四棱锥P-ABCD中,底面ABCD为梯形,AB//CD,AB垂直BC,PC垂直AD,PA垂直底面...答:设AB=1,(1)连结BD交AC于O,则PD‖OE,由PC⊥AD,得AC⊥AD,易求得CD=2,∴PE/BE=OD/OB=CD/AB=2,即PE=2EB. (2)过E作EF⊥AB于F,过F作FH⊥DH于H。∠EHF即为所求。易求得EF=1/3,FH=√2/3,∴tan∠EHF=√2/2
...面ABCD为直角梯形,AD‖BC,AB⊥BC,AB=AD=PB=3.点E在棱P答:过D点作y轴的垂线,垂足为M.设CM=x.∵AD⊥AB,AD=AB=3 ∴BD=三倍根号二 在三角形PBD中,由勾股定理得PD=根号下27.由已知得CD^2=x^2+9,PC^2=(x+3)^2+9 ∵PD⊥CD ∴由勾股定理得27+x^2+9=(x+3)^2+9 解得x=3,即CB=6.C(0,-6,0)∴向量PA=(3,0,-3),向量CD=(3,3...
如图,在四棱锥P-ABCD中,PA⊥底面ABCD,底面ABCD为梯形,AB∥DC,∠ABC=...答:(1) PE= PB (2) (1)在梯形ABCD中,由题知AB⊥BC,AB=BC,∴AC= AB,∠BAC= ,∴∠DCA=∠BAC= .又∠CAD=90°,∴△DAC为等腰直角三角形.∴DC= AC= ( AB)=2AB.连接BD,交AC于点M,连接ME, ∵AB∥DC,∴ = =2.∵PD∥平面EAC,又平面EAC∩平面PDB=ME,∴PD∥EM...