题目内容
(本小题满分13分)
已知:如图,长方体
中,
、
分别是棱
,
上的点,
,
.
(1) 求异面直线
与
所成角的余弦值;
(2) 证明![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352165238.gif)
平面![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352290207.gif)
;
(3) 求二面角
的正弦值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643528671484.gif)
已知:如图,长方体
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164351885407.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164351947220.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164351963218.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164351978242.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352056251.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352072352.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352103413.gif)
(1) 求异面直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352134237.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352150256.gif)
(2) 证明
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352165238.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352197215.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352290207.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352696276.gif)
(3) 求二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352711318.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643528671484.gif)
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353086238.gif)
(2)略
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353257276.gif)
解:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643533985991.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643534131471.gif)
法一:
如图所示,以点A为坐标原点,建立空间直角坐标系,
设
,
依题意得
,
,
,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353710392.gif)
(1)易得
,
,
于是![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353881783.gif)
所以异面直线
与
所成角的余弦值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353086238.gif)
(2)已知
,
,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354022446.gif)
于是
·
=0,
·
=0.
因此,
,
,又![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354381349.gif)
所以
平面![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352696276.gif)
(3)设平面
的法向量
,则
,即![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354661551.gif)
不妨令X=1,可得
。
由(2)可知,
为平面
的一个法向量。
于是
,从而
,
所以二面角
的正弦值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353257276.gif)
法二:
(1)设AB=1,可得AD=2,AA1=4,CF=1.CE=![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354911240.gif)
连接B1C,BC1,设B1C与BC1交于点M,易知A1D∥B1C,
由
,可知EF∥BC1.
故
是异面直线EF与A1D所成的角,
易知BM=CM=
,
所以
,
所以异面直线FE与A1D所成角的余弦值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353086238.gif)
(2)连接AC,设AC与DE交点N 因为
,
所以
,从而
,
又由于
,所以
,
故AC⊥DE,又因为CC1⊥DE且
,所以DE⊥平面ACF,从而AF⊥DE.
连接BF,同理可证B1C⊥平面ABF,从而AF⊥B1C,
所以AF⊥A1D因为
,所以AF⊥平面A1ED.
(3)连接A1N.FN,由(2)可知DE⊥平面ACF,
又NF
平面ACF, A1N
平面ACF,所以DE⊥NF,DE⊥A1N,
故
为二面角A1-ED-F的平面角.
易知
,所以
,
又
所以
,
在![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355863664.gif)
,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355894599.gif)
连接A1C1,A1F 在![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164356065683.gif)
。所以![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164356112424.gif)
所以二面角A1-DE-F正弦值为
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643533985991.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231643534131471.gif)
法一:
如图所示,以点A为坐标原点,建立空间直角坐标系,
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353460263.gif)
依题意得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353476311.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353663298.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353679320.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353710392.gif)
(1)易得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353725441.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353866371.gif)
于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353881783.gif)
所以异面直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352134237.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352150256.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353086238.gif)
(2)已知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353991341.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354006468.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354022446.gif)
于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354053246.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354084260.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354053246.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354271246.gif)
因此,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354349317.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354365296.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354381349.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354396254.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164352696276.gif)
(3)设平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354443259.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354599321.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354630466.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354661551.gif)
不妨令X=1,可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354677321.gif)
由(2)可知,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354053246.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354771277.gif)
于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354817635.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354833461.gif)
所以二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354849306.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353257276.gif)
法二:
(1)设AB=1,可得AD=2,AA1=4,CF=1.CE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354911240.gif)
连接B1C,BC1,设B1C与BC1交于点M,易知A1D∥B1C,
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164354927447.gif)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355005287.gif)
易知BM=CM=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355036363.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355051736.gif)
所以异面直线FE与A1D所成角的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353086238.gif)
(2)连接AC,设AC与DE交点N 因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355254434.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355285411.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355301359.gif)
又由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355317409.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355488411.gif)
故AC⊥DE,又因为CC1⊥DE且
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355582350.gif)
连接BF,同理可证B1C⊥平面ABF,从而AF⊥B1C,
所以AF⊥A1D因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355675355.gif)
(3)连接A1N.FN,由(2)可知DE⊥平面ACF,
又NF
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355691211.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355691211.gif)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355769293.gif)
易知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355785408.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355800393.gif)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355831303.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355847348.gif)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355863664.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355878377.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164355894599.gif)
连接A1C1,A1F 在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164356065683.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164356081941.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164356112424.gif)
所以二面角A1-DE-F正弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823164353257276.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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