...在四棱锥P-ABCD中,PA⊥底面ABCD,AD⊥AB,AB∥DC,AD=DC=AP=2,AB=...
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发布时间:2024-10-24 17:28
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时间:2024-10-28 06:16
证明:(I)∵PA⊥底面ABCD,AD⊥AB,
以A为坐标原点,建立如图所示的空间直角坐标系,
∵AD=DC=AP=2,AB=1,点E为棱PC的中点.
∴B(1,0,0),C(2,2,0),D(0,2,0),P(0,0,2),E(1,1,1)
∴BE=(0,1,1),DC=(2,0,0)
∵BE?DC=0,
∴BE⊥DC;
(Ⅱ)∵BD=(-1,2,0),PB=(1,0,-2),
设平面PBD的法向量m=(x,y,z),
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时间:2024-10-28 06:20
碰到这么标准的,直接以A为原点,AB,AD,AP为x,y,z轴建坐标系来做吧