Animated Solution for Mathematics - Vector Algebra: In a triangle $ABC$, $D$ 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.

Welcome, future engineers. Today, we are not just solving a geometry problem; we are learning to see the hidden structure of space. When you look at a triangle $ABC$ with points $D$ and $E$ scattered along its sides, it might look like a static drawing.

With the power of vectors, we can make this triangle dance. We are going to find the ratio $BP/PE$ where $P$ is the intersection of $AD$ and $BE$. This is a classic JEE problem, and the beauty lies in the elegance of the vector approach.

We set vertex $A$ as the origin, meaning $\vec{a}=\vec{0}$. This is a brilliant tactical move. By placing our origin at $A$, we effectively erase $\vec{a}$ from our equations, leaving us with only two basis vectors: $\vec{b}$ (for vertex $B$) and $\vec{c}$ (for vertex $C$).

Now, any point in this plane can be described as a linear combination of $\vec{b}$ and $\vec{c}$. It is like having a map where every location is defined by just two directions.

Now, let us locate $D$ and $E$. Point $D$ divides $BC$ in the ratio $2:1$. Using the section formula, the position vector $\vec{d}$ is given by:

Similarly, point $E$ divides $AC$ in the ratio $3:1$. Since $A$ is the origin, $\vec{e}$ is:

Notice how clean this is? We have defined the positions of $D$ and $E$ purely in terms of our basis vectors.

This is where the magic happens. Point $P$ is the intersection of $AD$ and $BE$. This means $P$ has a dual identity.

First, because $P$ lies on $AD$, it must be a scalar multiple of $\vec{d}$. Let $P$ divide $AD$ in ratio $k:1$. Then $\vec{p}=\frac{k}{k+1}\vec{d}$. Substituting our expression for $\vec{d}$, we get:

Second, because $P$ lies on $BE$, it must be a linear combination of $\vec{b}$ and $\vec{e}$. Let $P$ divide $BE$ in ratio $m:1$. Then $\vec{p}=\frac{1\vec{b}+m\vec{e}}{m+1}$. Substituting our expression for $\vec{e}$, we get:

We have two expressions for the same point $\vec{p}$. Since $\vec{b}$ and $\vec{c}$ are linearly independent, the coefficients must match. This gives us a system of equations:

Now, watch the beauty of the cancellation. If we divide the second equation by the first, the terms involving $k$ vanish entirely! We are left with:

Solving this, we find $3m=8$, or $m=\frac{8}{3}$. This $m$ is exactly the ratio $BP/PE$. The ratio is $8:3$. We have conquered the geometry using nothing but the fundamental laws of vectors.