Question

在三角形ABC中,AB=根号三,AC=2,若O为△ABC内部一点,且 向量OA+OB+OC=0向量,则向量AO*BC=?

Answer 1

BC = (BA+AC)

AO.BC = AO.(BA+AC)

= (OB+OC).(BA+AC) ( AO=OB+OC)

=( OA+AB+OA+AC)(BA+AC)

= 2OA.(BA+AC) + |AC|² -|AB|²

= 2OA.BC + 4-3

- OA.BC = 2OA.BC + 1

3OA.BC = -1

OA.BC = -1/3