D.
6OA
.
Lời giải
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u 169. Trong không gian với hệ trục tọa độ
Oxyz
, cho hai mặt phẳng song song
1
:2 2 1 0x y z
,
2
:2 2 5 0x y z
một điểm
1;1;1A
nằm trong khoảng giữa
của hai mặt phẳng đó. Gọi
S
mặt cầu đi qua A tiếp xúc với
12
,

. Biết rằng khi
S
thay đổi thì tâm
I
của nó nằm trên một đường tròn cố định
. Tính diện tích hình tròn giới hạn
bởi
.
A.
2
3
. B.
4
9
. C.
8
9
. D.
16
9
.
Lời giải
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u 170.(THPT Bình Minh 2018) Cho mt cu
1
S
tâm
1
3;2;2I
bán kính
1
2R
, mt cu
2
S
tâm
2
1;0;1I
bán kính
2
1R
. Phương trình mặt phng
P
đồng thi tiếp xúc vi
1
S
2
S
và cắt đoạn
12
II
có dng
20 x by cz d
. Tính
T b c d
.
A.
5
. B.
1
. C.
3
. D.
2
.
Lời giải
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u 171. Trong không gian vi h trc tọa độ
Oxyz
, cho mt cu
1
S
có tâm
2;1;1I
và bán kính
bng
4
, cho mt cu
2
S
có tâm
2;1;5J
và bán kính bng
2
. Gi
P
là mt phng tiếp xúc vi
hai mt cu
12
;SS
. Đặt
,Mm
lần lượt giá tr ln nht giá tr nh nht ca khong cách
t
P
đến
P
. Giá tr
Mm
bng
A.
83
. B.
8
. C.
9
. D.
15
.
Lời giải
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A . L THUY󰈸T
I. PHƯƠNG PHÁP CHUNG.
Bước 1: Chọn hệ trục tọa
.Oxyz
Xác định ba đường thẳng đồng quy đôi một cắt nhau trên sở sẵn của hình (như tam
diện vuông, hình hộp chữ nhật, hình chóp tứ giác đều …),
Hoặc dựa trên các mặt phẳng vuông góc dựng thêm đường phụ.
Bước 2: Tọa độ hóa các điểm của hình không gian.
Tính tọa độ điểm liên quan trực tiếp đến giả thiết và kết luận của bài toán.
Cơ sở tính toán chủ yếu dựa vào quan hệ song song, vuông góc cùng các dữ liệu của bài toán.
Bước 3: Chuyển giả thiết qua hình học giải tích.
Lập các phương trình đường, mặt liên quan.
Xác định tọa độ các điểm, véc tơ cần thiết cho kết luận.
Bước 4: Giải quyết bài toán.
Sử dụng các kiến thức hình học giải tích để giải quyết yêu cầu của bài toán hình không gian.
Chú ý các công thức về góc, khoảng cách, diện tích và thể tích …
II. MỘT SỐ CÁCH CHỌN HỆ TỌA ĐỘ CÁCNH KHÔNG GIAN
Hình hộp lập phương – nh hộp chữ nhật
. ' ' ' 'ABCD A B C D
1. Phương pháp.
Với hình lập phương
Chọn h trục tọa độ sao cho :
0;0;0 , ;0;0 , ; ;0 , 0; ;0A B a C a a D a
' 0;0; , ;0; , ' ; ; , ' 0; ;A a B a a C a a a D a a
Với hình hộp chữ nhật.
Chọn hệ trục tọa độ sao cho :
0;0;0 , ;0;0 , ; ;0 , 0; ;0A B a C a b D b
' 0;0; ; ' ;0; ; ' ; ; ; ' 0; ;A c B a c C a b c D b c
Chú ý: Tam diện vuông là một nửa của hình hộp chữ nhật nên ta chọn hệ trục tọa độ tương tự
như hình hộp chữ nhật.
2. Ví d minh ha .
dụ 1. Cho hình lập phương
. ' ' ' 'ABCD A B C D
cạnh bằng
a
. Chứng minh hai đường chéo
''BD
'AB
của hai mặt bên hai đường thẳng chéo nhau. Tìm khoảng cách giữa hai đường
thẳng chéo nhau
''BD
'AB
.
Lời giải
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§BI 5. NG DNG PHƯƠNG PHÁP TA ĐỘ
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Ví dụ 2. Trong không gian với hệ tọa độ
Oxyz
, cho hình hộp chữ nhật
. ' ' ' 'ABCD A B C D
, , , 'A O B Ox D Oy A Oz
1,AB
2,AD
'3AA
.
1). Tìm tọa độ các đỉnh của hình hộp;
2). Tìm điểm
E
trên đường thẳng
'DD
sao cho
''B E A C
3). Tìm điểm
M
thuộc
'AC
,
N
thuộc
BD
sao cho
,'MN BD MN A C
.
Từ đó tính khoảng cách giữa hai đường thẳng chéo nhau
'AC
BD
Lời giải.
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Ví dụ 3. Cho lăng trụ
. ' ' 'ABC A B C
các mặt bên đều hình vuông cạnh
a
. Gọi
,DF
lần lượt
trung điểm của các cạnh
, ' 'BC B C
. Tính khoảng cách giữa
2
đường thẳng
'AB
''BC
.
Lời giải
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 1.(Chuyên Phan Bi Châu 2019) Cho hình lập phương
.ABCD A B C D
cạnh bằng
2
.
Khoảng cách giữa hai mặt phẳng
AB D
BC D
bằng
A.
3
3
. B.
23
3
. C.
3
2
. D.
3
.
Lời giải
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u 2.(THPT Chuyên Quý Đôn 2018) Cho hình lập phương
.ABCD A B C D
cạnh bằng
a
. Gọi
K
là trung điểm
DD
. Tính khoảng cách giữa hai đường thẳng
CK
AD
.
A.
4
3
a
. B.
3
a
. C.
2
3
a
. D.
3
4
a
.
Lời giải
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u 3.(THPT Thăng Long 2018) Cho hình lập phương
.ABCD A B C D
độ dài cạnh bằng
1
. Gọi
M
,
N
,
P
,
Q
lần lượt trung điểm của các cạnh
AB
,
BC
,
CD

