$ABCD$ is a rhombus whose diagonals intersect at $E$. Then $\vec{EA} + \vec{EB} + \vec{EC} + \vec{ED}$ equals to

$\vec{0}$

The correct answer is Option (1) → $\vec{0}$ ##

Given, $ABCD$ is a rhombus whose diagonals bisect each other.

$|\vec{EA}| = |\vec{EC}|$

and $|\vec{EB}| = |\vec{ED}|$

but since they are opposite to each other so they are of opposite signs.

$\Rightarrow \vec{EA} = -\vec{EC}$

and $\vec{EB} = -\vec{ED}$

$\Rightarrow \vec{EA} + \vec{EC} = \vec{0} \quad \dots(i)$

and $\vec{EB} + \vec{ED} = \vec{0} \quad \dots(ii)$

Adding (i) and (ii), we get

$\vec{EA} + \vec{EB} + \vec{EC} + \vec{ED} = \vec{0}.$