题目内容
如图,在直三棱柱ABC—A1B1C1中,AB⊥BC,P为A1C1的中点,AB=BC=kPA。
(I)当k=1时,求证PA⊥B1C;
(II)当k为何值时,直线PA与平面BB1C1C所成的角的正弦值为
,并求此时二面角A—PC—B的余弦值。![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549539277217.jpg)
(I)当k=1时,求证PA⊥B1C;
(II)当k为何值时,直线PA与平面BB1C1C所成的角的正弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953896221.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549539277217.jpg)
(I)证明略
(II)二面角A—PC—B的余弦值是![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
(II)二面角A—PC—B的余弦值是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
(方法一)
(I)连接B1P,因为在直三棱柱ABC—A1B1C1中,P为A1C1的中点,
AB=BC,所以B1P⊥面A1C。
所以B1P⊥AP。
又因为当k=1时,
AB=BC=PA=PC,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953959636.gif)
∴AP⊥PC。
∴AP⊥平面B1PC,
∴PA⊥B1C。
(II)取线段AC中点M,线段BC中点N,
连接MN、MC1、NC1,
则MN//AB,∵AB⊥平面B1C,∴MN⊥平面B1C,
是直线PA与平面BB1C1C所成的角,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953990715.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954005775.gif)
设AB=a,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315495405272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954083770.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954099693.gif)
即
时,直线PA与平面BB1C1C所成的角的正弦值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954130232.gif)
此时,过点M作MH,垂足为H,连接BH,
,
由三垂线定理得BH⊥PC,
所以
是二面角A—PC—B的平面角。
设AB=2,则BC=2,PA=-4,
,
在直角三角形中AA1P中
,
连接MP,在直角三角形中
由
,
又由
,在直角三角形中BMH中,
解得
,
在直角三角形BMH中
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549543021220.gif)
所以二面角A—PC—B的余弦值是![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
(方法二)
以点B为坐标原点,分别以直线BA、BC、BB1为x轴、y轴建立空间直角坐标
系Oxyz,
(I)设AB=2,则AB=BC=PA=2
根据题意得:![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549543641077.gif)
所以![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954395836.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549544585512.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954489871.gif)
(II)设AB=2,则
,
根据题意:A(2,0,0),C(0,2,0)
又因为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954520624.gif)
所以
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549545511721.gif)
所以由题意得![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954583669.gif)
即![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549545981270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954614501.gif)
即
时,直线PA与平面BB1C1C所成的角的正弦值为![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954130232.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954661390.gif)
的法
向量![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954723435.gif)
设平面BPC的一个法向量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954739981.gif)
由
,得
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549548011395.gif)
所以此时二面角A—PC—B的余弦值是![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
(I)连接B1P,因为在直三棱柱ABC—A1B1C1中,P为A1C1的中点,
AB=BC,所以B1P⊥面A1C。
所以B1P⊥AP。
又因为当k=1时,
AB=BC=PA=PC,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953959636.gif)
∴AP⊥PC。
∴AP⊥平面B1PC,
∴PA⊥B1C。
(II)取线段AC中点M,线段BC中点N,
连接MN、MC1、NC1,
则MN//AB,∵AB⊥平面B1C,∴MN⊥平面B1C,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953974471.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953990715.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954005775.gif)
设AB=a,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315495405272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954083770.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954099693.gif)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954115294.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954130232.gif)
此时,过点M作MH,垂足为H,连接BH,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954146583.gif)
由三垂线定理得BH⊥PC,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954193423.gif)
设AB=2,则BC=2,PA=-4,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954208701.gif)
在直角三角形中AA1P中
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954224696.gif)
连接MP,在直角三角形中
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954239959.gif)
又由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954255464.gif)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954271411.gif)
在直角三角形BMH中
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549543021220.gif)
所以二面角A—PC—B的余弦值是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
(方法二)
以点B为坐标原点,分别以直线BA、BC、BB1为x轴、y轴建立空间直角坐标
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315495433372.gif)
(I)设AB=2,则AB=BC=PA=2
根据题意得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549543641077.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954395836.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549544585512.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954489871.gif)
(II)设AB=2,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954505432.gif)
根据题意:A(2,0,0),C(0,2,0)
又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954520624.gif)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954536849.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549545511721.gif)
所以由题意得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954583669.gif)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549545981270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954614501.gif)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954115294.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954130232.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954661390.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954692355.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315495405272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954723435.gif)
设平面BPC的一个法向量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954739981.gif)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954770627.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154954785715.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231549548011395.gif)
所以此时二面角A—PC—B的余弦值是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823154953943364.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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