Q57 / 150 · MHT CET 2022 (10 Aug Shift 1) · Mathematics

Let $O$ be the origin and let $P Q R$ be an arbitrary triangle. The point $S$ $\overline{O P} \cdot \overline{O Q}+\overline{O R} \cdot \overline{O S}=\overline{O R} \cdot \overline{O P}+\overline{O Q} \cdot \overline{O S}=\overline{O Q} \overline{O Q} \cdot \overline{O R}+\overline{O P} \cdot \overline{O S}$ that $\overline{O P} \cdot \overline{O Q}+\overline{O R} \cdot \overline{O S}=\overline{O R} \cdot \overline{O P}+\overline{O Q} \cdot \overline{O S}=\overline{O Q} \cdot \overline{O R}+\overline{O P} \cdot \overline{O S}$, then the triangle $P Q R$ has $S$ as its

  • A
    Incentre.
  • B
    Centroid.
  • Correct answer
    C
    Orthocentre.
  • D
    Circumcentre.

Solution

$\begin{aligned} & \overrightarrow{O P} \cdot \overrightarrow{O Q}+\overrightarrow{O R} \cdot \overrightarrow{O S}=\overrightarrow{O R} \cdot \overrightarrow{O P}+\overrightarrow{O Q} \cdot \overrightarrow{O S} \\ & \Rightarrow \overrightarrow{O P} \cdot(\overrightarrow{O Q}-\overrightarrow{O R})=\overrightarrow{O S} \cdot(\overrightarrow{O Q}-\overrightarrow{O R}) \\ & \Rightarrow \overrightarrow{O P} \cdot \overrightarrow{R Q}=\overrightarrow{O S} \cdot \overrightarrow{R Q} \\ & \Rightarrow \overrightarrow{R Q} \cdot(\overrightarrow{O P}-\overrightarrow{O S})=0 \\ & \Rightarrow \overrightarrow{R Q} \cdot \overrightarrow{P S}=0 \\ & \Rightarrow \overrightarrow{P S} \perp \overrightarrow{Q R}\end{aligned}$ Similarly $\overrightarrow{Q S} \perp \overrightarrow{P R}$ and $\overrightarrow{R S} \perp \overrightarrow{P Q}$ i.e., $S$ is the orthocenter

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

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