题目内容
若A(-3,0),B(0,| 3 |
| OC |
| OA |
| OB |
分析:∠AOC=60°,
=λ
+
,则
=(-3λ,0)+(0,
)=(-3λ,
),由此可求出实数λ的值.
| OC |
| OA |
| OB |
| OC |
| 3 |
| 3 |
解答:解:∵∠AOC=60°,
=λ
+
,
∴
=(-3λ,0)+(0,
)=(-3λ,
),
∴tan60°=
,∴tan60°=
,λ=
.
答案:
.
| OC |
| OA |
| OB |
∴
| OC |
| 3 |
| 3 |
∴tan60°=
| 3λ | ||
|
| ||
| 3λ |
| 1 |
| 3 |
答案:
| 1 |
| 3 |
点评:解题时要仔细审题,认真挖掘隐含条件.
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