问题 选择题
平面上O,A,B三点不共线,设
OA
=
a
OB
=
b
,则△OAB的面积等于(  )
A.
|
a
|
2
|
b
|
2
-(
a
b
)
2
B.
|
a
|
2
|
b
|
2
+(
a
b
)
2
C.
1
2
|
a
|
2
|
b
|
2
-(
a
b
)
2
D.
1
2
|
a
|
2
|
b
|
2
+(
a
b
)
2
答案

S△OAB=

1
2
|
a
|
|
b
|
sin<
a
b

=

1
2
|
a
|
|
b
|
1-cos2
a
b

=

1
2
|
a
|
|
b
|
1-
(
a
b
)
2
|
a
|2|
b
|2

=

1
2
|
a
|
2
|
b
|
2
-(
a
b
)
2

故选C.

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