题目内容
(本小题满分13分)如图,四面体
中,
是
的中点,
,
.(Ⅰ)求证:
平面
;(Ⅱ)求异面直线
与
所成角的大小;
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(Ⅲ)求二面角
的大小.
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(Ⅲ)求二面角
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(Ⅰ) 见解析 (Ⅱ)
(Ⅲ)
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(I)证明:
.
连接
.
,又
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即
平面
.
(II)方法1 取
的中点
,
的中点
,
为
的中点,
或其补角是
与
所成的角.∴连接
是
斜边
上的中线,
,
.在
中,由余弦定理得
,∴直线
与
所成的角为
.
(Ⅲ)方法l
平面
,过
作
于
,连接
,
是
在平面
上的射影,由三垂线定理得
.
是二面角
的平面角,
,又
.
在
中,
,
.
∴二面角
为
.
(II)方法2建立空间直角坐标系
.则
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.
.∴直线
与
所成的角为
.
(Ⅲ)方法2在坐标系中,平面
的法向量
.
设平面
的法向量
,则
,
求得
,
∴二面角
为
.
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连接
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即
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(II)方法1 取
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(Ⅲ)方法l
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在
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∴二面角
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(II)方法2建立空间直角坐标系
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(Ⅲ)方法2在坐标系中,平面
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设平面
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求得
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∴二面角
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