题目内容
如图,在四棱锥
中,顶点
在底面
内的射影恰好落在
的中点
上,又
,
且
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(1)求证:
;
(2)若
,求直线
与
所成角的余弦值;
(3)若平面
与平面
所成的角为
,求
的值。
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(1)求证:

(2)若
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(3)若平面

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(1)利用两直线的方向向量垂直证明线线垂直;(2)
;(3)
.
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试题分析:因为AB中点O为点P在平面ABCD内的射影,所以PO⊥底面ABCD.以O为坐标原点,AB所在直线为x轴,OP所在直线为z轴,建立空间直角坐标系o﹣xyz(如图).

(1)设BC=a,OP=h则依题意得:B(a,0,0),A(﹣a,0,0),P(0,0,h),C(a,a,0),D(﹣a,2a,0).
∴
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于是
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(2)由PO=BC,得h=a,于是P(0,0,a),5分
∵
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∴
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∴直线PD与AB所成的角的余弦值为

(3)设平面PAB的法向量为m,可得m=(0,1,0),
设平面PCD的法向量为n=(x,y,z),
由
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∴
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cos<m,n>=
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解得=
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点评:运用向量在解决立体几何问题主要集中在法向量的应用上,它可以证明空间线面的位置关系、求解空间角、距离.同时运用空间向量解答立体几何问题,淡化了传统立体几何中的“形”的推理方法,强化了代数运算,从而降低了思维难度
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