题目内容
| PQ |
| BC |
| OA |
| a |
| OB |
| b |
| OC |
| c |
| a |
| b |
| c |
| OP |
| OQ |
考点:平面向量的基本定理及其意义
专题:平面向量及应用
分析:由
=t,PQ∥BC,可得
=
=t,而
=t
,
=t
.
=
-
,
=
-
.分别代入
=
+
,
=
+
,即可得出.
| PQ |
| BC |
| AP |
| AB |
| AQ |
| AC |
| AP |
| AB |
| AQ |
| AC |
| AB |
| OB |
| OA |
| AC |
| OC |
| OA |
| OP |
| OA |
| AP |
| OQ |
| OA |
| AQ |
解答:
解:∵
=t,PQ∥BC,∴
=
=t,
∴
=t
,
=t
.
=
-
,
=
-
.
∴
=
+
=
+t(
-
)=(1-t)
+t
=(1-t)
+t
,
=
+
=
+t(
-
)=(1-t)
+t
=(1-t)
+t
.
| PQ |
| BC |
| AP |
| AB |
| AQ |
| AC |
∴
| AP |
| AB |
| AQ |
| AC |
| AB |
| OB |
| OA |
| AC |
| OC |
| OA |
∴
| OP |
| OA |
| AP |
| OA |
| OB |
| OA |
| OA |
| OB |
| a |
| b |
| OQ |
| OA |
| AQ |
| OA |
| OC |
| OA |
| OA |
| OC |
| a |
| c |
点评:本题考查了向量的三角形法则、向量共线定理、平行线的性质,考查了推理能力与计算能力,属于中档题.
练习册系列答案
相关题目
若直线ax+by=1与不等式
,表示的平面区域无公共点,则2a+3b的取值范围是( )
|
| A、(-7,1) | B、(-3,5) |
| C、(-7,3) | D、R |