题目内容
已知向量
=a=(
cosα,
sinα),
=b=(2cosβ,2sinβ),其中O为坐标原点,且
≤α<
<β≤
.
(1)若
⊥(
-
),求β-α的值;
(2)当
•(
-
)取最小值时,求△OAB的面积S.
| OA |
| 2 |
| 2 |
| OB |
| π |
| 6 |
| π |
| 2 |
| 5π |
| 6 |
(1)若
| a |
| b |
| a |
(2)当
| a |
| b |
| a |
(1)由
⊥(
-
)得
•(
-
)=0
即
•
-(
)2=0
又|
|=
,|
|=2,<
,
>=β-α∴2
•cos(β-α)-2=0cos(β-α)=
∵
≤α<
<β≤
π∴β-α=
…(6分)
(2)由(1)知
•(
-
)=2
cos(β-α)-2∵
≤α<
<β≤
π∴0<β-α≤
π
当β-α=
π时,
•(
-
)取最小值
此时S△OAB=
•2•sin
π=
…(12分)
| a |
| b |
| a |
| a |
| b |
| a |
即
| a |
| b |
| a |
又|
| a |
| 2 |
| b |
| a |
| b |
| 2 |
| ||
| 2 |
| π |
| 6 |
| π |
| 2 |
| 5 |
| 6 |
| π |
| 4 |
(2)由(1)知
| a |
| b |
| a |
| 2 |
| π |
| 6 |
| π |
| 2 |
| 5 |
| 6 |
| 2 |
| 3 |
当β-α=
| 2 |
| 3 |
| a |
| b |
| a |
此时S△OAB=
| 1 |
| 2 |
| 2 |
| 2 |
| 3 |
| ||
| 2 |
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