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 :
\[\overrightarrow{PA}+\overrightarrow{PB}=(\overrightarrow{PA}+\overrightarrow{AC})+(\overrightarrow{PB}+\overrightarrow{BC})-(\overrightarrow{AC}+\overrightarrow{BC})\] = \[\overrightarrow{PC}+\overrightarrow{PC}-(\overrightarrow{AC}-\overrightarrow{CB})=2\overrightarrow{PC}-0,\] \[(\because \,\,\,\overrightarrow{AC}=\overrightarrow{CB})\] \[\therefore \] \[\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}\].You need to login to perform this action.
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