Animated Solution for Mathematics - Vector Algebra: In a triangle $OAB$, $E$ is the midpoint of $BO$ and $D$ is a point on $AB$ such that $AD:DB=2:1$. If $OD$ and $AE$ intersect at $P$, determine the ratio $OP:PD$ using vector methods.

Unlocking the Triangle $OAB$

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey into the heart of vector geometry.

Many students fear vector problems, seeing them as a mess of symbols and coefficients. But I want you to see them differently. Think of vectors as the language of space itself.

When we use vectors, we are not just calculating; we are mapping the geometry of the universe. Let us dive into this triangle $OAB$ and uncover the hidden ratio $OP:PD$.

The Power of the Origin

The first step in any vector problem is the most critical: choosing your origin. In this problem, we have a triangle $OAB$.

It is natural, almost poetic, to place our origin at vertex $O$. By setting the origin at $O$, the position vector of $O$ becomes the zero vector, $\vec{0}$.

We define the position vector of $A$ as $\vec{a}$ and the position vector of $B$ as $\vec{b}$. Suddenly, the entire triangle is defined by just two vectors. This is the power of abstraction—we have reduced a geometric shape to a simple algebraic system.

Mapping the Landscape

Now, let us locate our points. We are told $E$ is the midpoint of $BO$.

Since $O$ is the origin, the position vector of $E$ is simply the average of the position vectors of $O$ and $B$:

Next, we look at point $D$ on the side $AB$. The problem tells us that $D$ divides $AB$ in the ratio $2:1$.

Here, we invoke the section formula. If a point divides a segment connecting $\vec{u}$ and $\vec{v}$ in ratio $m:n$, the position vector is $\frac{n\vec{u}+m\vec{v}}{m+n}$. Applying this to our segment $AB$, we get:

The Intersection of Two Worlds

Now, consider the point $P$. It is the intersection of $OD$ and $AE$. This is the crux of the problem.

Because $P$ lies on $OD$, we can express its position vector as a fraction of $\overrightarrow{OD}$. Let us assume $P$ divides $OD$ in the ratio $k:1$. Using the section formula again, we write:

But wait! $P$ also lies on $AE$. This gives us a second perspective. Let $P$ divide $AE$ in the ratio $m:1$. Applying the section formula on $AE$:

The Algebraic Dance

We now have two expressions for the same vector $\overrightarrow{OP}$. Because $\vec{a}$ and $\vec{b}$ are non-collinear, they are linearly independent.

This means the coefficients of $\vec{a}$ and $\vec{b}$ must be identical in both expressions. We equate them:

1. For $\vec{a}$: $\frac{k}{3(k+1)}=\frac{1}{m+1}$ 2. For $\vec{b}$: $\frac{2k}{3(k+1)}=\frac{m}{2(m+1)}$

Now, we solve this system. If we divide the second equation by the first, the complex terms cancel out beautifully:

Substituting $m=4$ back into our first equation:

The Final Ratio

We assumed the ratio $OP:PD$ was $k:1$. With $k=\frac{3}{2}$, the ratio is $\frac{3}{2}:1$.

Multiplying by $2$ to clear the fraction, we arrive at the final answer: $3:2$.

Do you see the beauty here? We started with a geometric puzzle, translated it into the language of vectors, performed a bit of algebraic manipulation, and arrived at a precise, undeniable truth.