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 $ \Delta 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 $$ \frac{\text{length of the line segment } EF}{\text{length of the line segment } ES} $$ is __________.

Hint:

In coordinate geometry vector problems, assigning convenient coordinates helps simplify vector calculations and distance evaluations.

Answer 1

Step 1: Express the given vector equation in terms of position vectors. Let \( \vec{P} = \vec{p}, \vec{Q} = \vec{q}, \vec{R} = \vec{r}, \vec{S} = \vec{s} \). The given equation becomes: \[ \vec{SP} + 5\vec{SQ} + 6\vec{SR} = \vec{0} \Rightarrow (\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \vec{0} \Rightarrow \vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \vec{0} \Rightarrow \vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} \] Step 2: Assign coordinates to simplify the geometry. Let \( P = (0, 0), Q = (2, 0), R = (0, 2) \). Then: \[ E = \text{Midpoint of } PR = \left( \frac{0 + 0}{2}, \frac{0 + 2}{2} \right) = (0, 1) \] \[ F = \text{Midpoint of } QR = \left( \frac{2 + 0}{2}, \frac{0 + 2}{2} \right) = (1, 1) \] \[ S = \frac{(0,0) + 5(2,0) + 6(0,2)}{12} = \frac{(10, 12)}{12} = \left( \frac{5}{6}, 1 \right) \] Step 3: Compute lengths. \[ EF = \text{Distance between } (0,1) \text{ and } (1,1) = 1 \] \[ ES = \text{Distance between } (0,1) \text{ and } \left( \frac{5}{6}, 1 \right) = \frac{5}{6} \] Step 4: Compute the required ratio. \[ \frac{EF}{ES} = \frac{1}{\frac{5}{6}} = \frac{6}{5} \]