题目内容
| AB |
| e1 |
| AD |
| e2 |
| MN |
-
+
| 1 |
| 3 |
| e1 |
| e2 |
-
+
(用| 1 |
| 3 |
| e1 |
| e2 |
| e1 |
| e2 |
分析:根据M、N分别是AB、CD的中点证出2
=
+
,再利用AB∥CD且AB=3CD证出
=
-
,代入化简可得
=
+
,再结合题意可得本题答案.
| MN |
| AD |
| BC |
| BC |
| AD |
| 2 |
| 3 |
| AB |
| MN |
| 1 |
| 3 |
| AB |
| AD |
解答:解:∵
=
+
+
,
=
+
+

∴2
=(
+
+
)+(
+
+
)
=(
+
)+(
+
)+
+
∵M,N分别是AB,CD的中点,
∴
+
=
+
=
,可得2
=
+
∵AB∥CD,且AB=3CD,
∴
=
-
=
+
-
=
-
因此,2
=
+
=2
-
,得
=
+
结合
=
,
=
,可得
=-
+
故答案为:-
+
| MN |
| MA |
| AD |
| DN |
| MN |
| MB |
| BC |
| CN |
∴2
| MN |
| MA |
| AD |
| DN |
| MB |
| BC |
| CN |
=(
| MA |
| MB |
| DN |
| CN |
| AD |
| BC |
∵M,N分别是AB,CD的中点,
∴
| MA |
| MB |
| DN |
| CN |
| 0 |
| MN |
| AD |
| BC |
∵AB∥CD,且AB=3CD,
∴
| BC |
| AC |
| AB |
| AD |
| DC |
| AB |
| AD |
| 2 |
| 3 |
| AB |
因此,2
| MN |
| AD |
| BC |
| AD |
| 2 |
| 3 |
| AB |
| MN |
| 1 |
| 3 |
| AB |
| AD |
结合
| AB |
| e1 |
| AD |
| e2 |
| MN |
| 1 |
| 3 |
| e1 |
| e2 |
故答案为:-
| 1 |
| 3 |
| e1 |
| e2 |
点评:本题在梯形中求向量的线性表达式,着重考查了梯形的性质、向量的线性运算法则和平面向量基本定理等知识,属于中档题.
练习册系列答案
相关题目