試題分析:(1)求四面體的體積,當高不好確定時候,可考慮等體積轉化,該題中
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610494571.png)
,高
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610510430.png)
,可求體積;(2)證明直線和直線垂直,可先證明直線和平面垂直,由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610541728.png)
,從而
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610557460.png)
面
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610572522.png)
,所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610463597.png)
,(3) 求二面角的平面角,可以利用幾何法,先找到二面角的平面角,然后借助平面圖形去計算,∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610603617.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610619482.png)
,所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610650552.png)
,進而可證
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610666587.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610681558.png)
就是
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610479582.png)
的平面角,二面角也可以利用空間向量法,建立適當的空間直角坐標系,把相關點的坐標表示出來,計算兩個半平面的法向量,進而求法向量的夾角,然后得二面角的余弦值.
試題解析:(1)解:在三棱錐D
1-DCE中,D
1D⊥平面DCE,D
1D=1
在△DCE中,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610713882.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610728890.png)
CD=2,CD
2=CE
2+DE
2 ∴CE⊥DE.
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240236108221197.png)
∴三棱錐D
1-DCE的體積
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240236108371009.png)
. =
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610962327.png)
4分
(2)證明:連結AD
1,由題可知:四邊形ADD
1A
1是正方形
∴A
1D⊥AD
1 又∵AE⊥平面ADD
1A
1,A
1D
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面ADD
1A
1∴AB⊥AD
1 又∵AB
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面AD
1E,AD
1![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面A D
1E AB
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023611025255.png)
AD
1=A
∴A
1D⊥平面AD
1E 又∵D
1E
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面AD
1E
∴A
1D⊥D
1E 8分
(3)根據題意可得:D
1D⊥平面ABCD
又因為CE
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面ABCD,所以D
1D⊥CE。
又由(1)中知,DE⊥CE,D
1D
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面D
1DE,DE
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面D
1DE,D
1D
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023611025255.png)
DE=D,
∴CE⊥平面D
1DE,又∵D
1E
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610978228.png)
平面D
1DE ∴CE⊥D
1E.
∴∠D
1ED即為二面角D
1―EC―D的一個平面角.
在Rt△D
1DE中,∠D
1DE=90°,D
1D="1," DE=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023611103339.png)
∴
∴二面角D
1―ED―D的正切值是
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023610479413.png)
12分