ABCD is a quadrilateral, and E and F are the mid-points of its diagonals AC and BD respectively. Then \(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4\,\overrightarrow{EF}\)
Let \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) be the position vectors of A, B, C, D. By the mid-point formula:
\( \displaystyle \vec{e} = \frac{\vec{a}+\vec{c}}{2}, \quad \vec{f} = \frac{\vec{b}+\vec{d}}{2} \)
Writing each side as (end) minus (start):
\(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD} = (\vec{b}-\vec{a})+(\vec{d}-\vec{a})+(\vec{b}-\vec{c})+(\vec{d}-\vec{c})\)
\( = 2(\vec{b}+\vec{d})-2(\vec{a}+\vec{c})\)
\( = 4\vec{f}-4\vec{e}\)
\( = 4\,\overrightarrow{EF}\)
Why the other options are incorrect
Option (1) \(2\,\overrightarrow{EF}\) — Incorrect. Each of \(\vec{b}+\vec{d}\) and \(\vec{a}+\vec{c}\) is twice a mid-point, so the factor is \(2 \times 2 = 4\).
Option (3) \(4\,\overrightarrow{FE}\) — Incorrect. The sum is \(4(\vec{f}-\vec{e})\), which points from E to F; \(4\,\overrightarrow{FE}\) is its negative.
Option (4) \(\vec{0}\) — Incorrect. The sum is zero only when E and F coincide, as in a parallelogram.