人教版高中数学必修42.4.2 平面向量数量积的坐标表示、夹角
1.a =(-4,3),b =(5,6),则3|a |2-4a·b 等于( )
A .23
B .57
C .63
D .83
解析:选D.∵|a |=(-4)2+32=5,a·b =-4×5+3×6=-2,∴3|a |2-4a·b =3×52-4×(-2)=83.故选D.
2.已知a =(2,3),b =(-4,7),则a 在b 方向上的投影为( )
A.13
B.135
C.655
D.65
解析:选C.|a |cos θ=
a ·
b |b |=-8+2165=655. 3.已知a =(-3,2),b =(-1,0)向量λa +b 与a -2b 垂直,则实数λ的值为( )
A .-17
B.17 C .-16 D.16
解析:选A.向量λa +b =(-3λ-1,2λ),a -2b =(-1,2),因为两个向量垂直,故有(-3λ
-1,2λ)·(-1,2)=0,即3λ+1+4λ=0,解得:λ=-17
,故选A. 4.(2012·高考重庆卷)设x ,y ∈R ,向量a =(x,1),b =(1,y ),c =(2,-4)且a ⊥c ,b ∥c ,则|a +b |=( ) A. 5 B.10 C .2 5 D .10
解析:选B.由a ⊥c 得2x +1×(-4)=0,所以x =2;由b ∥c 得1×(-4)=2y ,所以y =-2.从而a =(2,1),b =(1,-2)
∴a +b =(3,-1),∴|a +b |= 32+(-1)2=10.
5.已知平面向量a =(2,4),b =(-1,2),若c =a -(a ·b )b ,则|c |等于( )
A .4 2
B .2 5
C .8
D .8 2
解析:选D.易得a ·b =2×(-1)+4×2=6,所以c =(2,4)-6(-1,2)=(8,-8),所以|c |= 82+(-8)2=8 2.
6.设向量a 与b 的夹角为θ,且a =(3,3),2b -a =(-1,-1),则cos θ=________.
解析:法一:b =12a +12
(-1,-1)=(1,1),则a ·b =6. 又|a |=32,|b |=2,∴cos θ=a ·b |a |·|b |=66
=1. 法二:由已知得: b =(1,1).
又a =(3,3),∴a ∥b ,且同向.
故θ=0°,cos θ=1.
答案:1
7.已知a =(25,15),b =(15,-25
),则向量3a +b 与-2(3a -b )的夹角为________. 解析:设夹角为 θ,∵|a |=1,|b |=1,a ·b =0,
∴(3a +b )·[-2(3a -b )]=-4,
又|3a +b |=2,|-2(3a -b )|=4,∴θ=2π3
. 答案:2π3
8.(2012·高考安徽卷)设向量a =(1,2m ),b =(m +1,1),c =(2,m ).若(a +c )⊥b ,则|a |=________.
解析:a +c =(1,2m )+(2,m )=(3,3m ).
∵(a +c )⊥b ,
∴(a +c )·b =(3,3m )·(m +1,1)=6m +3=0, ∴m =-12
. ∴a =(1,-1),∴|a |= 2.
答案: 2
9.已知平面向量a =(1,x ),b =(2x +3,-x ),x ∈R .
(1)若a ⊥b ,求x 的值;
(2)若a ∥b ,求|a -b |.
解:(1)若a ⊥b ,则a ·b =(1,x )·(2x +3,-x )=1×(2x +3)+x (-x )=0,即x 2-2x -3=0,解得x =-1或x =3.
(2)若a ∥b ,则1×(-x )-x (2x +3)=0,即x (2x +4)=0.
解得x =0或x =-2.
当x =0时,a =(1,0),b =(3, 0),
|a -b |=|(1,0)-(3,0)|=| (-2,0)|=2.
当x =-2时,a =(1,-2),b =(-1,2),
|a -b |=|(1,-2)-(-1,2)|=|(2,-4)|=2 5.
10.设平面三点A (1,0),B (0,1),C (2,5),
(1)试求向量2AB →+AC →的模;
(2)若向量AB →与AC →的夹角为θ,求cos θ.
解:(1)∵A (1,0),B (0,1),C (2,5), ∴AB →=(0,1)-(1,0)=(-1,1),
AC →=(2,5)-(1,0)=(1,5),
∴2AB →+AC →=2(-1,1)+(1,5)=(-1,7), ∴|2AB →+AC →|= (-1)2+72=5 2.
(2)由(1)知AB →=(-1,1),AC →=(1,5),
∴cos θ=(-1,1)·(1,5)
(-1)2+12×12+52
=21313.