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

• A

  Incentre
• Correct answer
  B

  Orthocentre
• C

  Circumcentre
• D

  Centroid

Solution

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

Hence, the correct answer is option: 'B'.