2018秋新版高中数学人教A版选修2-1习题：第三章空间向量与立体几何 3.1.5

3.1.5　空间向量运算的坐标表示
课时过关·能力提升基础巩固
1已知向量a =(3,-2,1),b =(-2,4,0),则4a +2b 等于( )
A.(16,0,4)
B.(8,-16,4)
C.(8,16,4)
D.(8,0,4)
a +2
b =4(3,-2,1)+2(-2,4,0)=(12,-8,4)+(-4,8,0)=(8,0,4).
2若a =(2,-3,1),b =(2,0,3),c =(0,2,2),则a ·(b +c )的值为( )
A.4
B.15
C.7
D.3
b +
c =(2,2,5),
∴a ·(b +c )=4-6+5=3.
3已知向量a =(1,1,0),b =(-1,0,2),则|3a +b |为
( )A. B.4 C.5 D.1517
3a +b =(3,3,0)+(-1,0,2)=(2,3,2),
∴|3a +b |=.
22+32+22=17
4已知向量a =(1,1,0),b =(-1,0,2),且k a +b 与2a -b 相互垂直,则k 的值是(
)A.1 B. C. D.15357
5
a +
b =(k-1,k ,2),2a -b =(3,2,-2),
且(k a +b )·(2a -b )=3(k-1)+2k-4=0,
故k=.
75
5设M (5,-1,2),A (4,2,-1),若,则点B 应为
( )OM =AB A.(-1,3,-3) B.(9,1,1)
C.(1,-3,3)
D.(-9,-1,-1)
6已知向量a =(1,0,-1),则下列向量与a 成60°夹角的是( )
A.(-1,1,0)
B.(1,-1,0)
C.(0,-1,1)
D.(-1,0,1)
7已知向量a =(3,5,1),b =(2,2,3),c =(4,-1,-3),则向量2a -3b +4c 的坐标为 .
-19)
8已知a =(λ+1,0,2),b =(6,2μ-1,2),若a ∥b ,则λ= ,μ= .
a ∥
b ,∴存在实数m ,使a =m b ,
即
∴m=1,λ=5,μ=.{
λ+1=6m ,0=m (2μ-
1),2=2m ,12　1
29已知向量a =(3,1,5),b =(1,2,-3),试求一向量x ,使该向量与z 轴垂直,而且满足x ·a =9,x ·b =-4
.x =(t ,u ,v ),依题意及向量垂直的充要条件,
可得{
(t ,u ,v )·(0,0,1)=0
(t ,u ,v )·(3,1,5)=9
(t,u ,v )·(1,2,-3)=-4⇔{v =03t +u +5v =9t +2u -3v =-4⇔{v =0,
t =225,u =-215.故所求向量x =.
(225,-215,0)10已知空间四点A ,B ,C ,D 的坐标分别是(-1,2,1),(1,3,4),(0,-1,4),(2,-1,-2).若p =,q =,求下列各AB CD 式的值:
(1)p+2q ;(2)3p-q ;(3)(p-q )·(p+q ).
A (-1,2,1),
B (1,3,4),
C (0,-1,4),
D (2,-1,-2),
所以p ==(2,1,3),q ==(2,0,-6).
AB CD (1)p +2q =(2,1,3)+2(2,0,-6)
=(2,1,3)+(4,0,-12)=(6,1,-9).
(2)3p-q =3(2,1,3)-(2,0,-6)
=(6,3,9)-(2,0,-6)=(4,3,15).
(3)(p-q )·(p+q )=p 2-q 2=|p|2-|q|2=(22+12+32)-(22+02+62)=-26.能力提升
1已知A (3,4,5),B (0,2,1),O (0,0,0),若,则点C 的坐标是( )
OC =2
5AB A. B.(-65,-45,-85)(6
5,-4
5,-8
5)
C. D.(-65,-45,8
5)(6
5,45,8
5)
=(-3,-2,-4),
AB ∴.
25AB =(-6
5,-4
5,-8
5)设点C 的坐标为(x ,y ,z ),则=(x ,y ,z )=.故选A .
OC 25AB =(-6
5,-4
5,-8
5)
2已知a =(1,0,1),b =(-2,-1,1),c =(3,1,0),则|a -b +2c |等于(
)A.3 B.2 C. D.5
101010
a -
b +2
c =(3,1,0)+(6,2,0)=(9,3,0),
∴|a -b +2c |==3.
