Question

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

Answer 1

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