For any two points $M$ and $N$ in the $XY$-plane, let $\vec{MN}$ denote the vector from $M$ to $N$, and $\vec{0}$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\Delta PQR$ such that $$\vec{SP} + 5 \vec{SQ} + 6 \vec{SR}...

The correct numerical answer is 1.2. Click to view the complete step-by-step calculation.

JEE Advanced 2025 Paper 1 Mathematics Question 9 Solution

Detailed Solution

Let the position vectors of $P, Q, R, S$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}$ respectively.

From $\vec{SP} + 5 \vec{SQ} + 6 \vec{SR} = \vec{0}$ , we have: $(\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \vec{0}$ $\vec{p} + 5\vec{q} + 6\vec{r} = 12\vec{s} \implies \vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}$ .
$E$ is the midpoint of $PR$ , so $\vec{e} = \frac{\vec{p} + \vec{r}}{2}$ .
$F$ is the midpoint of $QR$ , so $\vec{f} = \frac{\vec{q} + \vec{r}}{2}$ .
Length $EF = |\vec{f} - \vec{e}| = |\frac{\vec{q} + \vec{r}}{2} - \frac{\vec{p} + \vec{r}}{2}| = |\frac{\vec{q} - \vec{p}}{2}| = \frac{1}{2} |\vec{PQ}|$ .
Vector $\vec{ES} = \vec{s} - \vec{e} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} - \frac{\vec{p} + \vec{r}}{2} = \frac{\vec{p} + 5\vec{q} + 6\vec{r} - 6\vec{p} - 6\vec{r}}{12} = \frac{5\vec{q} - 5\vec{p}}{12}$ .
Length $ES = |\vec{ES}| = \frac{5}{12} |\vec{q} - \vec{p}| = \frac{5}{12} |\vec{PQ}|$ .
The ratio $\frac{EF}{ES} = \frac{(1/2) |\vec{PQ}|}{(5/12) |\vec{PQ}|} = \frac{1}{2} \times \frac{12}{5} = \frac{6}{5} = 1.2$ . Final Answer: 1.2