题目内容
关于平面向量
,
,
,有下列四个命题( )
①若
∥
,
≠
则?λ∈R,使得
=λ
②
•
=0,则
=
或
=
③若
=(1,k),
=(-2,6),
∥
则,k=-3
④若
•
=
•
则
⊥(
-
),其中正确命题序号是( )
| a |
| b |
| c |
①若
| a |
| b |
. |
| a |
| 0 |
| b |
| a |
②
. |
| a |
. |
| b |
| a |
| o |
| b |
| 0 |
③若
. |
| a |
| b |
. |
| a |
| b |
④若
| a |
| b |
| a |
| c |
| a |
| b |
| c |
分析:①若
∥
,
≠
则?λ∈R,使得
=λ
;②
•
=0,有可能
和
都不是
;③
=(1,k),
=(-2,6),
∥
,则
=
,解得k=-3;④若
•
=
•
,则
⊥(
-
)不一定成立.
| a |
| b |
. |
| a |
| 0 |
| b |
| a |
. |
| a |
. |
| b |
| a |
| b |
| 0 |
. |
| a |
| b |
. |
| a |
| b |
| 1 |
| -2 |
| k |
| 6 |
| a |
| b |
| a |
| c |
| a |
| b |
| c |
解答:解:①若
∥
,
≠
则?λ∈R,使得
=λ
,故①成立;
②
•
=0,有可能
和
都不是
,故②不成立;
③∵
=(1,k),
=(-2,6),
∥
,
∴
=
,解得k=-3,故③成立;
④若
•
=
•
,当
=
时,
⊥(
-
)不成立,故④不成立.
故选B.
| a |
| b |
. |
| a |
| 0 |
| b |
| a |
②
. |
| a |
. |
| b |
| a |
| b |
| 0 |
③∵
. |
| a |
| b |
. |
| a |
| b |
∴
| 1 |
| -2 |
| k |
| 6 |
④若
| a |
| b |
| a |
| c |
| a |
| 0 |
| a |
| b |
| c |
故选B.
点评:本题考查向量的性质和应用,是基础题.解题时要认真审题,仔细解答,注意真假命题的判断.
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