Vectors 1 Question 5
6. Let $O$ be the origin and let $P Q R$ be an arbitrary triangle. The point $S$ is such that
$ \begin{aligned} OP \cdot OQ+OR \cdot OS & =OR \cdot OP+OQ \cdot OS \\ & =OQ \cdot OR+OP \cdot OS \end{aligned} $
Then the triangle $P Q R$ has $S$ as its
(2017 Adv.)
(a) centroid
(b) orthocentre
(c) incentre
(d) circumcentre
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Answer:
Correct Answer: 6. (b)
Solution:
- $OP \cdot OQ+OR \cdot OS=OR \cdot OP+OQ \cdot OS$
$\Rightarrow \quad \mathbf{O P}(\mathbf{O Q}-\mathbf{O R})+\mathbf{O S}(\mathbf{O R}-\mathbf{O Q})=0$
$\Rightarrow \quad(\mathbf{O P}-\mathbf{O S})(\mathbf{O Q}-\mathbf{O R})=0$
$\Rightarrow \quad \mathbf{S P} \cdot \mathbf{R Q}=0$
Similarly $\mathbf{S R} \cdot \mathbf{P Q}=0$ and $\mathbf{S Q} \cdot \mathbf{P R}=0$
$\therefore S$ is orthocentre.
