题目内容
以下有关平面向量的结论:
①
;
②
;
③
,且1+x+y=0?A,B,C三点共线;
④
;
⑤
,
其中正确的结论有
- A.1个
- B.2个
- C.3个
- D.4个
B
分析:根据平面向量的数量积运算、模的运算性质和平面向量基本定理,对五个选项逐个加以判断,并且结合举反例和直接证明的方法,可得只有选项②、③是正确的,由此可得本题的答案.
解答:因为当
且
时,有
=0,但不能得出
的结论,故①不正确;
由
,可得
-
=0,即
=
,所以
成立,故②正确;
根据
,得
,
∵1+x+y=0,∴
,
化简得
-
=x(
-
),即
=x
,可得A,B,C三点共线故③正确;
因为
=λ
,是一个与
共线的向量,而
=μ
,是一个与
共线的向量.
所以等式
不一定成立,故④不正确;
若
,说明向量
、
共线且同向,不一定相等,故⑤不正确.
故正确的选项只有②③,2个
故答案为:B
点评:本题给出平面向量的几个命题,叫我们找出其中的真命题,着重考查了平面向量的数量积运算、模的运算性质和平面向量基本定理等知识,属于基础题.
分析:根据平面向量的数量积运算、模的运算性质和平面向量基本定理,对五个选项逐个加以判断,并且结合举反例和直接证明的方法,可得只有选项②、③是正确的,由此可得本题的答案.
解答:因为当
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由
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根据
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∵1+x+y=0,∴
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化简得
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因为
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所以等式
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若
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故正确的选项只有②③,2个
故答案为:B
点评:本题给出平面向量的几个命题,叫我们找出其中的真命题,着重考查了平面向量的数量积运算、模的运算性质和平面向量基本定理等知识,属于基础题.
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