1
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
Open in App
Solution
The correct option is B Orthocenter
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS=−−→OQ⋅−−→OR+−−→OP⋅−−→OS
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS
⇒ −−→OP⋅(−−→OQ−−−→OR)+−−→OS⋅(−−→OR−−−→OQ)=0
⇒ −−→RQ⋅(−−→OP−−−→OS)=→0
⇒ −−→RQ⋅−→SP=→0
⇒ −−→RQ⊥−→SP
and similarly from −−→OR⋅−−→OP+−−→OQ⋅−−→OS=−−→OQ⋅−−→OR+−−→OP⋅−−→OS
−−→SR⊥−−→PQ
∴ S is the orthocentre.
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS=−−→OQ⋅−−→OR+−−→OP⋅−−→OS
−−→OP⋅−−→OQ+−−→OR⋅−−→OS=−−→OR⋅−−→OP+−−→OQ⋅−−→OS
⇒ −−→OP⋅(−−→OQ−−−→OR)+−−→OS⋅(−−→OR−−−→OQ)=0
⇒ −−→RQ⋅(−−→OP−−−→OS)=→0
⇒ −−→RQ⋅−→SP=→0
⇒ −−→RQ⊥−→SP
and similarly from −−→OR⋅−−→OP+−−→OQ⋅−−→OS=−−→OQ⋅−−→OR+−−→OP⋅−−→OS
−−→SR⊥−−→PQ
∴ S is the orthocentre.
Suggest Corrections
1
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
