题目内容

已知向量
OA
=
a
=(cosα,sinα)
OC
=
c
=(0,2)
OB
=
b
=(2cosβ,2sinβ)
,其中O为坐标原点,且0<α<
π
2
<β<π
(1)若
a
⊥(
b
-
a
)
,求β-α的值;
(2)若
OB
OC
=2,
OA
OC
=
3
,求△OAB的面积S.
(1)∵
a
⊥(
b
-
a
)
a
•(
b
-
a
)=0

∴2cosαcosβ+2sinαsinβ-1=0
cos(α-β)=
1
2

0<α<
π
2
<β<π
∴0<β-α<π∴β-α=
π
3

(2)∵
OB
OC
=2
OA
OC
=
3

sinβ=
1
2
sinα=
3
2

cosβ=
3
2
cosα=
1
2

OA
OB
=2cosαcosβ+2sinαsinβ=0

OA
OB

S=
1
2
|
OA
|•|
OB
|=
1
2
×1×2=1
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