题目内容
(本小题满分14分)下图为一简单组合体,其底面ABCD为正方形,
平面
,
,且
,
(1)求证:BE//平面PDA;
(2)若N为线段
的中点,求证:
平面
;
(3)若
,求平面PBE与平面ABCD所成的锐二面角的大小.![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555089391536.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508830265.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845301.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845435.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508861455.gif)
(1)求证:BE//平面PDA;
(2)若N为线段
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508876234.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508908369.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
(3)若
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923411.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555089391536.gif)
(1)证明略;
(2)证明略;
(3)45°
(2)证明略;
(3)45°
(1)证明:∵
,
平面
,
平面![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
∴EC//平面
,同理可得BC//平面
----------------------------------------2分
∵EC
平面EBC,BC
平面EBC且
∴平面
//平面
-----------------------------------------------------------------3分
又∵BE
平面EBC ∴BE//平面PDA-----------------------------------------------------4分
(2)证法1:连结AC与BD交于点F, 连结NF,
∵F为BD的中点,
∴
且
,--------------------------6分
又
且![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509298506.gif)
∴
且![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509344426.gif)
∴四边形NFCE为平行四边形-------------------------7分
∴![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509360435.gif)
∵![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315550939173.gif)
,
平面
,
面
∴
,
又![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509547405.gif)
∴
面
∴
面
----------------------------------------9分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555096561369.gif)
证法2:如图以点D为坐标原点,以AD所在的直线为x轴建立空间直角坐标系如图示:设该简单组合体的底面边长为1,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509672285.gif)
则![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509812813.gif)
,
--------------------------------6分
∴
,
,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509922411.gif)
∵
,
,
∴
---------------------------------8分
∵
、
面
,且![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510062405.gif)
∴
面
--------------------------------------------------------------------9分
(3)解法1:连结DN,由(2)知
面![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
∴
, ∵
,
∴
∴![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510327332.gif)
∴
为平面PBE的法向量,设
,则
∴
=
---11分
∵
为平面ABCD的法向量,
,---------------------------------------------12分
设平面PBE与平面ABCD所成的二面角为
,
则
------------------------------------------------13分
∴
即平面PBE与平面ABCD所成的二面角为45°--------------------14分
解法2:延长PE与DC的延长线交于点G,连结GB,
则GB为平面PBE与ABCD的交线--------------------10分
∵
∴![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510546399.gif)
∴D、B、G在以C为圆心、以BC为半径的圆上,
∴
-------------------11分
∵
平面
,
面
∴
且![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510670398.gif)
∴
面
∵
面
∴![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510748331.gif)
∴
为平面PBE与平面ABCD所成的二面角的平面角----------------------------13分
在
中 ∵![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510312325.gif)
∴
=45°即平面PBE与平面ABCD所成的二面角为45°----------------14分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555108891614.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555109041243.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845435.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508986266.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509017369.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
∴EC//平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
∵EC
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509079135.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509079135.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509173402.gif)
∴平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509188274.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509001272.gif)
又∵BE
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509079135.gif)
(2)证法1:连结AC与BD交于点F, 连结NF,
∵F为BD的中点,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509251447.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509266529.gif)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845435.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509298506.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509329438.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509344426.gif)
∴四边形NFCE为平行四边形-------------------------7分
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509360435.gif)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315550939173.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509407328.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508830265.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845301.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509469259.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845301.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509516327.gif)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509547405.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509563255.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509578271.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509610368.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555096561369.gif)
证法2:如图以点D为坐标原点,以AD所在的直线为x轴建立空间直角坐标系如图示:设该简单组合体的底面边长为1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509672285.gif)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509812813.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509828517.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509844576.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509875630.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509890408.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509922411.gif)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509937844.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509968848.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509984621.gif)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508876234.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510031265.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510062405.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509610368.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
(3)解法1:连结DN,由(2)知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155509610368.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510249452.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923411.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510280404.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510312325.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510327332.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510343263.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510374272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510390630.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510343263.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510421469.gif)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510436254.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510468467.gif)
设平面PBE与平面ABCD所成的二面角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510483200.gif)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555104991064.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510514269.gif)
解法2:延长PE与DC的延长线交于点G,连结GB,
则GB为平面PBE与ABCD的交线--------------------10分
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508861455.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510546399.gif)
∴D、B、G在以C为圆心、以BC为半径的圆上,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510561334.gif)
∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508830265.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845301.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510608264.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508845301.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510655333.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510670398.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510686266.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510717254.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155508923272.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510748331.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510780297.gif)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510826352.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510312325.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823155510780297.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555108891614.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231555109041243.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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