DD
. Tính thể tích khối t
diện
MNPQ
.
A.
3
8
. B.
1
8
. C.
1
12
. D.
1
24
.
Lời giải
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u 4.(THPT Chuyên Hạ Long 2018) Cho hình lập phương
.ABCD EFGH
cạnh bằng
a
. Khoảng
cách giữa hai đường thẳng
AH
BD
bằng
A.
3
6
a
. B.
3
4
a
. C.
3
3
a
. D.
2
3
a
.
Lời giải
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u 5. (Đề BGD Chính Thức 2018) Cho hình lập phương
.ABCD A B C D
tâm
O
. Gọi
I
tâm
của hình vuông
A B C D
điểm
M
thuộc đoạn
OI
sao cho
2MO MI
(tham khảo hình vẽ).
Khi đó sin của góc tạo bởi hai mặt phẳng
MC D

MAB
bằng
A.
6 13
65
. B.
7 85
85
. C.
17 13
65
. D.
6 85
85
.
Lời giải
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u 6.(THPT Chuyên Quốc Học 2018) Một khối đa diện
H
được tạo thành bằng cách từ một khối
lập phương cạnh bằng
3
, ta bỏ đi khối lập phương cạnh bằng
1
ở một “góc” của nó như hình vẽ.
Gọi
S
khối cầu thể tích lớn nhất chứa trong
H
tiếp xúc với các mặt phẳng
A B C D
,
BCC B


DCC D
. Tính bán kính của
S
.
A.
23
3
. B.
33
. C.
23
3
. D.
2
.
Lời giải
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u 7.(THPT Hồng Bàng 2018) Cho lăng trụ tam giác đều
.ABC A B C
cạnh đáy
AB a
, cạnh
bên
2
2
a
AA
. Khoảng cách giữa hai đường thẳng
BC
CA
bằng
A.
6
6
a
. B.
6
24
a
. C.
6
12
a
. D.
6
3
a
.
Lời giải
D
B
D'
A'
B'
C'
C
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u 8.(THPT Chuyên Ti Bình 2018) Cho hình lăng trụ đứng
.ABC A B C
AB AC a
, góc
120BAC 
,
AA a
. Gọi
M
,
N
lần lượt là trung điểm của
BC

