JEE Advanced 2025 Paper 1 · Q09 · Vector Algebra
In the XY-plane, let $P, Q, R$ be three distinct points and $S$ a point inside triangle $PQR$ with $\vec{SP} + 5\vec{SQ} + 6\vec{SR} = \vec{0}$ (zero vector). Let $E$ and $F$ be the midpoints of $PR$ and $QR$ respectively. Find the value of $\text{length}(EF) / \text{length}(ES)$.
Reveal answer + step-by-step solution
Correct answer:1.20
Solution
Take $S$ as origin. Then $\vec{P} + 5\vec{Q} + 6\vec{R} = \vec{0}$ (position vectors), so $\vec{R} = -(\vec{P} + 5\vec{Q})/6$. $\vec{E} = (\vec{P} + \vec{R})/2 = (\vec{P} + (-\vec{P}/6 - 5\vec{Q}/6))/2 = (5\vec{P}/6 - 5\vec{Q}/6)/2 = \frac{5}{12}(\vec{P} - \vec{Q})$. $\vec{F} = (\vec{Q} + \vec{R})/2 = (\vec{Q} + (-\vec{P}/6 - 5\vec{Q}/6))/2 = (-\vec{P}/6 + \vec{Q}/6)/2 = (\vec{Q} - \vec{P})/12$. $\vec{EF} = \vec{F} - \vec{E} = (\vec{Q}-\vec{P})/12 - \frac{5}{12}(\vec{P}-\vec{Q}) = (\vec{Q}-\vec{P})/12 + \frac{5}{12}(\vec{Q}-\vec{P}) = \frac{6}{12}(\vec{Q}-\vec{P}) = (\vec{Q}-\vec{P})/2$. (This matches the midsegment theorem: $EF$ parallel to $PQ$ and $EF = PQ/2$.) $\vec{ES} = -\vec{E} = -\frac{5}{12}(\vec{P} - \vec{Q}) = \frac{5}{12}(\vec{Q} - \vec{P})$. $|EF| / |ES| = (1/2) / (5/12) = 12/10 = 6/5 = 1.20$. Answer: $1.20$.
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