答案:B

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)

答案:B

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　　　　　　　.

答案:(16,0,-19)

8已知a=(λ+1,0,2),b=(6,2μ-1,2),若a∥b,则λ=　　　　　,μ=　　　　　.

解析:∵a∥b,∴存在实数m,使a=mb,

　　即{■(λ+1=6m"," @0=m"(" 2μ"-" 1")," @2=2m"," )┤∴m=1,λ=5,μ=1/2.

答案:5　1/2

9已知向量a=(3,1,5),b=(1,2,-3),试求一向量x,使该向量与 轴垂直,而且满足x·a=9,x·b=-4.

解:设向量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=0@3t+u+5v=9@t+2u"-" 3v="-" 4)┤⇔{■(v=0"," @t=22/5 "," @u="-" 21/5 "." )┤

　　故所求向量x=(22/5 ",-" 21/5 "," 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=(AB) ⃗,q=(CD) ⃗,求下列各式的值:

(1)p+2q;(2)3p-q;(3)(p-q)·(p+q).

解:因为A(-1,2,1),B(1,3,4),C(0,-1,4),D(2,-1,-2),

　　所以p=(AB) ⃗=(2,1,3),q=(CD) ⃗=(2,0,-6).

　　(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)=p2-q2=|p|2-|q|2=(22+12+32)-(22+02+62)=-26.