Triangle PQR has S as itsA- centroidB- orthocenterC- incentreD- circumcentre

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Dot Product of Two Vectors

triangle PQR ...

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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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.