题目内容
已知O为三角形ABC内一点,∠AOB=150°,∠BOC=90°,若
=
,
=
,
=
,且|
|=2,|
|=1,|
|=3,
(1)用
,
表示
(2)求
•(
-
)
| OA |
| a |
| OB |
| b |
| OC |
| c |
| a |
| b |
| c |
(1)用
| a |
| b |
| c |
(2)求
| a |
| b |
| c |
(1)如图,建立直角坐标系:易知:
=(-
,-1),
=(1,0),
=(0,3).
设
=x
+y
,则(0,3)=x(-
,-1)+y(1,0)=(-
x+y,-x),
解得x=-3,y=-3
,即
=-3
-3
.
(2)
•(
-
)=
•
-
•
=(-
)-(-3)=3-
.
| a |
| 3 |
| b |
| c |
设
| c |
| a |
| b |
| 3 |
| 3 |
解得x=-3,y=-3
| 3 |
| c |
| a |
| 3 |
| b |
(2)
| a |
| b |
| c |
| a |
| b |
| a |
| c |
| 3 |
| 3 |
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