????????y?2z?0?y?2z?n?PC?0????则????? 取z?1?n?(1,2,1) ?CD?0?2x?y?0?x?z??n?????AD?(2,0,0)是平面PAC的法向量

???????????????AD?n630 cos?AD,n?????????sin?AD,n??66ADn得：二面角A?PC?D的正弦值为30 6????11????????（3）设AE?h?[0,2]；则AE?(0,0,2)，BE?(,?,h),CD?(2,?1,0)

22????????????????BE?CD331010cos?BE,CD???????h? 即AE? ??????221010BECD10?20h12. 【解析】(1)在Rt?DAC中,AD?AC

得:?ADC?45?

??同理:?A1DC1?45??CDC1?90

得:DC1?DC,DC1?BD?DC1?面BCD?DC1?BC (2)DC1?BC,CC1?BC?BC?面ACC1A1?BC?AC 取A1B1的中点O,过点O作OH?BD于点H,连接C1O,C1H

AC11?B1C1?C1O?A1B1,面A1B1C1?面A1BD?C1O?面A1BD OH?BD?C1H?BD 得:点H与点D重合

且?C1DO是二面角A1?BD?C1的平面角 设AC?a,则C1O?2a?,C1D?2a?2C1O??C1DO?30 2?既二面角A1?BD?C1的大小为30

13. 解:设AC?BD?O,以O为原点,OC为x轴,OD为y轴建立空间直角坐标系,则A(?2,0,0),C(2,0,0),P(?2,0,2),设B(0,?a,0),D(0,a,0),E(x,y,z).

????2222????E(,0,)BE?(,a,)33, 所以PC?(22,0,?2),33,(Ⅰ)证明:由PE?2EC得

????????22????PC?BE?(22,0,?2)?(,a,)?0BD?(0,2a,0),所以33,

????????????????????????PC?BD?(22,0,?2)?(0,2a,0)?0.所以PC?BE,PC?BD,所以PC?平面

BED;

?????????(Ⅱ) 设平面PAB的法向量为n?(x,y,z),又AP?(0,0,2),AB?(2,?a,0),由?2????????????n?(1,,0)n?AP?0,n?AB?0得a,设平面PBC的法向量为m?(x,y,z),又??2????????????????????m?(1,?,2)BC?(2,a,0),CP?(?22,0,2),由m?BC?0,m?CP?0,得a,

????由于二面角A?PB?C为90,所以m?n?0,解得a?2. ??????所以PD?(2,2,?2),平面PBC的法向量为m?(1,?1,2),所以PD与平面

????????|PD?m|1???????????PBC所成角的正弦值为|PD|?|m|2,所以PD与平面PBC所成角为6.

14. 解:(1)由AC?BC,D为AB的中点,得CD?AB,又CD?AA1,故

CD?面A1ABB1,所以点C到平面A1ABB1的距离为CD?BC2?BD2?5 (2)如图,取D1为A1B1的中点,连结DD1,则DD1∥AA1∥CC1,又由(1)知

CD?面A1ABB1,故CD?A1DCD?DD1,所以?A1DD1为所求的二面角A1?CD?C1的平面角.

因A在面A1D为AC11ABB1上的射影,又已知AB1?AC1,由三垂线定理的逆定理得都与?B1AB互余,因此?A,所以AB1?A1D,从而?A1AB1,?A1DA1AB1??A1DA,因此,Rt?A?B1AD?Rt1A1AAA1A1B12,即AA?A1B1?8,得AA1?22. 1?AD?ADAA1,

所

以

,

在

从而

A1D?AA12?AD2?23Rt?A1DD1中,

cosA1DD1?

DD1AA16 ??A1DA1D3