SATHEE: Vectors 1 Question 5

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

$\begin{aligned} O P \cdot O Q + O R \cdot O S & = O R \cdot O P + O Q \cdot O S \\ = O Q \cdot O R + O P \cdot O S \end{aligned}$

Then the triangle $P Q R$ has $S$ as its

(2017 Adv.)

(a) centroid

(b) orthocentre

(c) incentre

(d) circumcentre

Show Answer

Answer:

Correct Answer: 6. (b)

Solution:

$O P \cdot O Q + O R \cdot O S = O R \cdot O P + O Q \cdot O S$

$\Rightarrow O P (O Q - O R) + O S (O R - O Q) = 0$

$\Rightarrow (O P - O S) (O Q - O R) = 0$

$\Rightarrow S P \cdot R Q = 0$

Similarly $S R \cdot P Q = 0$ and $S Q \cdot P R = 0$

$∴ S$ is orthocentre.

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Vectors 1 Question 5

6. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

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