题目内容
(本小题12分)四棱锥P-ABCD中,PA⊥底面ABCD,AB∥CD,AD=CD=1,∠BAD=120°,PA=
,∠ACB=90°。
(1)求证:BC⊥平面PAC;
(2)求二面角D-PC-A的大小的正切值;
(3)求点B到平面PCD的距离。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522696217.gif)
(1)求证:BC⊥平面PAC;
(2)求二面角D-PC-A的大小的正切值;
(3)求点B到平面PCD的距离。
(1)同解析(2)二面角D-PC-A的大小的正切值为2。(3)即点B到平面PCD的距离为
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231515227431188.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522712267.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231515227431188.gif)
解法一:(1)∵PA⊥底面ABCD,BC
平面AC,∴PA⊥BC
∵∠ACB=90°,∴BC⊥AC,又PA∩AC=A,∴BC⊥平面PAC
(2)∵AB∥CD,∠BAD=120°,∴∠ADC=60°,又AD=CD=1
∴ΔADC为等边三角形,且AC=1,取AC的中点O,则DO⊥AC,又PA⊥底面ABCD,
∴PA⊥DO,∴DO⊥平面PAC,过O作OH⊥PC,垂足为H,连DH
由三垂成定理知DH⊥PC,∴∠DHO为二面角D-PC-A的平面角
由OH=
,DO=
,∴tan∠DHO=
=2
∴二面角D-PC-A的大小的正切值为2。
(3)设点B到平面PCD的距离为d,又AB∥平面PCD
∴VA-PCD=VP-ACD,即![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522946706.gif)
∴
即点B到平面PCD的距离为
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231515230391333.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522883131.gif)
∵∠ACB=90°,∴BC⊥AC,又PA∩AC=A,∴BC⊥平面PAC
(2)∵AB∥CD,∠BAD=120°,∴∠ADC=60°,又AD=CD=1
∴ΔADC为等边三角形,且AC=1,取AC的中点O,则DO⊥AC,又PA⊥底面ABCD,
∴PA⊥DO,∴DO⊥平面PAC,过O作OH⊥PC,垂足为H,连DH
由三垂成定理知DH⊥PC,∴∠DHO为二面角D-PC-A的平面角
由OH=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522899255.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522915250.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522930317.gif)
∴二面角D-PC-A的大小的正切值为2。
(3)设点B到平面PCD的距离为d,又AB∥平面PCD
∴VA-PCD=VP-ACD,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522946706.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522961331.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151522712267.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231515230391333.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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