Question

在三角形ABC中,AB=3,AC=5,若O为三角形ABC的外心,则向量AO乘向量BC=

Answer 1

设圆的半径为R,角AOB为a,角AOC为b

AB^2=AO^2+BO^2-2AO BOcosa=2R^2-2R^2cosa

AC^2=AO^2+CO^2-2AO COcosb=2R^2-2R^2cosb

AO*BC=AO*(BO+OC)=AO*BO-AO*CO=R^2cosa-R^2cosb=(AC^2-AB^2)/2=8