Animated Solution for Mathematics - Vector Algebra: Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac{1}{2}$. Then $\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta ABC)}$ is equal to

A Masterclass in Elegance

Welcome, future engineer. Today, we are not merely solving a geometry problem; we are embarking on a journey to witness the sheer elegance of vector algebra.

When you first look at a triangle $PQR$ with points $A$, $B$, and $C$ dividing its sides in a ratio of $1:2$, your instinct might be to reach for the sine rule or to start subtracting areas of smaller triangles from the larger one. While those methods work, they are often prone to calculation errors and tedious bookkeeping.

Today, we will use the power of vectors to slice through this complexity.

The Strategic Origin

In vector geometry, the choice of origin is your most powerful tool. It is the anchor of your entire coordinate system.

By setting vertex $P$ as the origin (the zero vector, $\vec{0}$), we immediately simplify the position vectors of the other vertices. Let the position vector of $Q$ be $\vec{q}$ and the position vector of $R$ be $\vec{r}$.

Why do we do this? Because it allows us to express any point in the triangle as a linear combination of $\vec{q}$ and $\vec{r}$. It turns a spatial problem into a playground of algebra. We are no longer dealing with angles and side lengths; we are dealing with vectors $\vec{q}$ and $\vec{r}$.

The Section Formula

Now, let us locate our points $A$, $B$, and $C$. We are given that $\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac{1}{2}$. This is a classic application of the section formula.

For point $A$ on $QR$, dividing it in a $1:2$ ratio, the position vector $\vec{a}$ is given by:

Similarly, for point $B$ on $RP$, dividing it in a $1:2$ ratio, we get:

And for point $C$ on $PQ$, dividing it in the same ratio:

Look at how clean this is! We have defined the entire inner triangle $ABC$ using only the base vectors $\vec{q}$ and $\vec{r}$.

The Area Machine

We know that the area of the large triangle $PQR$ is given by the vector area formula:

Now, to find the area of the inner triangle $ABC$, we need two side vectors. Let us choose $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Now, we calculate the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$. This is where the magic happens. We factor out the constants:

Expanding this, we get:

The Grand Finale

Recall the fundamental properties of the cross product: $\vec{x}\times \vec{x}=\vec{0}$ and $\vec{r}\times \vec{q}=-\vec{q}\times \vec{r}$. Applying these, the expression simplifies beautifully:

Thus, the area of $\Delta ABC$ is:

Finally, the ratio of the area of $\Delta PQR$ to $\Delta ABC$ is:

There it is. The larger triangle is exactly three times the area of the smaller one. This result is not just a number; it is a testament to the power of vector methods. By translating geometry into algebra, we removed the ambiguity and arrived at the truth with absolute certainty.