Subject: Mathematics — Vectors
Question: For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow{MN}$ denote the vector from $M$ to $N$, and $\overrightarrow{0}$ 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 $\bigtriangleup PQR$ such that $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}.$
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
\_\_\_\_\_\_\_\_\_\_.
Answer: $1.20$
Solution: $\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \overrightarrow{0}$
$$\overrightarrow{P} - \overrightarrow{S} + 5\overrightarrow{Q} - 5\overrightarrow{S} + 6\overrightarrow{R} - 6\overrightarrow{S} = \overrightarrow{0}$$
$$\Rightarrow \overrightarrow{S} = \frac{\overrightarrow{P} + 5\overrightarrow{Q} + 6\overrightarrow{R}}{12}.$$
[IMAGE] $$\overrightarrow{E} = \frac{\overrightarrow{P} + \overrightarrow{R}}{2},\quad\quad\overrightarrow{F} = \frac{\overrightarrow{Q} + \overrightarrow{R}}{2}.$$
$$\overrightarrow{F} - \overrightarrow{E} = \frac{\overrightarrow{Q} - \overrightarrow{P}}{2},$$
$$\overrightarrow{S} - \overrightarrow{E} = \frac{5\overrightarrow{Q} - 5\overrightarrow{P}}{12}.$$
$$\frac{|EF|}{|ES|} = \frac{\left| \overrightarrow{Q} - \overrightarrow{P} \right|/2}{5\left| \overrightarrow{Q} - \overrightarrow{P} \right|/12} = \frac{6}{5} = 1.20.$$
##