题目内容
如果| a |
| b |
| c |
| a |
| b |
| c |
| c |
| a |
| b |
分析:先由
-
=
,
+2
=3
,可得
=
,
=
.则有
=
,从而判断
与
平行.或直接消去c向量得到a、b向量的关系.
| a |
| b |
| c |
| a |
| b |
| c |
| b |
| 2 |
| 3 |
| c |
| a |
| 5 |
| 3 |
| c |
| a |
| 5 |
| 2 |
| b |
| a |
| b |
解答:解:方法1:∵
-
=
①,
+2
=3
②,
②-①得3
=2
,
=
,
将
=
代入①得
=
.
∴
=
,
∴
∥
.
方法2:∵
-
=
①,
+2
=3
②,
①×3-②得2
-5
=0,
即
=
,
∴
∥
.
| a |
| b |
| c |
| a |
| b |
| c |
②-①得3
| b |
| c |
| b |
| 2 |
| 3 |
| c |
将
| b |
| 2 |
| 3 |
| c |
| a |
| 5 |
| 3 |
| c |
∴
| a |
| 5 |
| 2 |
| b |
∴
| a |
| b |
方法2:∵
| a |
| b |
| c |
| a |
| b |
| c |
①×3-②得2
| a |
| b |
即
| a |
| 5 |
| 2 |
| b |
∴
| a |
| b |
点评:本题考查了平面向量的平行的判定.向量的数量积的性质:a=kb<=>a∥b.
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