【答案】
分析:解法一:在含有直線與平面垂直垂直的條件的棱柱、棱錐、棱臺(tái)中,也可以建立空間直角坐標(biāo)系,設(shè)定參量求解.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(duì)位置的有關(guān)定理,因?yàn)檫@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?yàn)橹恍璁媯€(gè)草圖以建立坐標(biāo)系和觀察有關(guān)點(diǎn)的位置即可.設(shè)AB=a,則A
1(0,0,2a),C(0,a,0),C
1(0,a,2a),D(a,0,a)
(Ⅰ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/0.png)
=(a,-a,-a),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/1.png)
=(0,a,-2a)
(Ⅱ)又∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/2.png)
=(a,0,-a),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/3.png)
=(0,a,0),∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/4.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/6.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/7.png)
,∴A
1D⊥平面ACD
解法二:
(Ⅰ)求異面直線所成的角,可用幾何法,其基本解題思路是“異面化共面,認(rèn)定再計(jì)算”,即利用平移法和補(bǔ)形法將兩條異面直線轉(zhuǎn)化到同一個(gè)三角形中,結(jié)合余弦定理來求.連接AC
1交A
1C于點(diǎn)E,取AD中點(diǎn)F,連接EF,則EF∥C
1D,∴直線EF與A
1C所成的角就是異面直線C
1D與A
1C所成的角.
(Ⅱ)欲證平面A
1DC⊥平面ADC,先證直線與平面垂直,由題意可得:AC⊥A
1D,AD⊥A
1D,∴A
1D⊥平面ACD,又A
1D?平面A
1CD,∴平面A
1DC⊥平面ADC
解答:解:解法一:(Ⅰ)建立如圖所示的空間直角坐標(biāo)系設(shè)AB=a,
則A
1(0,0,2a),C(0,a,0),C
1(0,a,2a),D(a,0,a)(2分)
于是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/8.png)
=(a,-a,-a),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/9.png)
=(0,a,-2a)
∵cos<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/11.png)
>=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/14.png)
,(6分)
∴異面直線C
1D與A
1C所成的角為arccos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/15.png)
(7分)
(Ⅱ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/16.png)
=(a,0,-a),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/17.png)
=(0,a,0),
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/18.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/19.png)
=a
2+0-a
2=0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/20.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/21.png)
=0(10分)
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/22.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/24.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/25.png)
∴A
1D⊥平面ACD(12分)
又A
1D?平面A
1CD,
∴平面A
1DC⊥平面ADC(14分)
解法二:
(Ⅰ)連接AC
1交A
1C于點(diǎn)E,取AD中點(diǎn)F,連接EF,則EF∥C
1D
∴直線EF與A
1C所成的角就是異面直線C
1D與A
1C所成的角(2分)
設(shè)AB=a,
則C
1D=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/27.png)
a,
A
1C=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/29.png)
a,AD=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/31.png)
a.
△CEF中,CE=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/32.png)
A
1C=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/33.png)
a,EF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/34.png)
C
1D=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/35.png)
a,
直三棱柱中,∠BAC=90°,則AD⊥AC(4分)
CF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/38.png)
a(4分)
∵cos∠CEF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/39.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/40.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/41.png)
,(6分)
∴異面直線C
1D與A
1C所成的角為arccos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/42.png)
(7分)
(Ⅱ)直三棱柱中,∠BAC=90°,∴AC⊥平面ABB
1A
1,則AC⊥A
1D(9分)
又AD=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/43.png)
a,A
1D=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180430791407435/SYS201310241804307914074013_DA/44.png)
a,AA
1=2a,
則AD
2+A
1D
2=AA
12,于是AD⊥A
1D(12分)
∴A
1D⊥平面ACD又A
1D?平面A
1CD,
∴平面A
1DC⊥平面ADC(14分)
點(diǎn)評(píng):本小題主要考查空間線面關(guān)系、面面關(guān)系、二面角的度量,考查數(shù)形結(jié)合、化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及空間想象能力、推理論證能力和運(yùn)算求解能力.