题目内容

已知空间四边形OABC各边及对角线长都相等,E,F分别为AB,OC的中点,则异面直线OE与BF所成角的余弦值为(  )
A、
1
3
B、-
1
3
C、
2
3
D、-
2
3
考点:异面直线及其所成的角
专题:空间角
分析:
OA
=
a
OB
=
b
OC
=
c
,且|
a
|=|
b
|=|
c
|=1,则∠AOB=∠BOC=∠AOC=
π
3
a
b
=
b
c
=
c
a
=
1
2
,由此能求出异面直线OE与BF所成角的余弦值.
解答: 解:如图所示,设
OA
=
a
OB
=
b
OC
=
c

且|
a
|=|
b
|=|
c
|=1,
∴∠AOB=∠BOC=∠AOC=
π
3

a
b
=
b
c
=
c
a
=
1
2

OE
=
1
2
(
OA
+
OB
)
=
1
2
(
a
+
b
)

BF
=
OF
-
OB
=
1
2
OC
-
OB
=
1
2
c
-
b

|
OE
|=|
BF
|=
3
2

OE
BF
=
1
2
(
a
+
b
)•(
1
2
c
-
b
)

=
1
4
a
c
+
1
4
b
c
-
1
2
a
b
-
1
2
b
2
=-
1
2

∴cos<
OE
BF
>=
-
1
2
3
2
3
2
=-
2
3

∴异面直线OE与BF所成角的余弦值为
2
3

故选:C.
点评:本题考查异面直线所成角的求法,解题时要认真审题,是中档题,解题时要注意向量法的合理运用.
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