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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 triangle PQR has S as its Centroid, Circumcentre, Orthocenter, Incentre?

Circumcentre

Orthocenter

Incentre

Centroid

Solution:

The correct option is B Orthocenter
Given →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 ⊥ →QR
Similarly →QS ⊥ →PR
→RS ⊥ →PQ
So S is orthocentre.