...PB⊥BC,PD⊥CD,且PA=2,E为PD中点.(Ⅰ)求证:PA
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发布时间：2024-10-22 07:46
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时间：2024-10-23 03:19
解：（Ⅰ）证明：∵底面ABCD为正方形，
∴BC⊥AB，又BC⊥PB，
∴BC⊥平面PAB，
∴BC⊥PA．（2分）
同理CD⊥PA，（4分）
∴PA⊥平面ABCD．
（Ⅱ）解：设M为AD中点，连接EM，
又E为PD中点，
可得EM∥PA，从而EM⊥底面ABCD．
过M作AC的垂线MN，垂足为N，连接EN．
由三垂线定理有EN⊥AC，
∴∠ENM为二面角E-AC-D的平面角．（7分）
在Rt△EMN中，可求得EM＝1，MN＝22，
∴tanENM＝EMMN＝2．（9分）
∴二面角E-AC-D的大小为arctan2．（10分）
（Ⅲ）解：由E为PD中点可知，
要使得点E到平面PAF的距离为255，即要点D到平面PAF的距离为455．
过D作AF的垂线DG，垂足为G，
∵PA⊥平面ABCD，∴平面PAF⊥平面ABCD，
∴DG⊥平面PAF，即 DG为点D到平面PAF的距离．
∴DG＝455，∴AG＝2已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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