题目内容
已知| a |
| b |
| c |
| a |
| b |
| c |
分析:先求出(
+
)的值,再由(
+
)⊥
,知(
+
)•
=0,由此能求出k.
| a |
| b |
| a |
| b |
| c |
| a |
| b |
| c |
解答:解:∵
=(1,-2),
=(2,k),
=(2,-1),(
+
)⊥
,
∴(
+
)•
=[(1,-2)+(2,k)]•(2,-1)
=(3,k-2)•(2,-1)
=6-k+2=0,
∴k=8.
答案为:8.
| a |
| b |
| c |
| a |
| b |
| c |
∴(
| a |
| b |
| c |
=(3,k-2)•(2,-1)
=6-k+2=0,
∴k=8.
答案为:8.
点评:本题考查利用数量积判断两个平面向量的垂直关系,先求出(
+
)的值,再由(
+
)⊥
,知(
+
)•
=0,由此能求出k.
| a |
| b |
| a |
| b |
| c |
| a |
| b |
| c |
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