题目内容
已知|
|=1,|
|=k,∠AOB=
,点C在∠AOB内,
•
=0,若
=2m
+m
,|
|=2
,则k=( )
| OA |
| OB |
| 2π |
| 3 |
| OC |
| OA |
| OC |
| OA |
| OB |
| OC |
| 3 |
分析:由已知|
|=1,|
|=k
•
=0,
=2m
+m
,|
|=2
两边同时平方可得m,k的关系式,然后再由
-2m
=m
两边平方可得关于m,k的关系,从而可求k
| OA |
| OB |
| OC |
| OA |
| OC |
| OA |
| OB |
| OC |
| 3 |
| OC |
| OA |
| OB |
解答:解:|
|=1,|
|=k
由
•
=0可得∠AOC=90°
∵∠AOB=
,点C在∠AOB内
∴
•
=|
||
|cos
=-
k,且∠BOC=30°
∵
=2m
+m
,|
|=2
∴
2=4m2
2+4m2
•
+m2
2
∴12=4m2+4m2×(-
k)+(km)2
∵
=2m
+m
∴
-2m
=m
同上平方可得,12+4m2=m2k2
两式联立可得,k=4
故选D
| OA |
| OB |
由
| OC |
| OA |
∵∠AOB=
| 2π |
| 3 |
∴
| OA |
| OB |
| OA |
| OB |
| 2π |
| 3 |
| 1 |
| 2 |
∵
| OC |
| OA |
| OB |
| OC |
| 3 |
∴
| OC |
| OA |
| OA |
| OB |
| OB |
∴12=4m2+4m2×(-
| 1 |
| 2 |
∵
| OC |
| OA |
| OB |
∴
| OC |
| OA |
| 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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