Question

$ABC$ is a triangle, the point $P$ is on side $BC$ such that $3\overrightarrow{BP}=2\overrightarrow{PC}$, the point $Q$ is on the line $\overrightarrow{CA}$ such that $4\overrightarrow{CQ}=\overrightarrow{QA}$. If $R$ is the common point of $\overrightarrow{AP}$ & $\overrightarrow{BQ}$, then the ratio in which the line joining $CR$ divides $\overrightarrow{AB}$ is

• A. $2:5$
• B. $3:8$
• C. $4:1$
• D. $6:1$

Answer 1

The correct option is D $6:1$

We have,

$\begin{array}{c}\frac{BP}{PC}=\frac{2}{3}PVofP=\frac{2\overrightarrow{c}+3\overrightarrow{b}}{5}\\ \frac{QA}{CQ}=4PVofQ=\frac{4\overrightarrow{c}}{5}\end{array}$

Let R divides $BQ$ in $\lambda :1$ and AP in $\mu :1$

$\begin{array}{c}\frac{\lambda \frac{4\overrightarrow{c}}{5}+\overrightarrow{b}}{\lambda +1}=\frac{\mu \left(\frac{2\overrightarrow{c}}{5}+\frac{3}{5}\overrightarrow{b}\right)}{\mu +1}\\ \frac{4\lambda }{5\left(\lambda H\right)}\overrightarrow{c}+\frac{1}{\lambda H}\overrightarrow{b}=\frac{2\mu }{5\left(\mu H\right)}\overrightarrow{c}+\frac{3\mu }{5\left(\mu H\right)}\overrightarrow{b}\\ \frac{4\lambda }{5\left(\lambda H\right)}=\frac{2\mu }{5\left(\mu H\right)}\\ \frac{\lambda }{\lambda H}=\frac{3\mu }{5\left(\mu H\right)}\\ \frac{4\lambda }{5}\frac{3\mu }{5\left(\mu H\right)}=\frac{2\mu }{5\left(\mu H\right)}\\ \frac{4\lambda }{5}\times 3=2\\ 2\lambda =\frac{5}{3}\\ \lambda =\frac{5}{6}\\ PVofR=\frac{\frac{4}{5}\times \frac{5}{6}\overrightarrow{c}+\overrightarrow{b}}{\frac{11}{6}}=\frac{\frac{2}{3}\overrightarrow{c}+\overrightarrow{b}}{\frac{11}{6}}\\ PVofR=\frac{12}{33}\overrightarrow{c}+\frac{6}{11}\overrightarrow{b}\end{array}$

Let E divides BA in $r:1$ and CR in $\beta :1$

$\begin{array}{c}\frac{\overrightarrow{b}}{rH}=\frac{\beta \overrightarrow{c}-\frac{12}{33}\overrightarrow{c}-\frac{6}{11}\overrightarrow{b}}{\beta -1}\\ \frac{\overrightarrow{b}}{rH}=\frac{\left(\beta \overrightarrow{c}-\frac{4}{33}\right)\overrightarrow{c}-\frac{6}{11}\overrightarrow{b}}{\beta -1}\\ \beta =\frac{4}{11}\\ \frac{1}{rH}=\frac{-6}{11}\times \frac{1}{\frac{4}{11}-1}\\ =\frac{-6}{11}\times \frac{11}{-7}\\ \frac{1}{rH}=\frac{6}{7}\\ rH=\frac{7}{6}\\ e=\frac{1}{6}\\ \frac{BE}{AE}=\frac{1}{6}\\ \frac{AE}{BE}=\frac{6}{1}\end{array}$

Then,

Option $D$ is correct answer.