题目内容
(本小题满分12分)
如图,四棱锥
的底面为正方形,侧棱
底面
,且
,
分别是线段
的中点.
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(Ⅰ)求证:
//平面
;
(Ⅱ)求证:
平面
;
(Ⅲ)求二面角
的大小.
如图,四棱锥
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(Ⅰ)求证:
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(Ⅱ)求证:
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(Ⅲ)求二面角
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(1)、(2)见解析;(3)
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(1)因为
分别是线段
的中点,所以
,由线面平行的判定定理得
//平面
;(2)由已知易证
平面
,所以
,又
,
分别是线段
的中点,得
,根据线面垂直的判定定理得
平面
;(3)由二面角的定义知
就是所求二面角的平面角,等于
.另解:因为四棱锥
的底面为正方形,侧棱
底面
,可建立空间直角坐标系,写出需要的各点坐标.(1)只需证出
与
共线;(2)只需证
与平面
内的任意两个不共线向量垂直;(3)需求出平面
的法向量和平面
的法向量,把二面角转化为两个法向量的夹角,注意角的范围.
建立如图所示的空间直角坐标系
,
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,
,
,
,
.…………1分
(Ⅰ)证明:
∵
,
,
∴
,
∵
平面
,且
平面
,
∴
//平面
.………………………………4分
(Ⅱ)证明:
,
,
,
,
又
,
∴
平面
. ………………………………………………8分
(Ⅲ)设平面
的法向量为
,
因为
,
,
则
取
又因为平面
的法向量为
所以
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所以二面角
的大小为
.…………………………………12分
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建立如图所示的空间直角坐标系
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(Ⅰ)证明:
∵
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∴
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∵
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∴
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(Ⅱ)证明:
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又
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∴
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(Ⅲ)设平面
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因为
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则
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又因为平面
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所以
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所以二面角
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