问题 选择题
已知空间向量
a
b
满足|
a
|=|
b
|=1
,且
a
, 
b
的夹角为
π
3
,O为空间直角坐标系的原点,点A、B满足
OA
=2
a
+
b
OB
=3
a
-
b
,则△OAB的面积为(  )
A.
5
2
3
B.
5
4
3
C.
7
4
3
D.
11
4
答案

由题意可得|

OA
|=
(2
a
+
b
)
2
=
4
a
2
+2
a
b
+
b
2
=
12+4×1×1×
1
2
+12
=
7

同理可得|

OB
|=
(3
a
-
b
)
2
=
9
a
2
-6
a
b
+
b
2
=
12-6×1×1×
1
2
+12
=
7

OA
OB
=(2
a
+
b
)•(3
a
-
b
)=6
a
2
+
a
b
-
b
2
=6×12+1×1×
1
2
-12=
11
2

故cos∠BOA=

OA
OB
|
OA
||
OB
|
=
11
2
7
7
=
11
14
,可得sin∠BOA=
1-(
11
14
)2
=
5
3
14

所以△OAB的面积S=

1
2
|
OA
||
OB
|sin∠BOA=
1
2
×
7
×
7
×
5
3
14
=
5
3
4

故选B

解答题
单项选择题