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Let O be the origin and let PQR be an ar...

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that `bar(OP)*bar(OQ)+bar(OR)*bar(OS)=bar(OR)*bar(OP)+bar(OQ)*bar(OS)=bar(OQ)*bar(OR)+bar(OP)*bar(OS)` Then the triangle PQR has S as its

A

centriod

B

circumectre

C

incente

D

orthocenter

Text Solution

Verified by Experts

The correct Answer is:
D
`vec(OP) . vec(OQ) + vec(OR).vec(OS)= vec(OR) -vec(OP)+vec(OQ).vec(OS).`
`=vec(OP).(vec(OQ)-vec(OR))=vec(OS).(vec(OQ)-vec(OR))`
`implies (vec(OP)-vec(OS)).(vec(RQ))=0implies vec(SP).vec(RQ)=0`
`implies vec(SP)botvec(RQ)`
Similarly ,`vec(SR)bot vec(OP) and vec(SQ) botvec(PR).`
hence , S is orthocenter.
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Knowledge Check

  • 0.bar(34)+0.bar(34)=

    A
    `0.6bar(87)`
    B
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    D
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  • If bar(PR)=2hat(i)+hat(j)+hat(k),bar(QS)=hat(i)+3hat(j)+2hat(k), then the area of the quadrilateral PQRS is

    A
    `(5sqrt(3))/(2)`
    B
    `(1)/(2)|bar(PR)xxbar(QS)|`
    C
    `(1)/(2)(bar(PR)xxbar(QS))`
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