题目内容

已知|
OA
|=1,|
OB
|=k,∠AOB=
3
,点C在∠AOB内,
OC
OA
=0,若
OC
=2m
OA
+m
OB
,|
OC
|=2
3
,则k=(  )
分析:由已知|
OA
|=1,|
OB
|=k
OC
OA
=0,
OC
=2m
OA
+m
OB
,|
OC
|=2
3
两边同时平方可得m,k的关系式,然后再由
OC
-2m
OA
=m
OB
两边平方可得关于m,k的关系,从而可求k
解答:解:|
OA
|=1,|
OB
|=k
OC
OA
=0可得∠AOC=90°
∠AOB=
3
,点C在∠AOB内
OA
OB
=|
OA
||
OB
|cos
3
=-
1
2
k
,且∠BOC=30°
OC
=2m
OA
+m
OB
,|
OC
|=2
3

OC
2
=4m2
OA
2
+4m2
OA
OB
+m2
OB
2

∴12=4m2+4m2×(-
1
2
k)
+(km)2
OC
=2m
OA
+m
OB

OC
-2m
OA
=m
OB

同上平方可得,12+4m2=m2k2
两式联立可得,k=4
故选D
点评:本题主要考查了向量的数量积的运算性质的应用,解题的关键是对已知式子两边同时平方结合数量积的定义进行求解.属于向量知识的综合应用
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