If $ABCD$ be a parallelogram and $M$ be the point of intersection of the diagonals. If $O$ is any point, then $\overrightarrow{\mathbf{OA}} + \overrightarrow{\mathbf{OB}} + \overrightarrow{\mathbf{OC}} + \overrightarrow{\mathbf{OD}}$ is

Question

Dr. Abubacker · Accepted Answer

(b)
We know that the diagonals of a parallelogram bisect each other. Therefore, $M$ is the mid point of $AC$ and $BD$ both.
$	herefore\ \overrightarrow{\mathbf{OA}} + \overrightarrow{\mathbf{OC}} = 2\overrightarrow{\mathbf{OM}}$
and $\overrightarrow{\mathbf{OB}} + \overrightarrow{\mathbf{OD}} = 2\ \overrightarrow{\mathbf{OM}}$
$\Longrightarrow \ \overrightarrow{\mathbf{OA}} + \overrightarrow{\mathbf{OB}} + \overrightarrow{\mathbf{OC}} + \overrightarrow{\mathbf{OD}} = 4\ \overrightarrow{\mathbf{OM}}$

Q. If $ABCD$ be a parallelogram and $M$ be the point of intersection of the diagonals. If $O$ is any point, then $\overrightarrow{\mathbf{OA}} + \overrightarrow{\mathbf{OB}} + \overrightarrow{\mathbf{OC}} + \overrightarrow{\mathbf{OD}}$ is

Explanation

Reviewed by Dr. Abubacker

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