【答案】
分析:(I)根據(jù)正方體的幾何特征及面面平行的性質(zhì)定理,易證得EG∥D
1F;
(Ⅱ)以D為原點分別以DA、DC、DD
1為x、y、z軸,建立空間直角坐標(biāo)系,分別求出平面D
1EGF的法向量和平面ABCD的法向量,代入向量夾角公式,即可得到答案;
(III)幾何體ABGEA
1-DCFD
1由正方體ABCD-A
1B
1C
1D
1減去一個棱臺D
1FC
1-EGB
1得到,分別求出正方體ABCD-A
1B
1C
1D
1的體積和棱臺D
1FC
1-EGB
1的體積,即可得到答案.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/images0.png)
證明:(Ⅰ)在正方體ABCD-A
1B
1C
1D
1中,∵平面ABB
1A
1∥平面DCC
1D
1平面D
1EGF∩平面ABB
1A
1=EG,平面D
1EGF∩平面DCC
1D
1=D
1F,
∴EG∥D
1F.(3分)
解:(Ⅱ)如圖,以D為原點分別以DA、DC、DD
1為
x、y、z軸,建立空間直角坐標(biāo)系,則有
D
1(0,0,2),E(2,1,2),F(xiàn)(0,2,1),
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/0.png)
=(2,1,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/1.png)
=(0,2,-1)
設(shè)平面D
1EGF的法向量為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/2.png)
=(x,y,z)
則由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/3.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/4.png)
=0,和
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/5.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/6.png)
=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/7.png)
,
取x=1,得y=-2,z=-4,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/8.png)
=(1,-2,-4)(6分)
又平面ABCD的法向量為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/9.png)
(0,0,2)
以二面角C
1-D
1E-F的平面角為θ,
則cosθ=|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/10.png)
|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/11.png)
故截面D
1EGF與底面ABCD所成二面角的余弦值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/12.png)
.(9分)
解:(Ⅲ)設(shè)所求幾何體ABGEA
1-DCFD
1的體積為V,
∵△EGB
1∽△D
1FC
1,D
1C
1=2,C
1F=1,
∴EB
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/13.png)
D
1C
1=1,B
1G=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/14.png)
C
1F=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/15.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/17.png)
EB
1•B
1G=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/18.png)
•1•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/20.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/22.png)
D
1C
1•C
1F=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/23.png)
•2•1=1(11分)
故V
棱臺D1FC1-EGB1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/24.png)
∴V=V
正方體-V
棱臺D1FC1-EGB1=2
3-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182322074771109/SYS201310241823220747711017_DA/26.png)
.(14分)
點評:本題考查的知識點是用空間向量求平面間的夾角,組合體的體積,線線平行的判定,其中(1)的關(guān)鍵是熟練掌握線線平行、線面平行、面面平行之間的相互轉(zhuǎn)化,(2)的關(guān)系是求出平面D
1EGF的法向量和平面ABCD的法向量,(3)的關(guān)鍵是分析出幾何體ABGEA
1-DCFD
1由正方體ABCD-A
1B
1C
1D
1減去一個棱臺D
1FC
1-EGB
1得到.