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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
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
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Solution
The correct option is D orthocenter
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS⇒−−→OP⋅(−−→OQ−−−→OR)=−−→OS⋅(−−→OQ−−−→OR)⇒(−−→OP−−−→OS)⋅(−−→OQ−−−→OR)=0⇒−→SP⋅−−→RQ=0⇒−→SP⊥−−→RQ
Similarly, −−→SR⊥−−→QP and −−→SQ⊥−−→PR
Hence, S is the orthocentre of triangle PQR
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS⇒−−→OP⋅(−−→OQ−−−→OR)=−−→OS⋅(−−→OQ−−−→OR)⇒(−−→OP−−−→OS)⋅(−−→OQ−−−→OR)=0⇒−→SP⋅−−→RQ=0⇒−→SP⊥−−→RQ
Similarly, −−→SR⊥−−→QP and −−→SQ⊥−−→PR
Hence, S is the orthocentre of triangle PQR
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