【答案】
分析:方法一:(1)取OB中點(diǎn)E,連接ME,NE,證明平面MNE∥平面OCD,方法是兩個(gè)平面內(nèi)相交直線互相平行得到,從而的到MN∥平面OCD;
(2)∵CD∥AB,∴∠MDC為異面直線AB與MD所成的角(或其補(bǔ)角)作AP⊥CD于P,連接MP
∵OA⊥平面ABCD,∴CD⊥MP菱形的對(duì)角相等得到∠ABC=∠ADC=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/0.png)
,
利用菱形邊長(zhǎng)等于1得到DP=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/1.png)
,而MD利用勾股定理求得等于
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/2.png)
,在直角三角形中,利用三角函數(shù)定義求出即可.
(3)AB∥平面OCD,∴點(diǎn)A和點(diǎn)B到平面OCD的距離相等,連接OP,過(guò)點(diǎn)A作AQ⊥OP于點(diǎn)Q,
∵AP⊥CD,OA⊥CD,∴CD⊥平面OAP,∴AQ⊥CD,
又∵AQ⊥OP,∴AQ⊥平面OCD,線段AQ的長(zhǎng)就是點(diǎn)A到平面OCD的距離,求出距離可得.
方法二:(1)分別以AB,AP,AO所在直線為x,y,z軸建立坐標(biāo)系,分別表示出A,B,O,M,N的坐標(biāo),
求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/5.png)
的坐標(biāo)表示.設(shè)平面OCD的法向量為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/6.png)
=(x,y,z),則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/7.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/8.png)
,∴MN∥平面OCD
(2)設(shè)AB與MD所成的角為θ,表示出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/9.png)
和
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/10.png)
,利用a•b=|a||b|cosα求出叫即可.
(3)設(shè)點(diǎn)B到平面OCD的距離為d,則d為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/11.png)
在向量
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/12.png)
上的投影的絕對(duì)值,由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/13.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/14.png)
.所以點(diǎn)B到平面OCD的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/15.png)
.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/images16.png)
解:方法一(綜合法)
(1)取OB中點(diǎn)E,連接ME,NE
∵M(jìn)E∥AB,AB∥CD,∴ME∥CD
又∵NE∥OC,∴平面MNE∥平面OCD∴MN∥平面OCD
(2)∵CD∥AB,∴∠MDC為異面直線AB與MD所成的角(或其補(bǔ)角)
作AP⊥CD于P,連接MP
∵OA⊥平面ABCD,∴CD⊥MP
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/16.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/18.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/19.png)
所以AB與MD所成角的大小為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/20.png)
.
(3)∵AB∥平面OCD,
∴點(diǎn)A和點(diǎn)B到平面OCD的距離相等,連接OP,過(guò)點(diǎn)A作AQ⊥OP于點(diǎn)Q,
∵AP⊥CD,OA⊥CD,
∴CD⊥平面OAP,∴AQ⊥CD.
又∵AQ⊥OP,∴AQ⊥平面OCD,線段AQ的長(zhǎng)就是點(diǎn)A到平面OCD的距離,
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/21.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/22.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/23.png)
,所以點(diǎn)B到平面OCD的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/24.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/images26.png)
方法二(向量法)
作AP⊥CD于點(diǎn)P,如圖,分別以AB,AP,AO所在直線為x,y,z軸建立坐標(biāo)系:
A(0,0,0),B(1,0,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/26.png)
,
O(0,0,2),M(0,0,1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/27.png)
(1)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/28.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/29.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/30.png)
設(shè)平面OCD的法向量為n=(x,y,z),則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/31.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/32.png)
=0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/33.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/34.png)
=0
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/35.png)
取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/36.png)
,解得
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/37.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/38.png)
=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/39.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/40.png)
,-1)•(0,4,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/41.png)
)=0,
∴MN∥平面OCD.
(2)設(shè)AB與MD所成的角為θ,
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/42.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/43.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/44.png)
,AB與MD所成角的大小為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/45.png)
.
(3)設(shè)點(diǎn)B到平面OCD的距離為d,則d為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/46.png)
在向量
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/47.png)
=(0,4,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/48.png)
)上的投影的絕對(duì)值,
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/49.png)
,得d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/50.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/51.png)
所以點(diǎn)B到平面OCD的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466235510/SYS201310241810384662355016_DA/52.png)
.
點(diǎn)評(píng):培養(yǎng)學(xué)生利用多種方法解決數(shù)學(xué)問(wèn)題的能力,考查學(xué)生利用空間向量求直線間的夾角和距離的能力.