Question

In $\Delta ABC$, $P,Q,R$ are points on $BC,CA,AB$ respectively, dividing them in the ratio $1:4$, $3:2$ and $3:7$. The point $S$ divides $AB$ in the ratio $1:3$. Then $\frac{|\overline{AP}+\overline{BQ}+\overline{CR}|}{|\overline{CS}|}$ is

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{5}{2}$
• D. $\frac{7}{10}$

Answer 1

Answer: (B)

$\frac{2}{5}$

P.v of P $=\frac{\overrightarrow{c}+4\overrightarrow{b}}{5}$
P.v of R $=\frac{7\overrightarrow{a}+3\overrightarrow{b}}{10}$
P.v of Q$=\frac{3\overrightarrow{a}+2\overrightarrow{b}}{5}$
P.v of S $=\frac{\overrightarrow{b}+3\overrightarrow{a}}{4}$
$\overrightarrow{AP}=\frac{\overrightarrow{c}+4\overrightarrow{b}-5\overrightarrow{a}}{5}\overrightarrow{BQ}=\frac{3\overrightarrow{a}+2\overrightarrow{c}-5\overrightarrow{b}}{5}$
$\overrightarrow{CR}=\frac{7\overrightarrow{a}+3\overrightarrow{b}-10\overrightarrow{a}}{10}\overrightarrow{CS}=\frac{\overrightarrow{b}+3\overrightarrow{a}-4\overrightarrow{c}}{4}$
$\therefore$ $\frac{\frac{|2\overrightarrow{c}+4\overrightarrow{c}-10\overrightarrow{c}+8\overrightarrow{b}-10\overrightarrow{b}+3\overrightarrow{b}-10\overrightarrow{a}+6\overrightarrow{a}+7\overrightarrow{a}|}{10}}{\left|\frac{\overrightarrow{b}+3\overrightarrow{a}-4\overrightarrow{c}}{4}\right|}$
$=\frac{4}{10}$