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The Centroid

Let O be th...

Question

Let $O$ be the origin and let $P Q R$ be an arbitrary triangle. The point $S$ is such that $¯ ¯¯¯¯¯¯ ¯ O P . ¯ ¯¯¯¯¯¯¯ ¯ O Q + ¯ ¯¯¯¯¯¯ ¯ O R . ¯ ¯¯¯¯¯¯ ¯ O S = ¯ ¯¯¯¯¯¯ ¯ O R . ¯ ¯¯¯¯¯¯ ¯ O P + ¯ ¯¯¯¯¯¯¯ ¯ O Q . ¯ ¯¯¯¯¯¯ ¯ O S = ¯ ¯¯¯¯¯¯¯ ¯ O Q . ¯ ¯¯¯¯¯¯ ¯ O R + ¯ ¯¯¯¯¯¯ ¯ O P . ¯ ¯¯¯¯¯¯ ¯ O S$ . Then the $Δ$ PQR has $S$ as its

centroid

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circumcentre

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incentre

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orthocentre

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Solution

The correct option is D orthocentre
Given, $\to O P . \to O Q + \to O R . \to O S = \to O R . \to O P + \to O Q . \to O S = \to O Q . \to O R + \to O P . \to O S$

Consider, $\to O P . \to O Q + \to O R . \to O S = \to O R . \to O P + \to O Q . \to O S$

$\to O P . \to O Q - \to O R . \to O P = \to O Q . \to O S - \to O R . \to O S$

$\to O P (\to O Q - \to O R) = \to O S (\to O Q - \to O R)$

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

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

$\to Q R . \to P S = 0$

We know that

If dot product of 2 vectors is equal to $0$ , then the 2 vectors are perpendicular.

$∴ \to Q R$ is perpendicular to $\to P S$ ---------(i)

Consider,
$\to O P . \to O Q + \to O R . \to O S = \to O Q . \to O R + \to O P . \to O S$

Following the same steps as above, we get that

$\to Q S . \to P R = 0$

$∴ \to Q S$ is perpendicular to $\to P R$ ---------(ii)

Consider,
$\to O P . \to O Q + \to O R . \to O S = \to O Q . \to O R + \to O P . \to O S$

Following the same steps as above, we get that

$\to R S . \to P Q = 0$

$∴ \to R S$ is perpendicular to $\to P Q$ ---------(iii)

$∴ S$ is the intersection of perpendiculars.
Hence it's the orthocentre of PQR

Option D.

Suggest Corrections

Similar questions

Q. Let

O

be the origin and let

P Q R

be an arbitrary triangle. The point

S

is such that

O P . O Q + O R . O S = O R . O P + O Q . O S = O Q . O R + O P . O S

Then the triangle

P Q R

has

S

as its

Q. Let

O

be the origin and let

P Q R

be an arbitrary triangle. The point

S

is such that

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

Then the triangle

P Q R

has

S

as its

Q. Let

O

be the origin and let

P Q R

be an arbitrary triangle. The point

S

is such that

- - \to O P \cdot - - \to O Q + - - \to O R \cdot - - \to O S = - - \to O R \cdot - - \to O P + - - \to O Q \cdot - - \to O S = - - \to O Q . - - \to O R + - - \to O P . - - \to O S

Then the triangle

P Q R

has

S

as its

Q. Let O(0,0),P(3,4),Q(6,0) be the verticals of triangle OPQ .The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The point S is such that OS = PS = QS.Then RS =

Q. PQR is a triangle. S is a point on the side QR of

Δ

PQR such that

∠

PSR

= ∠

QPR/ Given

Q P = 8

cm, PR

= 6

cm and SR

= 3

cm.
(I) Prove

Δ

PQR

\sim Δ

SPR
(II) Find the length of QR and PS
(III)

\frac{a r e a o f Δ P Q R}{a r e a o f Δ S P R}

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