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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

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
$\to O P \cdot \to O Q + \to O R \cdot \to O S = \to O R \cdot \to O P + \to O Q \cdot \to O S = \to O Q \cdot \to O R + \to O P \cdot \to O S$
Then the triangle $P Q R$ has $S$ as its

Circumcentre

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Orthocenter

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Incentre

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Centroid

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Open in App

Solution

The correct option is C Orthocenter
Given $\to O P \cdot \to O Q + \to O R \cdot \to O S = \to O R \cdot \to O P + \to O Q \cdot \to O S$
$\to p \cdot \to q + \to r \cdot \to s = \to r \cdot \to p + \to q \cdot \to s$
$\to p \cdot (\to q - \to r) = \to s \cdot (\to q - \to r)$
$(\to p - \to s) \cdot (\to q - \to r) = 0$
$\to S P ⊥ \to Q R$
Similarly
$\to Q S ⊥ \to P R$
$\to R S ⊥ \to P Q$
So S is orthocentre.

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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

Δ

PQR 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

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 . - - \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. The point O lies inside a triangle PQR such that

△

OPQ,

△

OQR and

△

ORPare equal in area. Then, the point O is called as

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Let O be the origin and let PQR be an arbitrary triangle. The point S is such that →OP⋅→OQ+→OR⋅→OS=→OR⋅→OP+→OQ⋅→OS=→OQ⋅→OR+→OP⋅→OSThen the triangle PQR has S as its

The correct option is C OrthocenterGiven →OP⋅→OQ+→OR⋅→OS=→OR⋅→OP+→OQ⋅→OS→p⋅→q+→r⋅→s=→r⋅→p+→q⋅→s→p⋅(→q−→r)=→s⋅(→q−→r)(→p−→s)⋅(→q−→r)=0→SP⊥→QRSimilarly→QS⊥→PR→RS⊥→PQSo S is orthocentre.

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