JEE Main · 2005 · 08 May 2005 · Offline · Vector Algebra

Expressing each vector through $C$, we have $\overrightarrow{PA}=\overrightarrow{PC}+\overrightarrow{CA}$ and $\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{CB}$. Adding these, $\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}+(\overrightarrow{CA}+\overrightarrow{CB})$. Because $C$ is the midpoint of $AB$, the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are equal in length and opposite in direction, so their sum is zero. This leaves $\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}$.