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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 ΔPQR has S as its

A
centroid
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B
circumcentre
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C
incentre
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D
orthocentre
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Solution

The correct option is D orthocentre
Given, OP.OQ+OR.OS=OR.OP+OQ.OS=OQ.OR+OP.OS

Consider, OP.OQ+OR.OS=OR.OP+OQ.OS

OP.OQOR.OP=OQ.OSOR.OS

OP(OQOR)=OS(OQOR)

OP(OQOR)OS(OQOR)=0

(OQOR).(OPOS)=0

QR.PS=0

We know that

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

QR is perpendicular to PS ---------(i)

Consider,
OP.OQ+OR.OS=OQ.OR+OP.OS

Following the same steps as above, we get that

QS.PR=0

QS is perpendicular to PR ---------(ii)

Consider,
OP.OQ+OR.OS=OQ.OR+OP.OS

Following the same steps as above, we get that

RS.PQ=0

RS is perpendicular to PQ ---------(iii)

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

Option D.

937645_1003647_ans_268d222d74254c91b664d738f656f358.JPG

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