question_answer

P is the point of intersection of the diagonals of the parallelogram ABCD. If O is any point, then \[\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\]                    [RPET 1989; J & K 2005]

A) \[\overrightarrow{OP}\]       

B) \[2\,\,\overrightarrow{OP}\]

C) \[3\,\,\overrightarrow{OP}\]

D) \[4\,\,\overrightarrow{OP}\]

Correct Answer: D

Solution :

We know that P will be the midpoint of AC and BD \[\therefore \]   \[\,\overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OP}\]                               ......(i) and \[\overrightarrow{OB}+\overrightarrow{OD}=2\overrightarrow{OP}\]                           ?..(ii) Adding (i) and (ii), we get, \[\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}.\]