题目内容
(本小题12分)
如图,正方体ABCD—A1B1C1D1中,M、N分别为AB、BC的中点.
(Ⅰ)求证:平面B1MN⊥平面BB1D1D;
(II)当点P为棱DD1中点时,求直线MB1与平面A1C1P所成角的正弦值;
如图,正方体ABCD—A1B1C1D1中,M、N分别为AB、BC的中点.
(Ⅰ)求证:平面B1MN⊥平面BB1D1D;
(II)当点P为棱DD1中点时,求直线MB1与平面A1C1P所成角的正弦值;
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231627211745160.jpg)
(1)略
(2)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721189348.gif)
(法一)(I)正方体ABCD—A1B1C1D1中,
平面ABCD,
平面ABCD,
,连结AC,
M、N分别为AB、BC
的中点,
MN//AC,又四边形ABCD是正方形,
,
平面BB1D1D,
又
平面B1MN,
平面B1MN
平面BB1D1D (6分)![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/2014082316272142373.gif)
(II)设正方体棱长为2,取CD中点H,连C1H、MH,由于MH∥C1B1,MH=C1B1,所以四边形C1HM B1为平行四边形, M B1∥C1H,所以直线C1H与平面A1C1P所成角θ即为直线MB1与平面A1C1P所成角。设H到平面的距离为h,∵P为DD1中点,所以A1P=C1P=
,
,
,由
得,h=
,所以sinθ=
; ( 12分)
(法二)
以D为原点,DA、DC、DD1所在直线分别为x轴、y轴、z轴建立空间直角坐标系,如图所示,则D(0,0,0),A(2,0,0),B(2,2,0),C(0,2,0),D1(0,0,2),A1(2,0,2),B1(2,2,2),C1(0,2,2),P(0,0,x),M(2,1,0),N(1,2,0)
(I)
,
,
,∴
,
,
,
,
,
平面BB1D1D,又
平面B1MN,
平面B1MN
平面BB1D1D (6分)![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231627220016598.jpg)
(II)设
为平面平面A1C1P的一个法向量,
P为DD1中点,P(0,0,1),
,
,则
,也就是
,
,令
,
又
,设MB1与平面A1C1P所成角为θ,
则![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722188693.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722203383.gif)
, ( 12分
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721205276.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721221391.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721236504.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721267183.gif)
的中点,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721283128.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721314655.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721330663.gif)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721345416.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721283128.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721408108.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/2014082316272142373.gif)
(II)设正方体棱长为2,取CD中点H,连C1H、MH,由于MH∥C1B1,MH=C1B1,所以四边形C1HM B1为平行四边形, M B1∥C1H,所以直线C1H与平面A1C1P所成角θ即为直线MB1与平面A1C1P所成角。设H到平面的距离为h,∵P为DD1中点,所以A1P=C1P=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721439172.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721455754.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721486904.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721501544.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721517282.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721533336.gif)
(法二)
以D为原点,DA、DC、DD1所在直线分别为x轴、y轴、z轴建立空间直角坐标系,如图所示,则D(0,0,0),A(2,0,0),B(2,2,0),C(0,2,0),D1(0,0,2),A1(2,0,2),B1(2,2,2),C1(0,2,2),P(0,0,x),M(2,1,0),N(1,2,0)
(I)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721267183.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721579556.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721595426.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721626442.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721798545.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721813567.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721845494.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721860468.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721891419.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721923404.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721345416.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721283128.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721408108.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231627220016598.jpg)
(II)设
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722016494.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721267183.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722063557.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722079563.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722094764.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722110700.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722125543.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722141238.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722157468.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722172577.gif)
则
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722188693.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162722203383.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823162721189348.gif)
![](http://thumb.zyjl.cn/images/loading.gif)
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