Question

在三角形ABC中,∠BAC=120°AB=2,AC=1,D是边上一点,DC=2BD,求向量AD×向量BC的值

Answer 1

BC^2=AB^2+AC^2-2AB*AC*cos∠BAC=5+2=7
BC=根号7
AC/sinB=BC/sin∠BAC
sinB=(AC/BC)sin∠BAC=(1/2)(根号3)/2=(根号3)/4
cosB=(根号13)/4
向量AD×向量BC=(向量AB-向量BD)*向量BC
=向量AB*向量BC-BD*BC
=AB*BC*cosB-(1/3)BC^2
=2*(根号7)*(根号13)/4-(7/3)
=((根号91)/2)-(7/3)