Question

In a $△ABC,D$ 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 the position vectors of $A,B$ and $C$ are $\overrightarrow{a},\vec{b}$ and $\vec{c}$ respectively, since the point $D$ divides $BC$ in the ratio of $2:1$, the position vector of $D$ will be

$D=\left(\frac{2\overrightarrow{c}+\overrightarrow{b}}{3}\right)$
and the point $E$ divides $AC$ in the ratio $3:1$, therefore $E\equiv \left(\frac{3\overrightarrow{c}+\overrightarrow{a}}{4}\right)$.
Now, let $P$ divides $BE$ in the ratio $l:m$ and $AD$ in the ratio $x:y$.
Hence, the position vector of $P$ getting from $BE$ and $AD$ must be the same.
Hence, we have
$\frac{l\left(\frac{3\overrightarrow{c}+\overrightarrow{a}}{4}\right)+m\overrightarrow{b}}{l+m}=\frac{x\left(\frac{2\overrightarrow{c}+\overrightarrow{b}}{3}\right)+y\overrightarrow{a}}{x+y}$
$\Rightarrow \frac{\frac{3l\overrightarrow{c}}{4}+\frac{l\overrightarrow{a}}{4}+m\overrightarrow{b}}{l+m}=\frac{\frac{2\overrightarrow{c}x}{3}+\frac{\overrightarrow{b}x}{3}+y\overrightarrow{a}}{x+y}$
$\Rightarrow \frac{3l}{4(l+m)}\overrightarrow{c}+\frac{l}{4(l+m)}\overrightarrow{a}+\frac{m\overrightarrow{b}}{l+m}$
$=\frac{2x}{3(x+y)}\overrightarrow{c}+\frac{x}{3(x+y)}\overrightarrow{b}+\frac{y}{(x+y)}\overrightarrow{a}$
Now, comparing the coefficients, we get
$\frac{3l}{4(l+m)}=\frac{2x}{3(x+y)}....$(i)
$\frac{l}{4(l+m)}=\frac{y}{x+y}.....$(ii)
and $\frac{m}{l+m}=\frac{x}{3(x+y)}.....$(iii)
On dividing Eq. (i) by Eq. (iii), we get
$\frac{\frac{3l}{4(l+m)}}{\frac{m}{l+m}}=\frac{\frac{2x}{3(x+y)}}{\frac{x}{3(x+y)}}$
$\Rightarrow \frac{3}{4}\cdot \frac{l}{m}=2$
$\Rightarrow \frac{l}{m}=\frac{8}{3}=\frac{BP}{PE}$