【答案】
分析:(1)取AD的中點(diǎn)G,連接BD、GH、GF,利用正方形的性質(zhì)結(jié)合三角形中位線定理,可證出四邊形EFGH為平行四邊形,從而EH∥FG,結(jié)合EH⊥平面ABCD,得到FG⊥平面ABCD,最后根據(jù)面面垂直的判定定理,得到平面ADF丄平面ABCD;
(2)在平面ABCD內(nèi)過點(diǎn)H作直線IJ∥AD,分別交AB、CD于I、J.由(1)的證明過程,可得三棱柱ADF-IJE是直三棱柱,從而得到它的體積為:S
△IJE×EF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/0.png)
IJ×EH×EF=1.又因?yàn)樗睦忮FE-IJCB的體積為:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/1.png)
S
IJCB×EH=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/2.png)
,相加即得五面體EF-ABCD的體積.
(3)以G為原點(diǎn),AD所在直線為x軸,建立如圖坐標(biāo)系,分別得出B、C、E、N各點(diǎn)的坐標(biāo),設(shè)M(x,y,0),若MN⊥平面BCE,則MN⊥EB且MN⊥EC,利用向量數(shù)量積為0,聯(lián)列方程組,解之得x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/3.png)
,y=1.從而得到向量
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/4.png)
的坐標(biāo),利用向量模的公式,可得MN的長.
解答:解:(1)由題意,得:EF∥AB,且EF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/5.png)
AB,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/images6.png)
取AD的中點(diǎn)G,連接BD、GH、GF,
∵H是正方形ABCD的中心,
∴H是BD的中點(diǎn),得到△ABD中,GH是中位線,
∴GH∥AB,GH=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/6.png)
AB,
∴EF∥GH且EF=GH,可得四邊形EFGH為平行四邊形,
∴EH∥FG,
又∵EH⊥平面ABCD,∴FG⊥平面ABCD,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/images8.png)
∵FG?平面ADF,∴平面ADF丄平面ABCD;
(2)在平面ABCD內(nèi)過點(diǎn)H作直線IJ∥AD,分別交AB、CD于I、J.
由(1)的證明過程,得EF∥AI∥DJ,且EF=AI=DJ=1
∵EF⊥平面ADF,∴三棱柱ADF-IJE是直三棱柱
∴V
三棱柱ADF-IJE=S
△IJE×EF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/7.png)
IJ×EH×EF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/8.png)
×2×1×1=1.
又∵V
四棱錐E-IJCB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/9.png)
S
IJCB×EH=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/10.png)
×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/11.png)
S
ABCD×EH=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/12.png)
.
∴五面體EF-ABCD的體積為V=V
三棱柱ADF-IJE+V
四棱錐E-IJCB=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/14.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/images17.png)
(3)以G為原點(diǎn),AD所在直線為x軸,建立如圖坐標(biāo)系,則
B(1,2,0),C(-1,2,0),E(0,1,1),N(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/17.png)
),
設(shè)M(x,y,0),可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/20.png)
若MN⊥平面BCE,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/21.png)
,
解之得:x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/22.png)
,y=1.
∴向量
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/23.png)
,
因此
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185756742350360/SYS201310241857567423503018_DA/25.png)
=
點(diǎn)評:本題給出一個(gè)由直三棱柱和四棱錐拼接而成的五面體,通過證明面面垂直和求體積,著重考查了組合幾何體的體積公式,以及平面與平面垂直的判定等知識點(diǎn),屬于中檔題.