Question

In a triangle $ABCD$ and E are points on BC and AC respectively, such that $BD=2DC$ and $AE=3EC$. Let $P$ be the point of intersection of AD and BE. Find $BP/PE$ using vector methods.

Answer 1

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ & $\overrightarrow{e}$ be the position vector of A, B, C, D & E points respectively.

$3\overrightarrow{d}=2\overrightarrow{c}+\overrightarrow{b}$ ….$\left(1\right)$

$4\overrightarrow{c}=\overrightarrow{a}+3\overrightarrow{b}$ ….$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$ we get

$9\overrightarrow{d}+2\overrightarrow{a}=6\overrightarrow{c}+3\overrightarrow{b}+2\overrightarrow{a}=8\overrightarrow{e}+3\overrightarrow{b}=$

P. V of point P

Point $\frac{Bp}{PD}=\frac{8}{3},\frac{AP}{PD}=\frac{9}{2}$.