JEE Main 202310 Apr, Evening shiftMathematicsHardVector Algebra184× since 2002If the points P\mathrm{P}P and Q\mathrm{Q}Q are respectively the circumcenter and the orthocentre of a △ABC\triangle \mathrm{ABC}△ABC, then PA→+PB→+PC→\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}PA+PB+PC is equal to :AOption A. QP→\overrightarrow {QP}QPBOption B. PQ→\overrightarrow {PQ}PQCOption C. 2PQ→2\overrightarrow {PQ}2PQDOption D. 2QP→2\overrightarrow {QP}2QPCheckSkip · show solution
