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22:4918、(1)证明: FA =1, AD=2, ABI/CD. FB=1, BC=2又AEI/ BC,CD=3√2. DE=2√2, EC=√2ECB=45°.. BE² =BC² +CE²-2BC·CE·cos45°=√2.1分又AB=√2,AE=FC=23: BE²+AB²=AE², BE ⊥ AB:面DAE⊥面AFCE且交于AE,又DA⊥AE:DA⊥面AFCE:DA⊥BE.....4分:DA∩AB= A:BE⊥面DAB:BE面BDE:面BDE⊥面ABD(2)解:DA⊥面AFCE,AF⊥AE.建立如图所示空间直角坐标系A(0,0,0), F(1,0,0),E(0,2,0), D(0,0,2),B(1,1,0), C(1,3,0) .....7分→→←DF = (1,0,-2),ED =(0,- 2,2),DC = (1,3,- 2)设面EDC的法向量为m=(x,y,z),则{-2y+2z=0. m = (-1,1,1).9分[x+3y-2z=0设直线DF与面EDC所成的角为θ,则←[1-2]_√15sinθ=cos
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