题目内容
已知向量
=(2,-1),
=(1+k,2+k-k2),若
⊥
,则实数k为( )
| a |
| b |
| a |
| b |
| A.-1 | B.0 | C.-1或0 | D.-1或4 |
∵
⊥
,
∴
•
=0
即2(1+k)-(2+k-k2)=0,化简得,k2=0
解得,k=0
故选B
| a |
| b |
∴
| a |
| b |
即2(1+k)-(2+k-k2)=0,化简得,k2=0
解得,k=0
故选B
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