题目内容
平行四边形ABCD中,AB=2,AD=
,且
,以BD为折线,把
折起,使平面
,连AC。
(1)求异面直线AD与BC所成角大小;
(2)求二面角B-AC-D平面角的大小;
(3)求四面体ABCD外接球的体积。
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(1)求异面直线AD与BC所成角大小;
(2)求二面角B-AC-D平面角的大小;
(3)求四面体ABCD外接球的体积。
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(1)异面直线AD与BC所成角为
(2)二面角B-AC-D的大小是
;
(3)
。
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(2)二面角B-AC-D的大小是
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(3)
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本试题主要是考查了立体几何中异面直线所成的角和二面角的大小以及球的体积的求解的综合运用。
(1)在
中,
,易得
,
在四面体ABCD中,以D为原点,DB为
轴,DC为
轴,过D垂直于平面BDC的射线为
轴,建立如图空间直角坐标系,那么利用向量的夹角得到异面直线的角。
(2)利用法向量与法向量的夹角得到二面角的平面角。
(3)由于
均为直角三角形,故四面体ABCD的外接球球心在AD中点,
又
,所以球半径
,从而得到结论。
解:在
中,
,易得
,
在四面体ABCD中,以D为原点,DB为
轴,DC为
轴,过D垂直于平面BDC的射线为
轴,建立如图空间直角坐标系。
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则D(0,0,0),B(2,0,0),C(0,2,0),A(2,0,2)
(1)由于
,设AD与BC所成角为
,则
,即异面直线AD与BC所成角为
(2)设平面ABC的法向量为
,而
,
由
得:
,取
。
再设平面DAC的法向量为
,而
,
由
得:
,取
,
所以
,所以二面角B-AC-D的大小是
(3)……由于
均为直角三角形,故四面体ABCD的外接球球心在AD中点,
又
,所以球半径
,得
。
(1)在
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在四面体ABCD中,以D为原点,DB为
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(2)利用法向量与法向量的夹角得到二面角的平面角。
(3)由于
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又
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解:在
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在四面体ABCD中,以D为原点,DB为
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则D(0,0,0),B(2,0,0),C(0,2,0),A(2,0,2)
(1)由于
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(2)设平面ABC的法向量为
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由
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再设平面DAC的法向量为
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由
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所以
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(3)……由于
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又
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