如图,建立空间直角坐标系A-xyz.∵PA⊥平面ABCD,PB与平面ABC成60°,∴∠PBA=60°,∴PA=ABtan60°= 3 .取AB=1,则A(0,0,0),B(1,0,0),C(1,1,0),P(0,0, 3 ),D(0,2,0).(1)∵ AC =(1,1,0), AP =(0,0, 3 ), CD =(-1,1,0),∴ AC ? CD =-1+1+0=0,