Vectors
PYP_JEE_ADV_2025_P1
Grade None
Question:
For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow{MN}$ 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 $\triangle PQR$ such that
$$\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0}.$$
Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$, respectively. Then the value of
$$\dfrac{\text{length of the line segment } EF}{\text{length of the line segment } ES}$$
is ___.
Step-by-Step Solution
Key Concept: Barycentric coordinates from vector condition; midline of triangle
From $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0}$:
Let position vectors be $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s}$.
$(\mathbf{p}-\mathbf{s}) + 5(\mathbf{q}-\mathbf{s}) + 6(\mathbf{r}-\mathbf{s}) = \vec{0}$
$\mathbf{p} + 5\mathbf{q} + 6\mathbf{r} = 12\mathbf{s}$
$\mathbf{s} = \dfrac{\mathbf{p} + 5\mathbf{q} + 6\mathbf{r}}{12}$
So $S$ divides with barycentric coordinates $(1:5:6)$ w.r.t. $P, Q, R$.
$E$ = midpoint of $PR$: $\mathbf{e} = \dfrac{\mathbf{p}+\mathbf{r}}{2}$
$F$ = midpoint of $QR$: $\mathbf{f} = \dfrac{\mathbf{q}+\mathbf{r}}{2}$
$EF$ is the midline of $\triangle PQR$ parallel to $PQ$, so $EF = \dfrac{PQ}{2}$.
Actually $\overrightarrow{EF} = \mathbf{f} - \mathbf{e} = \dfrac{\mathbf{q}-\mathbf{p}}{2}$, so $|EF| = \dfrac{|PQ|}{2}$.
$\overrightarrow{ES} = \mathbf{s} - \mathbf{e} = \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12} - \dfrac{\mathbf{p}+\mathbf{r}}{2} = \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r} - 6\mathbf{p}-6\mathbf{r}}{12} = \dfrac{-5\mathbf{p}+5\mathbf{q}}{12} = \dfrac{5(\mathbf{q}-\mathbf{p})}{12}$
$|ES| = \dfrac{5|PQ|}{12}$
$\dfrac{|EF|}{|ES|} = \dfrac{|PQ|/2}{5|PQ|/12} = \dfrac{1/2}{5/12} = \dfrac{6}{5} = 1.2$
Correct Answer: 1.2