题目内容
已知O为坐标原点,向量
=(sinα,1),
=(cosα,0),
=(-sinα,2),点P满足
=
.
(Ⅰ)记函数f(α)=
•
,求函数f(α)的最小正周期;
(Ⅱ)若O,P,C三点共线,求|
+
|的值.
| OA |
| OB |
| OC |
| AB |
| BP |
(Ⅰ)记函数f(α)=
| PB |
| CA |
(Ⅱ)若O,P,C三点共线,求|
| OA |
| OB |
(Ⅰ)∵
=(sinα,1),
=(cosα,0),
=(-sinα,2)
∴
=(cosα-sinα,-1),
=(2sinα,-1)
设
=(x,y),则
=(x-cosα,y),
由
=
得,
,
故
=(2cosα-sinα,-1),则
=(sinα-cosα,1),
∴f(α)=(sinα-cosα,1)•(2sinα,-1)
=2sin2α-2sinαcosα-1
=-(sin2α+cos2α)
=-
sin(2α+
)
∴f(α)的最小正周期T=π.
(Ⅱ)由O,P,C三点共线可得:
∥
则(-1)×(-sinα)=2×(2cosα-sinα),
解得tanα=
,
∴sin2α=
=
=
,
∴|
+
|=
=
=
.
| OA |
| OB |
| OC |
∴
| AB |
| CA |
设
| OP |
| BP |
由
| AB |
| BP |
|
故
| OP |
| PB |
∴f(α)=(sinα-cosα,1)•(2sinα,-1)
=2sin2α-2sinαcosα-1
=-(sin2α+cos2α)
=-
| 2 |
| π |
| 4 |
∴f(α)的最小正周期T=π.
(Ⅱ)由O,P,C三点共线可得:
| OP |
| 0C |
则(-1)×(-sinα)=2×(2cosα-sinα),
解得tanα=
| 4 |
| 3 |
∴sin2α=
| 2sinαcosα |
| sin2α+cos2α |
| 2tanα |
| 1+tan2α |
| 24 |
| 25 |
∴|
| OA |
| OB |
| (sinα+cosα)2+1 |
=
| 2+sin2α |
| ||
| 5 |
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