Question

In a $\triangle A B C, D$ and $E$ are points on $B C$ and $A C$ respectively, such that $B D=2 D C$ and $A E=3 E C$. Let $P$ be the point of intersection of $A D$ and $B E$. Find $B P / P E$ 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 $B C$ 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 $A C$ in the ratio $3: 1$, therefore $E \equiv\left(\frac{3 \overrightarrow{ c }+\overrightarrow{ a }}{4}\right)$.
Now, let $P$ divides $B E$ in the ratio $l: m$ and $A D$ in the ratio $x: y$.
Hence, the position vector of $P$ getting from $B E$ and $A D$ 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{3 l \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{3 l}{4(l+m)} \overrightarrow{ c }+\frac{l}{4(l+m)} \overrightarrow{ a }+\frac{m \overrightarrow{ b }}{l+m} $
$=\frac{2 x}{3(x+y)} \overrightarrow{ c }+\frac{x}{3(x+y)} \overrightarrow{ b }+\frac{y}{(x+y)} \overrightarrow{ a }$
Now, comparing the coefficients, we get
$\frac{3 l}{4(l+m)}=\frac{2 x}{3(x+y)}....$(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
$\frac{\frac{3 l}{4(l+m)}}{\frac{m}{l+m}}=\frac{\frac{2 x}{3(x+y)}}{\frac{x}{3(x+y)}}$
$\Rightarrow \frac{3}{4} \cdot \frac{l}{m}=2$
$\Rightarrow \frac{l}{m}=\frac{8}{3}=\frac{B P}{P E}$