题目内容
(2012•天津)已知△ABC为等边三角形,AB=2.设点P,Q满足
=λ
,
=(1-λ)
,λ∈R.若
•
=-
,则λ=( )
| AP |
| AB |
| AQ |
| AC |
| BQ |
| CP |
| 3 |
| 2 |
分析:根据向量加法的三角形法则求出
=
+
=
+(1-λ)
,
=
+
=
+λ
进而根据数量级的定义求出
•
再根据
•
=-
即可求出λ.
| BQ |
| BA |
| AQ |
| BA |
| AC |
| CP |
| CA |
| AP |
| CA |
| AB |
| BQ |
| CP |
| BQ |
| CP |
| 3 |
| 2 |
解答:解:∵
=λ
,
=(1-λ)
,λ∈R
∴
=
+
=
+(1-λ)
,
=
+
=
+λ
∵△ABC为等边三角形,AB=2
∴
•
=
•
+λ
•
+(1-λ)
•
+λ(1-λ)
•
=2×2×cos60°+λ×2×2×cos180°+(1-λ)×2×2×cos180°+λ(1-λ)×2×2×cos60°
=2-4λ+4λ-4+2λ-2λ2,
=-2λ2+2λ-2
∵
•
=-
∴4λ2-4λ+1=0
∴(2λ-1)2=0
∴λ=
故选A
| AP |
| AB |
| AQ |
| AC |
∴
| BQ |
| BA |
| AQ |
| BA |
| AC |
| CP |
| CA |
| AP |
| CA |
| AB |
∵△ABC为等边三角形,AB=2
∴
| BQ |
| CP |
| BA |
| CA |
| BA |
| AB |
| AC |
| CA |
| AC |
| AB |
=2×2×cos60°+λ×2×2×cos180°+(1-λ)×2×2×cos180°+λ(1-λ)×2×2×cos60°
=2-4λ+4λ-4+2λ-2λ2,
=-2λ2+2λ-2
∵
| BQ |
| CP |
| 3 |
| 2 |
∴4λ2-4λ+1=0
∴(2λ-1)2=0
∴λ=
| 1 |
| 2 |
故选A
点评:本题主要考查了平面向量数量级的计算,属常考题,较难.解题的关键是根据向量加法的三角形法则求出
,
然后再结合数量级的定义和条件△ABC为等边三角形,AB=2,
•
=-
即可求解!
| BQ |
| CP |
| BQ |
| CP |
| 3 |
| 2 |
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