题目内容
| AB |
| a |
| AD |
| b |
| PQ |
| a |
| b |
分析:首先根据题意求得:EP∥AD,EQ∥AB,EP=
AD,EQ=
AB=
AB,由
=
,
=
,即可求得
与
的值,然后根据三角形法则,求得
.
| 1 |
| 4 |
| 2 |
| 4 |
| 1 |
| 2 |
| AB |
| a |
| AD |
| b |
| EP |
| EQ |
| PQ |
解答:
解:根据题意得:EP∥AD,EQ∥AB,EP=
AD,EQ=
AB=
AB,
∴
=
=
,
=
=
,
∴
=
-
=
-
.
故答案为:
-
.
| 1 |
| 4 |
| 2 |
| 4 |
| 1 |
| 2 |
∴
| EP |
| 1 |
| 4 |
| AD |
| 1 |
| 4 |
| b |
| EQ |
| 1 |
| 2 |
| AB |
| 1 |
| 2 |
| a |
∴
| PQ |
| EQ |
| EP |
| 1 |
| 2 |
| a |
| 1 |
| 4 |
| b |
故答案为:
| 1 |
| 2 |
| a |
| 1 |
| 4 |
| b |
点评:此题考查了平面向量的知识.解题的关键是掌握三角形法则与数形结合思想的应用.
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