Exam

JEE Advanced

Year

2017

Paper

Paper 2

Subject

Mathematics

Chapter

Vector Algebra

JEE Advanced 2017 · Paper 2 · Mathematics · Vector Algebra

Let O be the origin and let PQR be an arbitrary triangle. The point...

Let $O$ be the origin and let $PQR$ be an arbitrary triangle. The point$S$ is such that $\overset{\to}{OP} . \overset{\to}{OQ} + \overset{\to}{OR} . \overset{\to}{OS} = \overset{\to}{OR} . \overset{\to}{OP} + \overset{\to}{OQ} . \overset{\to}{OS} = \overset{\to}{OQ} . \overset{\to}{OR} + \overset{\to}{OP} . \overset{\to}{OS}$ then triangle $PQR$ has $S$ as its

Step-by-step solution

Let position vector of  $P\left(\overset{\to}{p}\right), Q\left(\overset{\to}{q}\right), R\left(\overset{\to}{r}\right)andS\left(\overset{\to}{r}\right)$  with respect to $O\left(\overset{\to}{o}\right)$ Now, $\overset{\to}{OP} . \overset{\to}{OQ} + \overset{\to}{OR} . \overset{\to}{OS} = \overset{\to}{OR} . \overset{\to}{OP} + \overset{\to}{OQ} . \overset{\to}{OS}$ $⇒ \overset{\to}{p} . \overset{\to}{q} + \overset{\to}{r} . \overset{\to}{s} = \overset{\to}{r} . \overset{\to}{p} + \overset{\to}{q} . \overset{\to}{s}$ $⇒ \left(\overset{\to}{p} - \overset{\to}{s}\right) . \left(\overset{\to}{q} - \overset{\to}{r}\right) = 0$ .....(i) Also, $\overset{\to}{OR} . \overset{\to}{OP} + \overset{\to}{OQ} . \overset{\to}{OS} = \overset{\to}{OQ} . \overset{\to}{OR} + \overset{\to}{OP} . \overset{\to}{OS}$ $⇒ \overset{\to}{r} . \overset{\to}{p} + \overset{\to}{q} . \overset{\to}{s} = \overset{\to}{q} . \overset{\to}{r} + \overset{\to}{p} . \overset{\to}{s}$ $⇒ \left(\overset{\to}{r} - \overset{\to}{s}\right) . \left(\overset{\to}{p} - \overset{\to}{q}\right) = 0$ .....(ii) Also $\overset{\to}{OP} . \overset{\to}{OQ} + \overset{\to}{OR} . \overset{\to}{OS} = \overset{\to}{OQ} . \overset{\to}{OR} + \overset{\to}{OP} . \overset{\to}{OS}$ $⇒ \overset{\to}{p} . \overset{\to}{q} + \overset{\to}{r} . \overset{\to}{s} = \overset{\to}{q} . \overset{\to}{r} + \overset{\to}{p} . \overset{\to}{s}$ $⇒ \left(\overset{\to}{q} - \overset{\to}{s}\right) . \left(\overset{\to}{p} - \overset{\to}{r}\right) = 0$ .....(iii) $⇒$ Triangle PQR has S as its orthocenter $∴$ option (Orthocentre) is correct.