已知平面内三个点A(0,-3),B(3,3)C,(1,-1),则向量AB与BC的夹角为

题目

已知平面内三个点A(0,-3),B(3,3)C,(1,-1),则向量AB与BC的夹角为
A.0 B.π/2 C.π/3 D.π
求详解,谢蛤a~

答案

解AB=(3,6)BC=(-2,-4)∴/AB/=√3²+6²=√45=3√5/BC/=√(-2)²+(-4)²=√20=2√5又AB*BC=3×(-2)+6×(-4)=-6-24=-30∴AB与BC的夹角为cos=AB*BC/(/AB//BC/)=(-30)/(3√5×2√5)=-1∴是π选D...