题目内容
| OA |
| a |
| OB |
| b |
| OC |
| c |
| AC |
| CB |
| c |
| a |
| b |
考点:向量加减混合运算及其几何意义
专题:平面向量及应用
分析:由题意可得
=
,可得
=
+
=
+
=
+
(
-
),代入已知可得向量
的方程,变形可得.
| BC |
| 1 |
| 3 |
| AC |
| c |
| OB |
| BC |
| OB |
| 1 |
| 3 |
| AC |
| OB |
| 1 |
| 3 |
| OC |
| OA |
| c |
解答:
解:∵
=-3
,
∴
=
∴
=
=
+
=
+
=
+
(
-
)=
+
-
,
∴
=
-
,
∴
=-
+
故答案为:-
+
| AC |
| CB |
∴
| BC |
| 1 |
| 3 |
| AC |
∴
| c |
| OC |
| OB |
| BC |
| OB |
| 1 |
| 3 |
| AC |
=
| OB |
| 1 |
| 3 |
| OC |
| OA |
| b |
| 1 |
| 3 |
| c |
| 1 |
| 3 |
| a |
∴
| 2 |
| 3 |
| c |
| b |
| 1 |
| 3 |
| a |
∴
| c |
| 1 |
| 2 |
| a |
| 3 |
| 2 |
| b |
故答案为:-
| 1 |
| 2 |
| a |
| 3 |
| 2 |
| b |
点评:本题考查向量加减混合运算,涉及向量的数乘,属基础题.
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