题目内容
(本小题满分13分)
设
为坐标原点,
,
(1)若四边形
是平行四边形,求
的大小;
(2)在(1)的条件下,设
中点为
,
与
交于
,求
.
设
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
(1)若四边形
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
(2)在(1)的条件下,设
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


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
(1)
(2)

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试题分析:(1)有题意:
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
所以
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又
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所以

(2)
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
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
又由
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

所以
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因为


得

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点评:题中利用平行四边形的性质转化为向量关系,进而代入点的坐标进行计算,当遇到三点共线时,转化为三点确定的两向量共线

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