9010
3已知a =(1-t ,1-t ,t ),b =(2,t ,t ),则|b -a |的最小值是( )
A. B. C. D.55555355115
b -a =(1+t ,2t-1,0),
∴|b -a |2=(1+t )2+(2t-1)2+02=5t 2-2t+2
=5.
(t -1
5)2+9
5∴|b -a .∴|b -a |min =.
|2min =9535
5
4已知A (1,-2,11),B (4,2,3),C (6,-1,4),则△ABC 的面积是(
)
A. B.5542
242
C.5
D.5314
=(5,1,-7),=(2,-3,1),AC BC ∴=10-3-7=0.
AC ·BC ∴,即AC ⊥BC.
AC ⊥BC ∴S △ABC =|·||
12|AC BC =.
1
225+1+49×4+9+1=542
2
5如图,在空间直角坐标系中有直三棱柱ABC-A 1B 1C 1,CA=CC 1=2CB ,则直线BC 1与直线AB 1夹角的余弦值为( )
A. B. C. D.5
55325535
CA=CC 1=2CB=2,则A (2,0,0),B (0,0,1),B 1(0,2,1),C 1(0,2,0),
则=(-2,2,1),=(0,2,-1),
AB 1BC 1从而cos <AB 1,BC 1AB BC |AB 1||BC 1|=,(-2)×0+2×2+1×(-1)9×5=5故直线BC 1与直线
AB 1夹角的余弦值为.5
5
6已知空间三个向量a =(1,-2,z ),b =(x ,2,-4),c =(-1,y ,3),若它们两两垂直,则x= ,y= ,z= .
a ⊥
b ,得x-4-4z=0;
由a ⊥c ,得-1-2y+3z=0;
由b ⊥c ,得-x+2y-12=0.
故x=-64,y=-26,z=-17.
64　-26　-17
7若a =(2,3,-1),b =(-2,1,3),则以a ,b 为邻边的平行四边形的面积是 .
a |=,
4+9+1=14
|b |=.
4+1+9=14设θ为a 与b 的夹角,
则cos θ==-,a ·b |a ||b |=-4+3-3142
7所以sin θ=.
357所以S
平行四边形=|a ||b |sin θ=14×=6.
35
75658已知空间三点A (-2,0,2),B (-1,1,2),C (-3,0,4).设a =,b =.
AB AC (1)若|c |=3,c ∥,求c ;
BC (2)若k a +b 与k a -2b 互相垂直,求k ;
(3)若向量k a +b 与a +k b 平行,求
k.
∵=(-2,-1,2),且c ∥,
BC BC ∴设c =λ=(-2λ,-λ,2λ),λ∈R .
BC ∴|c |==3|λ|=3.
(-2λ)2+(-λ)2+(2λ)2解得λ=±1.∴c =(-2,-1,2)或c =(2,1,-2).
(2)∵a ==(1,1,0),b ==(-1,0,2),
AB AC ∴k a +b =(k-1,k ,2),k a -2b =(k+2,k ,-4).
∵(k a +b )⊥(k a -2b ),∴(k a +b )·(k a -2b )=0.
即(k-1,k ,2)·(k+2,k ,-4)=2k 2+k-10=0.
解得k=2或k=-.
52(3)∵k a +b =(k-1,k ,2),
a +k
b =(1,1,0)+(-k ,0,2k )=(1-k ,1,2k ),
又k a +b 与a +k b 平行,
∴k a +b =λ(a +k b )(λ∈R ),
即(k-1,k ,2)=λ(1-k ,1,2k ).
∴{
k -1=λ(1-k ),k =λ·1,2=λ·2k ,∴{k =-1,λ=-1或{k =1,λ=1.
∴k 的值为±1.
★9在正方体ABCD-A 1B 1C 1D 1中,M 是AA 1的中点,问当点N 位于线段AB 何处时,MN ⊥MC 1?
A 为坐标原点,棱A
B ,AD ,AA 1所在直线分别为x 轴,y 轴,z 轴,建立空间直角坐标系,如图.设正方体的棱长为a ,
则M ,C 1
(a ,a ,a ).(0,0,a
2)设N (x ,0,0),则
,MC 1=(a ,a ,a 2).MN =(x ,0,-a
2)由=xa-=0,
MN ·MC 1a 2
4得
x=.a
4所以点N 的坐标为,即N 为线段AB 的四等分点且靠近点A 时,MN ⊥MC 1.(a 4,0,0)。