Let O be the origin and let P Q R be an arbitrary triangle. The point S is such that O P → . O Q…

1. Incentre
2. Orthocentre
3. Circumcentre
4. 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.

Asked in: JEE Advanced 2017 (Paper 2)

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