CC
.
Số đo góc giữa mặt phẳng
AMN
và mặt phẳng
ABC
bằng
A.
60
. B.
30
. C.
3
arcsin
4
. D.
3
arccos
4
.
Lời giải
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u 9.(THPT Bc Ninh 2020) Cho hình lăng trụ đứng
.ABC A B C
đáy
ABC
tam giác vuông
cân,
AB AC a
,
AA h
,0ah
. Tính khoảng cách giữa hai đường thẳng chéo nhau
AB
BC
theo
a
,
h
.
A.
22
5
ah
ah
. B.
22
5
ah
ah
. C.
22
2
ah
ah
. D.
22
ah
ah
.
Lời giải
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u 10.(Tp Chí Toán Hc 2020) Cho hình lăng trụ đứng
.ABC A B C
đáy
ABC
tam giác cân
tại
C
,
2AB a
,
AA a
, góc giữa
BC
ABB A

bằng
60
. Gọi
N
trung điểm
AA
và
M
trung điểm
BB
. Tính khoảng cách từ điểm
M
đến mặt phẳng
BC N
.
A.
2 74
37
a
. B.
74
37
a
. C.
2 37
37
a
. D.
37
37
a
.
Lời giải
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u 11. (THPT Chuyên Lam Sơn 2018) Cho hình lập phương
1a
cạnh bằng
1a
. Một đường
thẳng
d
đi qua đỉnh
D
tâm
I
của mặt bên
BCC B

. Hai điểm
M
,
N
thay đổi lần lượt thuộc
các mặt phẳng
BCC B

ABCD
sao cho trung điểm
K
của
MN
thuộc đường thẳng
d
(tham khảo hình vẽ). Giá trị bé nhất của độ dài đoạn thẳng
MN
A.
1a
. B.
1a
. C.
1a
. D.
1a
.
Lời giải
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u 12.(THPT Chuyên Tĩnh 2018) Cho hình lăng trụ
.ABC A B C
.A ABC
tứ diện đều
cạnh
a
. Gọi
M
,
N
lần lượt trung điểm của
AA
BB
. Tính tan của góc giữa hai mặt phẳng
ABC
CMN
.
A.
2
5
. B.
32
4
. C.
22
5
. D.
42
13
.
Lời giải
A
A
B
B
C
C
D
D
N
K
M
d
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u 13.(THPT Thanh Miện 2018) Cho hình lập phương
.ABCD A B C D
cnh bng
a
. Lấy điểm
M
thuộc đon
AD
, điểm
N
thuộc đoạn
BD
sao cho
AM DN x
,
2
0
2
a
x





. Tìm
x
theo
a
để đon
MN
ngn nht.
A.
2
3
a
x
. B.
2
4
a
x
. C.
3
a
x
. D.
2
a
x
.
Lời giải
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Với hình hộp đứng có đáy là hình thoi
.ABCD A B C D
1. Phương pháp.
Chọn htrục tọa đ sao cho :
Gốc tọa độ trùng với giao điểm
O
của hai
đường chéo của hình thoi
ABCD
.
Trục
Oz
đi qua 2 tâm của 2 đáy.
Nếu
, , 'AC a BD b AA c
thì
0; ;0 , ;0;0 ,
22
ab
AB
0; ;0
2
a
C



,
;0;0 ,
2
b
D



' 0; ; , ' ;0;
22
ab
A c B c
' 0; ; , ' ;0;
22
ab
C c D c
.
Chú ý: Với lăng trụ đứng
. ' ' 'ABC A B C
có đáy
ABC
là tam giác cân tại
B
thì ta chọn hệ tọa độ
tương tự như trên với gốc tọa độ là trung điểm
AC
,
,B Ox C Oy
còn trục
Oz
đi qua trung
điểm hai cạnh
, ' 'AC A C
.
2. Ví d minh ha .
d4. (THPT Kinh Môn 2019) Cho hình hộp đứng
.ABCD A B C D
đáy hình thoi, tam
giác
ABD
đều. Gi
,MN
lần lượt là trung điểm
BC
CD

