ABCD is a quadrilateral, and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Then \(\overrightarrow{PQ}+\overrightarrow{SR}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\overrightarrow{AC}\)
Let \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) be the position vectors of A, B, C, D. By the mid-point formula:
\( \displaystyle \vec{p} = \frac{\vec{a}+\vec{b}}{2}, \quad \vec{q} = \frac{\vec{b}+\vec{c}}{2}, \quad \vec{r} = \frac{\vec{c}+\vec{d}}{2}, \quad \vec{s} = \frac{\vec{d}+\vec{a}}{2} \)
\( \displaystyle \overrightarrow{PQ} = \vec{q}-\vec{p} = \frac{\vec{c}-\vec{a}}{2} = \frac{1}{2}\,\overrightarrow{AC} \)
\( \displaystyle \overrightarrow{SR} = \vec{r}-\vec{s} = \frac{\vec{c}-\vec{a}}{2} = \frac{1}{2}\,\overrightarrow{AC} \)
\( \overrightarrow{PQ}+\overrightarrow{SR} = \overrightarrow{AC} \)
Why the other options are incorrect
Option (1) \( \displaystyle \frac{1}{2}\,\overrightarrow{AC} \) — Incorrect. This is \(\overrightarrow{PQ}\) alone; \(\overrightarrow{SR}\) adds another \(\frac{1}{2}\,\overrightarrow{AC}\).
Option (3) \(\overrightarrow{BD}\) — Incorrect. Both \(\overrightarrow{PQ}\) and \(\overrightarrow{SR}\) contain only \(\vec{a}\) and \(\vec{c}\), so the sum is along the diagonal AC, not BD.
Option (4) \(\vec{0}\) — Incorrect. \(\overrightarrow{PQ}\) and \(\overrightarrow{SR}\) are equal vectors, so they add up to twice \(\overrightarrow{PQ}\); it is \(\overrightarrow{PQ}+\overrightarrow{RS}\) that is \(\vec{0}\).