1.已知向量a=(1,1,0),b=(-1,0,2),且ka+b与2a-b互相垂直,则k值是( )
A.1 B.
C. D.
[答案] D
[解析] ka+b=k(1,1,0)+(-1,0,2)=(k-1,k,2),
2a-b=2(1,1,0)-(-1,0,2)=(3,2,-2),
∵两向量垂直,∴3(k-1)+2k-2×2=0,∴k=.
2.a=(cosα,1,sinα),b=(sinα,1,cosα),则a+b与a-b的夹角为( )
A.0° B.30°
C.60° D.90°
[答案] D
[解析] |a|=,|b|=,
(a+b)(a-b)=|a|2-|b|2=0,
∴(a+b)⊥(a-b).