, biết rng
MN B D
. Gi
là góc
to bởi đường thng
MN
và mặt đáy
ABCD
, khi đó giá trị
cos
bng
A.
1
cos
3
. B.
3
cos
2
. C.
1
cos
10
. D.
1
cos
2
.
Lời giải
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Hình cp
.S ABCD
,SA ABCD SA h
1. Phương pháp.
Nếu đáy là hình chữ nhật ta chọn hệ trục sao cho
, , ,A O B Ox D Oy S Oz
Nếu đáy là hình thoi, ta chọn hệ trục sao cho
O
là tâm của đáy,
,B Ox C Oy
Oz SA
.
Chú ý: Cho hình chóp
.S ABC
SA ABC
Nếu đáy
ABC
là tam giác vuông tại
A
thì cách chọn hệ trục hoàn toàn tương tự như hình
chóp
.S ABCD
có đáy là hình chữ nhật.
Nếu đáy
ABC
là tam giác cân tại
B
thì ta chọn hệ trục tọa độ như hình chóp
.S ABCD
đáy là hình thoi, khi đó gốc tọa độ là trung điểm cạnh
AC
.
2. Ví d minh ha .
Ví dụ 5. Cho hình tứ diện
ABCD
có cạnh
AD
vuông góc với mặt phẳng
ABC
biết
3AB cm
4AC AD cm
5BC cm
.
1). Tính khoảng cách từ điểm
A
đến mặt phẳng
BCD
.
2). Gọi
,MN
lần lượt là trung điểm các cạnh
,BD BC
.
Tính góc và khoảng cách giữa hai đường thẳng
CM
AN
.
Lời giải.
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Ví dụ 6. Trên các tia
,,Ox Oy Oz
của góc tam diện vuông
Oxyz
lần lượt lấy các điểm
,,A B C
sao
cho
, 2, ,OA a OB a OC c
( , 0)ac
.Gọi
D
đỉnh đối diện với
O
của hình chữ nhật
AOBD
M
trung điểm của đoạn
.BC
Mặt phẳng
()
qua
,AM
cắt mặt phẳng
()OCD
theo một
đường thẳng vuông góc với đường thẳng
.AM
1). Gọi
E
là giao điểm của
()
với đường thẳng
.OC
Tính độ dài đoạn thẳng
OE
;
2). Tính tỷ số thể tích của hai khối đa diện được tạo thành khi cắt khối chóp
.C AOBD
bởi mặt
phẳng
()
. Tính khoảng cách từ điểm
C
đến mặt phẳng
()
Lời giải.
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Ví dụ 7. Cho hình chóp
.O ABC
,,OA a OB b OC c
đôi một vuông c. Điểm
M
cố định
thuộc tam giác
ABC
có khoảng cách lần lượt đến các
mp OBC
,
mp OCA
,
mp OAB
1, 2, 3
.
Tính
, , a b c
để thể tích
.O ABC
nhỏ nhất.
Lời giải.
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3. Câu hi trc nghim.
Mức độ 3. Vận dụng
u 14.(Kì Thi THQG 2018) Cho tứ diện
OABC
,,OA OB OC
đôi một vuông góc với nhau,
;2OA OB a OC a
. Gọi
M
trung điểm của
AB
. Khoảng cách giữa hai đường thẳng
OM
AC
bằng.
A.
2
3
a
. B.
25
5
a
. C.
2
2
a
. D.
2
3
a
.
Lời giải
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u 15. Cho hình chóp
.S ABCD
có đáy
ABCD
là hình vuông cạnh
a
,
SA a
SA
vuông góc với
đáy. Gọi
M
trung điểm
SB
N
điểm thuộc cạnh
SD
sao cho
2SN ND
. Tính thể tích
khối tứ diện
ACMN
.
A.
3
1
12
Va
. B.
3
1
8
Va
. C.
3
1
6
Va
. D.
3
1
36
Va
.
Lời giải
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u 16.(THPT Chuyên Lê Hồng Phong 2019) Cho hình chóp
.S ABCD
có đáy là hình vuông cạnh
a
,
cạnh bên
2SA a
vuông góc với đáy. Gọi
M
là trung điểm cạnh
SD
.
Tính
cos
của góc tạo bởi hai mặt phẳng
AMC
SBC
.
A.
5
5
