question_answer

If C is the middle point of AB and P is any point outside AB, then                 [MNR 1991; UPSEAT 2000; AIEEE 2005]

A) \[\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}\]          

B) \[\overrightarrow{PA}+\overrightarrow{PB}=2\,\overrightarrow{PC}\]

C) \[\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=0\]    

D) \[\overrightarrow{PA}+\overrightarrow{PB}+2\,\overrightarrow{PC}=0\]

Correct Answer: B

Solution :

\[\overrightarrow{PA}+\overrightarrow{PB}=(\overrightarrow{PA}+\overrightarrow{AC})+(\overrightarrow{PB}+\overrightarrow{BC})-(\overrightarrow{AC}+\overrightarrow{BC})\] = \[\overrightarrow{PC}+\overrightarrow{PC}-(\overrightarrow{AC}-\overrightarrow{CB})=2\overrightarrow{PC}-0,\] \[(\because \,\,\,\overrightarrow{AC}=\overrightarrow{CB})\]       \[\therefore \] \[\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}\].