【答案】
分析:解法1:
(1)因為多面體是由底面為ABCD的長方體被截面AEC
1F所截面而得到的,所以AF∥EC
1,AE∥FC
1,過E作EH∥BC交CC
1于H,則CH=BE=1,所以DF=C
1H=2.故BF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/0.png)
=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/1.png)
.
(2)在立體幾何中,求點到平面的距離是一個常見的題型,同時求直線到平面的距離、平行平面間的距離及多面體的體積也常轉(zhuǎn)化為求點到平面的距離.本題采用的是“找垂面法”:即找(作)出一個過該點的平面與已知平面垂直,然后過該點作其交線的垂線,則得點到平面的垂線段.延長C
1E與CB交于G,連AG,則平面AEC
1F與平面ABCD相交于AG.過C作CM⊥AG,垂足為M,連C
1M,由三垂線定理可知AG⊥C
1M.由于AG⊥面C
1MC,且AG?面AEC
1F,所以平面AEC
1F⊥面C
1MC.在Rt△C
1CM中,作CQ⊥MC
1,垂足為Q,則CQ的長即為C到平面AEC
1F的距離.
解法2:
以D為坐標(biāo)原點,分別以DA、DC、DF為x、y、z軸,建立空間直角坐標(biāo)系O-xyz,則D(0,0,0),B(2,4,0),A(2,0,0),C(0,4,0),E(2,4,1),C
1(0,4,3).設(shè)F(0,0,z).這種解法的好處就是:(1)解題過程中較少用到空間幾何中判定線線、面面、線面相對位置的有關(guān)定理,因為這些可以用向量方法來解決.(2)即使立體感稍差一些的學(xué)生也可以順利解出,因為只需畫個草圖以建立坐標(biāo)系和觀察有關(guān)點的位置即可.
(1)由AEC
1F為平行四邊形,運用向量的模的計算方法,可得BF的長度;
(2)運用向量坐標(biāo)運算計算點到平面的距離,可以先設(shè)出此平面的法向量,設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/2.png)
為平面AEC
1F的法向量,顯然
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/3.png)
不垂直于平面ADF,故可設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/4.png)
=(x,y,1).進(jìn)一步可以求得C到平面AEC
1F的距離.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/images5.png)
解法1:(Ⅰ)過E作EH∥BC交CC
1于H,則CH=BE=1,EH∥AD,且EH=AD.
又∵AF∥EC
1,∴∠FAD=∠C
1EH.
∴Rt△ADF≌Rt△EHC
1.∴DF=C
1H=2.∴BF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/5.png)
=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/6.png)
.
(Ⅱ)延長C
1E與CB交于G,連AG,
則平面AEC
1F與平面ABCD相交于AG.
過C作CM⊥AG,垂足為M,連C
1M,
由三垂線定理可知AG⊥C
1M.由于AG⊥面C
1MC,且
AG?面AEC
1F,所以平面AEC
1F⊥面C
1MC.在Rt△C
1CM中,作CQ⊥MC
1,垂足為Q,則CQ的長即為C到平面AEC
1F的距離.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/8.png)
可得,BG=1,從而AG=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/10.png)
.
由∠GAB=∠MCG知,CM=3cosMCG=3cosGAB=3×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/12.png)
,∴CQ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/15.png)
解法2:(I)建立如圖所示的空間直角坐標(biāo)系,則D(0,0,0),B(2,4,0),A(2,0,0),C(0,4,0),E(2,4,1),C
1(0,4,3).設(shè)F(0,0,z).
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/images17.png)
∵AEC
1F為平行四邊形,∴由AEC
1F為平行四邊形,∴由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/17.png)
得,(-2,0,z)=(-2,0,2),∴z=2.∴F(0,0,2).∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/18.png)
=(-2,-4,2).于是|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/19.png)
|=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/20.png)
,即BF的長為2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/21.png)
.
(II)設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/22.png)
為平面AEC
1F的法向量,顯然
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/23.png)
不垂直于平面ADF,故可設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/24.png)
=(x,y,1).
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/25.png)
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/26.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/27.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/28.png)
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/29.png)
=(0,0,3),設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/30.png)
與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/31.png)
的夾角為a,則cosα=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/32.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/33.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/34.png)
.
∴C到平面AEC
1F的距離為d=|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/35.png)
|cosα=3×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212712205980491/SYS201310232127122059804019_DA/37.png)
.
點評:本小題主要考查空間中的線面關(guān)系、點到面的距離等基本知識,同時考查空間想象能力和推理、運算能力.