. B.
5
3
. C.
3
2
. D.
2
3
.
Lời giải
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u 17.(THPT Chuyên Lào Cai 2018) Cho hình chóp
.S ABCD
đáy
ABCD
hình vuông cạnh
bằng
a
, biết
SO a
SO
vuông góc với mặt đáy
ABCD
. Gọi
,MN
là trung điểm của
,SA BC
.
Gọi
là góc giữa đường thẳng
MN
và mặt phẳng
SBD
. Tính
cos
.
A.
2
7
. B.
21
7
. C.
5
10
. D.
2
5
.
Lời giải
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u 18.(THPT Chuyên Hạ Long 2018) Cho hình chóp
.S ABCD
đáy
ABCD
hình vuông cạnh
a
, cạnh bên
SA
vuông góc với mặt phẳng đáy,
2SA a
. Gọi
M
,
N
lần lượt là hình chiếu vuông
góc của điểm
A
trên các cạnh
SB
,
SD
. Góc giữa mặt phẳng
AMN
và đường thẳng
SB
bằng
A.
o
45
. B.
o
90
. C.
o
120
. D.
o
60
.
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Lời giải
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u 19.(THPT Chuyên Tiền Giang 2018) Cho hình chóp
.S ABCD
có đáy là hình thang vuông tại
A
B
,
AB BC a
,
2AD a
,
SA
vuông góc với mặt đáy
ABCD
,
SA a
. Gọi
M
,
N
lần lượt
trung điểm của
SB
,
CD
. Tính cosin của góc giữa
MN
SAC
.
A.
2
5
. B.
55
10
. C.
35
10
. D.
1
5
.
Lời giải
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u 20.(SGD&ĐT Bắc Giang 2018) Cho hình chóp
.S ABCD
đáy
ABCD
hình chữ nhật,
AB a
,
3BC a
,
SA a
SA
vuông góc với đáy
ABCD
. Tính
sin
, với
là góc tạo bởi giữa
đường thẳng
BD
và mặt phẳng
SBC
.
A.
7
sin
8
. B.
3
sin
2
. C.
2
sin
4
. D.
3
sin
5
.
Lời giải
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u 21.(THPT Trần Phú 2018) Cho hình chóp
.S ABCD
đáy
ABCD
hình vuông độ dài
đường chéo bằng
2a
SA
vuông góc với mặt phẳng
ABCD
. Gọi
góc giữa hai mặt
phẳng
SBD
ABCD
. Nếu
tan 2
thì góc giữa hai mặt phẳng
SAC
SBC
bằng
A.
30
. B.
60
. C.
45
. D.
90
.
Lời giải
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u 22.(THPT Nguyn Khuyến 2020) Cho hình chóp
.S ABCD
có đáy
ABCD
là hình vuông cnh
a
,
cnh bên
2SA a
vuông góc vi mt phẳng đáy. Gọi
M
trung điểm cnh
SD
. Tang ca góc
to bi hai mt phng
()AMC
()SBC
bng
A.
3
2
. B.
23
3
. C.
5
5
. D.
25
5
Lời giải
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u 23.
(THPT S1 Nghĩa 2019) Cho tứ diện
SABC
SA
vuông góc với mặt phẳng
ABC
,
3,SA AB cm
5BC cm
diện ch tam giác
SAC
bằng
2
6cm
. Một mặt phẳng
thay đổi
qua trọng tâm
G
của tứ diện cắt các cạnh
,,SA AB AC
lần lượt tại
,,M N P
. Tính giá trị nhỏ nhất
m
T
của biểu thức
2 2 2
1 1 1
T
AM AN AP
.
A.
8
17
m
T
. B.
41
144
m
T
. C.
1
10
m
T
. D.
1
34
m
T
.
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Lời giải
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Hình cp đều
1. Phương pháp.
Hình chóp tam giác đều
.S ABC
,
,AB a
SH h
,
Ta chọn
Hệ tọa độ sao cho
O
là trung điểm
BC
.
,A Ox B Oy
. Khi đó
3
;0;0 , 0; ;0 ,
22
3
0; ;0 , ;0;
26
aa
AB
aa
C S h










