Question

In triangle $ABC$. $D$ and $E$ are point on $BC$ and $AC$ respectively such that $\overline{BD}=\overline{2DC}$ and \bar { AE }=\bar { 3EC } Let $p$ be the point of intersection of $AD$ and $BE$ then $BP/PE$

• A. $\frac{9}{8}$
• B. $\frac{3}{8}$
• C. $\frac{8}{3}$
• D. $\frac{8}{9}$

Answer 1

The correct option is C $\frac{8}{3}$

Let position vectors of $A,B$ be $\overrightarrow{a}$ & $\overrightarrow{b}$ respectively

Equation of $\overrightarrow{AD}:\overrightarrow{r}=\overrightarrow{a}+t(\overrightarrow{b}-\overrightarrow{a})$

Equation of $\overrightarrow{BE}:\overrightarrow{r}=\overrightarrow{b}+s(\overrightarrow{a}-\overrightarrow{b})$

If they intersect at $p$ then $\overrightarrow{AD}=\overrightarrow{BE}$

Comparing coefficient of $\overrightarrow{a},\overrightarrow{b}$

$1-t=\frac{s}{4}---\left(1\right)\frac{t}{3}=1-s---\left(2\right)$

on solving $t=\frac{9}{11},s=\frac{8}{11},\Rightarrow \overrightarrow{r}=\frac{2\overrightarrow{a}+3\overrightarrow{b}}{11}$

If $P$ divides $BE$ in ratio $p:1$, then

$\frac{2\overrightarrow{a}+3\overrightarrow{b}}{11}=\frac{p\frac{a}{4}+\overrightarrow{b}}{p+1}$

On comparing, we get $11p=8(p+1),11=3(p+1)$

$\Rightarrow p=1/3,p=8/3$

$\therefore$ Option $C$ is correct