如图,在RTΔABC中,已知BC=a,若长为2a的线段PQ以点A为中点,向量PQ与BC的夹角θ取何值时,BP（向量）*C

CQ=CA+AQ
BP=BA+AP
BP*CQ=CA*BA+AQ*BA+CA*AP+AQ*AP
=0+1/2PQ*BA-1/2PQ*(BA-BC)-a^2
=1/2PQ*BC-a^2
=0.5*2a*a*cosθ-a^2
θ=0 或180