Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP.OQ + OR.OS = OR.OP + OQ.OS = OQ.OR + OP.OS Then the triangle PQR has S as its
OP.OQ + OR.OS = OR.OP + OQ.OS
⇒OP(OQ-OR)+OS(OR-OQ) = 0
⇒SP.RQ = 0
Similarly SR.PQ = 0 and SQ.PR = 0
∴S is orthocentre.