【答案】
分析:法一:(Ⅰ)如圖,連接CA
1、AC
1、CM,要證CD⊥平面BDM,只需證明直線CD垂直平面BDM內(nèi)兩條相交直線A
1B、DM即可;
(Ⅱ)設(shè)F、G分別為BC、BD的中點,連接B
1G、FG、B
1F,說明∠B
1GF
是所求二面角的平面角,然后解三角形,求面B
1BD與面CBD所成二面角的大�。�
法二:(Ⅰ)建立空間直角坐標系,求出相關(guān)向量計算
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/0.png)
即得證,
(Ⅱ)求出面B
1BD與面CBD的法向量,利用向量的數(shù)量積求解可得答案.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/images1.png)
解:法一:(I)如圖,連接CA
1、AC
1、CM,則CA
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/1.png)
,
∵CB=CA
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/2.png)
,∴△CBA
1為等腰三角形,
又知D為其底邊A
1B的中點,∴CD⊥A
1B,
∵A
1C
1=1,C
1B
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/3.png)
,∴A
1B
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/4.png)
,
又BB
1=1,∴A
1B=2,
∵△A
1CB為直角三角形,D為A
1B的中點,CD=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/5.png)
A
1B=1,CD=CC
1.
又DM=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/6.png)
AC
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/7.png)
,DM=C
1M,∴△CDN≌△CC
1M,∠CDM=∠CC
1M=90°,即CD⊥DM,
因為A
1B、DM為平面BDM內(nèi)兩條相交直線,所以CD⊥平面BDM.
(II)設(shè)F、G分別為BC、BD的中點,連接B
1G、FG、B
1F,
則FG∥CD,F(xiàn)G=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/8.png)
CD.∴FG=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/9.png)
,F(xiàn)G⊥BD.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/images11.png)
由側(cè)面矩形BB
1A
1A的對角線的交點為D,知BD=B
1D=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/10.png)
A
1B=1,
所以△BB
1D是邊長為1的正三角形,于是B
1G⊥BD,B
1G=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/11.png)
,
∴∠B
1GF是所求二面角的平面角.
又B
1F
2=B
1B
2+BF
2=1+(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/12.png)
)
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/13.png)
.
∴cos∠B
1GF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/14.png)
.
即所求二面角的大小為π-arccos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/15.png)
.
法二:如圖以C為原點建立坐標系.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/images18.png)
(I)B(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/16.png)
,0,0),B
1(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/17.png)
,1,0),A
1(0,1,1),D(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/20.png)
),
M(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/21.png)
,1,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/22.png)
=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/25.png)
),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/26.png)
=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/27.png)
,-1,-1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/28.png)
=(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/29.png)
,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/30.png)
),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/31.png)
,
∴CD⊥A
1B,CD⊥DM.
因為A
1B、DM為平面BDM內(nèi)兩條相交直線,
所以CD⊥平面BDM.
(II)設(shè)BD中點為G,連接B
1G,
則G
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/32.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/33.png)
=(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/35.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/36.png)
),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/38.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/39.png)
,∴BD⊥B
1G,
又CD⊥BD,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/40.png)
與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/41.png)
的夾角θ等于所求二面角的平面角,
cos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/42.png)
所以所求二面角的大小為π-arccos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810019_DA/43.png)
.
點評:本題考查直線與平面的垂直判定,二面角的求法,考查空間想象能力,邏輯思維能力,是中檔題.