Hình chóp từ giác đều
.S ABCD
,
,AB a
SH h
Ta chọn hệ tọa độ sao cho
O
là tâm đáy
,,B Ox C Oy S Oz
.
Khi đó
22
0; ;0 , ;0;0 ,
22
aa
AB
2
0; ;0
2
a
C




2
;0;0 , 0;0;
2
a
D S h




.
Chú ý: Ngoài cách chọn hệ trục như trên ta có thể chọn hệ trục bằng cách khác.
Chẳng hạn với hình chóp tam giác đều ta có thể chọn
HO
.
Trục
Oy
đi qua
H
và song song với
BC
.
2. Ví d minh ha.
d8.(THPT Lương Thế Vinh 2019) Cho hình chóp
.S ABCD
đáy
ABCD
hình chữ nhật,
,AB a
3,BC a
SA a
SA
vuông góc với đáy
ABCD
. Tính
sin
với
góc tạo bởi
đường thẳng
BD
và mặt phẳng
()SBC
.
A.
2
sin
4
. B.
7
sin
8
. C.
3
sin
5
. D.
3
sin
2
.
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Lời giải
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3. Câu hi trc nghim.
u 24. (THPT Chuyên Sơn La 2019) Cho hình chóp t giác đều
.S ABCD
đáy
ABCD
hình
vuông cnh
a
, tâm
O
. Gi
M
N
lần lượt trung điểm ca hai cnh
SA
BC
, biết
6
2
a
MN
. Khi đó giá trị sin ca góc giữa đường thng
MN
và mt phng
SBD
bng
A.
2
5
. B.
3
3
. C.
5
5
. D.
3
.
Lời giải
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Hình cp
.S ABC
SAB ABC
1. Phương pháp.
Đường cao
SH h
của
SAB
là đường cao của hình chóp.
Nếu tam giác
ABC
vuông tại
A
,
,AB a AC b
ta
chọn hệ trục sao cho
Gc tọa độ
, , ,A O B Oy C Ox
//Oz SH
.
Khi đó
0;0;0 , 0; ;0 , ;0;0A B a C b
,
AH c
.
0; ;0 , 0; ;H c S c h
Chú ý:
Nếu vuông tại
B
ta chọn
BO
, vuông tại
C
chọn
CO
.
Nếu tam giác
ASB
cân tại
S
,
ABC
cân tại
C
thì ta chọn
, , ,H O C Ox B Oy S Oz
2. Ví d minh ha.
Ví dụ 9. Cho hình chóp
.S ABC
đáy
ABC
tam giác vuông cân tại
,2B AB BC a
; hai mặt
phẳng
SAB
SAC
cùng vuông góc với mặt phẳng
ABC
. Gọi
M
trung điểm của
AB
;
mặt phẳng
SM
và song song với
BC
, cắt
AC
tại
N
. Biết góc giữa hai mặt phẳng
SBC
ABC
bẳng 60
o
.
1). Tính thể tích khối chóp
.S BCNM
.
2). Khoảng cách giữa hai đường thẳng
AB
SN
theo a
Đề thi ĐH khối A 2011
Lời giải.
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dụ 10. Cho hình chóp
.S ABCD
đáy
ABCD
hình vuông cạnh
2a
,
SA a
,
3SB a
mặt phẳng
()SAB
vuông góc với mặt phẳng đáy. Gọi
,MN
lần lượt trung điểm của các cạnh
,AB BC
. Tính theo
a
thể tích của khối chóp
.S BMDN
và tính cosin của góc giữa hai đường thẳng
, SM DN
Lời giải.
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