Vector Algebra Question 239
Question: 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]
Options:
A) $ \overrightarrow{OP} $
B) $ 2\overrightarrow{OP} $
C) $ 3\overrightarrow{OP} $
D) $ 4\overrightarrow{OP} $
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Answer:
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}. $
BETA
