ABCD is a rhombus, whose diagonals intersect at E. Then \(\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED}\) equals to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\vec{0}\)
A rhombus is a parallelogram, so its diagonals bisect each other: E is the midpoint of both AC and BD.
So \(\overrightarrow{EC}\) has the same length as \(\overrightarrow{EA}\) and points the opposite way, and the same holds for \(\overrightarrow{ED}\) and \(\overrightarrow{EB}\):
\(\overrightarrow{EC} = -\overrightarrow{EA}\)
\(\overrightarrow{ED} = -\overrightarrow{EB}\)
Hence
\(\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED} = \overrightarrow{EA}+\overrightarrow{EB}-\overrightarrow{EA}-\overrightarrow{EB}\)
\( = \vec{0}\)