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Q.

ABCD is a rhombus whose diagonals intersect at E. Then

$\vec{EA} + \vec{EB} + \vec{EC} + \vec{ED}$ equals

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a

\vec{0}

b

\vec{AD}

c

2

\vec{BC}

d

2

\vec{AD}

answer is A.

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Detailed Solution

Concept- The fact that a rhombus's sides are all equal and its diagonal cuts through it at the right will be our starting point. Finally, we will use this fact of the rhombus to create a pair of vectors and solve them. If two vectors have the same magnitude but different directions, their resultant is zero.
Now that we know that the diagonals of the rhombus ABCD intersect at E, we must determine the value of

\vec{EA} + \vec{EB} + \vec{EC} + \vec{ED}

Question Image

Now that we are aware that a parallelogram's diagonals are perpendicular and cut by a line, we may calculate
EA=EC _____(1)
EB=ED ______(2)
We now know that the sum of two vectors with the same magnitude but different directions is zero. As a result, we have (1) and (2)

\vec{EA}

+

\vec{EB}

=0 ______(3)

\vec{EB} + \vec{ED} = 0______(4)

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

Hence, the correct answer is option 1)

\vec{0}

.

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ABCD is a rhombus whose diagonals intersect at E. Then EA⃗+EB⃗+EC⃗+ED⃗ equals