Let O be the origin and let PQR be an arbitrary triangle. The point S is such that `bar(OP)*bar(OQ)+bar(OR)*bar(OS)=bar(OR)*bar(OP)+bar(OQ)*bar(OS)=bar(OQ)*bar(OR)+bar(OP)*bar(OS)` Then the triangle PQR has S as its

Options

A. centriod
B. circumectre
C. incente
D. orthocenter

Answer

D

Explanation

`vec(OP) . vec(OQ) + vec(OR).vec(OS)= vec(OR) -vec(OP)+vec(OQ).vec(OS).`
`=vec(OP).(vec(OQ)-vec(OR))=vec(OS).(vec(OQ)-vec(OR))`
`implies (vec(OP)-vec(OS)).(vec(RQ))=0implies vec(SP).vec(RQ)=0`
`implies vec(SP)botvec(RQ)`
Similarly ,`vec(SR)bot vec(OP) and vec(SQ) botvec(PR).`
hence , S is orthocenter.