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Standard XII

Mathematics

Dot Product of Two Vectors

Let O be the ...

Question

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.OS Then the triangle PQR has S as its

centroid

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orthocenter

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incentre

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circumcentre

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Solution

The correct option is B
orthocenter

OP.OQ + OR.OS = OR.OP + OQ.OS
⇒OP(OQ-OR)+OS(OR-OQ) = 0
⇒(OP-OS)(OQ-OR)=0
⇒SP.RQ = 0
Similarly SR.PQ = 0 and SQ.PR = 0
∴S is orthocentre.

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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

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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

\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

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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

△

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△

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△

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.OS Then the triangle PQR has S as its

The correct option is B orthocenter OP.OQ + OR.OS = OR.OP + OQ.OS ⇒OP(OQ-OR)+OS(OR-OQ) = 0 ⇒(OP-OS)(OQ-OR)=0 ⇒SP.RQ = 0 Similarly SR.PQ = 0 and SQ.PR = 0 ∴S is orthocentre.

The correct option is B
orthocenter

OP.OQ + OR.OS = OR.OP + OQ.OS
⇒OP(OQ-OR)+OS(OR-OQ) = 0
⇒(OP-OS)(OQ-OR)=0
⇒SP.RQ = 0
Similarly SR.PQ = 0 and SQ.PR = 0
∴S is orthocentre.