18【解析】(1)∵平面PAD⊥平面ABCD,△PAD为等腰直角Tn=b1+b2+…+bn形,PA=PD,E为AD的中点,∴PE⊥AD,则PE⊥平面ABCD,所以PE⊥AE,PE⊥BE,又DC∥BE,AD⊥CD,∴BE⊥AD,∴可以EA、EB、EP分别为x、y、z轴建立空间C直角坐标系E-xyz,如图所示,则A(a,0,0),D(-a,0,0),B(0,a,0),C(-a,a,0),P(0,0,a),Q(0,,a).B35,(E0)ADQ=(a,t,a),EC=(-a,a,0),∴t=a,于是Q(0,a,a),∴四边形PQBE为矩形,(4)=(D-+(-1+)长∵DQ⊥EC,∴DQ=0,即-a2+at=0,分故|PQ|=a.……18(2)设点F为AB中点,连结EF,QB⊥平面ABCD,∴QB⊥EF,而△AEB为等腰直角三角形,EF⊥AB,∴EF⊥平面ABQ,E为平面ABQ的一个法向量,而EF=+(1-54):发(=)人升设n=(x,y,x)为平面ADQ的一个法向量,而DA=(2a,0,0),D=(a,a,a),1)(n·DA=0(2ax=00又即令z=-1,则y=1,更为n·D=0ax+ay+az=0(y+z=0离等于知位∵n=(0,1(8-,0)宝女用Qqa1(x)(1)【设二面角B-AQ-D的平面角为0,则cos0√2两单√2二面角B-AQ-D的平面角为6=60°0)(),立0≤()0≤(x)3,8≥

书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!

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