题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-21118.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-83244.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-90509.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-99765.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-85471.jpg)
(Ⅰ)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-33093.gif)
(Ⅱ)设线段
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-55134.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-57276.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-59424.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-49668.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-65822.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-49668.gif)
(Ⅲ)求二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-27775.gif)
答案
(Ⅱ)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-49668.gif)
(Ⅲ)arctan
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-21086.gif)
解析
解法一:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-87067.jpg)
(Ⅰ)因为平面
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-83244.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-21118.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-75272.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-83832.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-21118.gif)
平面
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-83244.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-43026.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-21118.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-32892.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-75272.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035907-83244.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-75272.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-58233.gif)
因为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-14924.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-15176.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-92440.gif)
又因为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-37920.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-10450.gif)
即
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-58233.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-93345.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-68070.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-58233.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-23654.gif)
(Ⅱ)存在点
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-49668.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035908-49668.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-23654.gif)
取BE的中点N,连接AN,MN,则MN∥=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035910-43580.gif)
所以PMNC为平行四边形,所以PM∥CN
因为CN在平面BCE内,PM不在平面BCE内,
所以PM∥平面BCE………………………………8分
(Ⅲ)由EA⊥AB,平面ABEF⊥平面ABCD,易知,EA⊥平面ABCD
作FG⊥AB,交BA的延长线于G,则FG∥EA。从而,FG⊥平面ABCD
作GH⊥BD于G,连结FH,则由三垂线定理知,BD⊥FH
因此,∠AEF为二面角F-BD-A的平面角
因为FA="FE," ∠AEF=45°,
所以∠AFE=90°,∠FAG=45°.
设AB=1,则AE=1,AF=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-37868.gif)
FG=AF·sinFAG=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-83169.gif)
在Rt△FGH中,∠GBH=45°,BG=AB+AG=1+
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-83169.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-52417.gif)
GH=BG·sinGBH=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-52417.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-37868.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-48987.gif)
在Rt△FGH中,tanFHG=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-66413.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-21086.gif)
故二面角F-BD-A的大小为arctan
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-21086.gif)
解法二:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-42926.jpg)
(Ⅰ)因为△ABE为等腰直角三角形,AB=AE,
所以AE⊥AB.
又因为平面ABEF⊥平面ABCD,AE
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-83832.gif)
平面ABEF∩平面ABCD=AB,
所以AE⊥平面ABCD.
所以AE⊥AD.
因此,AD,AB,AE两两垂直,以A为坐标原点,建立如图所示的直角坐标系A-xyz.
设AB=1,则AE=1,B(0,1,0),D (1, 0, 0 ) ,
E ( 0, 0, 1 ), C ( 1, 1, 0 ).
因为FA="FE," ∠AEF = 45°,
所以∠AFE= 90°.
从而,
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-48752.gif)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-51885.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-66019.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-93887.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-99162.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-25443.gif)
所以EF⊥BE, EF⊥BC.
因为BE
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035909-83832.gif)
所以EF⊥平面BCE.
(Ⅱ) M(0,0,
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-83169.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-83169.gif)
从而
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-43555.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-64985.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-43219.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035911-83169.gif)
于是
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035912-54642.gif)
所以PM⊥FE,又EF⊥平面BCE,直线PM不在平面BCE内,
故PM∥平面BCE………………………………8分
(Ⅲ) 设平面BDF的一个法向量为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-52052.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-52052.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-18492.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-38474.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-85079.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-41384.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-86708.gif)
去y=1,则x=1,z=3,从
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035913-52052.gif)
取平面ABD的一个法向量为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035914-90004.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035914-46560.gif)
故二面角F-BD-A的大小为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035914-79148.gif)
核心考点
试题【(本小题满分12分)如图,正方形所在平面与平面四边形所在平面互相垂直,△是等腰直角三角形,。(Ⅰ)求证:;(Ⅱ)设线段的中点为,在直线上是否存在一点,使得?若存】;主要考察你对空间几何体的结构特征等知识点的理解。[详细]
举一反三
如图3,在正三棱柱
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035855-85930.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035855-28835.gif)
点E在AC上,且DE
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035855-98766.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035855-63339.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035856-66682.jpg)
(Ⅰ)证明:平面
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035856-48846.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035855-98766.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035856-60004.gif)
(Ⅱ)求直线AD和平面
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035856-48846.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035851-15236.jpg)
(1)求证:M为PC的中点;
(2)求证:面ADM⊥面PBC。
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035844-87180.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035837-93499.gif)
(I)求证:PA//平面EFG;
(II)若M为线段CD上的一个动点,问当M在什么位置时,MF与平面EFG所成角最大。
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035838-12597.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035825-17264.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035826-26368.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035826-71209.gif)
且
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035826-13183.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035826-33721.gif)
(Ⅰ) 求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035827-62848.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035827-30572.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035827-73906.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035827-31288.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035827-36830.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035828-15171.gif)
(Ⅲ)求四面体
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021035828-49611.gif)
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