Animated Solution for Mathematics - Vector Algebra: $A_1,A_2,\dots ,A_n$ are the vertices of a regular plane polygon with $n$ sides and $O$ is its centre. Show that ${\sum }_{i=1}^{n-1}({\overrightarrow{OA}}_i\times {\overrightarrow{OA}}_{i+1})=(1-n)({\overrightarrow{OA}}_2\times {\overrightarrow{OA}}_1)$.

Unlocking the Regular Polygon

Welcome, future engineer. Today, we are not just solving a vector problem; we are peeling back the layers of geometric perfection.

When you look at a regular polygon, do not just see a shape. See a system in perfect equilibrium where every vertex, every angle, and every radial vector is dancing to the same rhythm. This problem is a beautiful exercise in recognizing that rhythm.

The Geometry of the Fan

Imagine you are standing at the center $O$ of a regular $n$-sided polygon. You look out at the vertices $A_1,A_2,\dots ,A_n$.

Because the polygon is regular, it is inscribed in a circle of radius $a$. This is our first anchor point, where every radial vector ${\overrightarrow{OA}}_i$ has the exact same magnitude: $|{\overrightarrow{OA}}_i|=a$.

Now, think about the angular spacing. The total rotation around the center is $2\pi$ radians, which is divided perfectly into $n$ equal slices.

Therefore, the angle $\theta$ between any two consecutive radial vectors, say ${\overrightarrow{OA}}_i$ and ${\overrightarrow{OA}}_{i+1}$, is fixed at:

This is the 'fan' of the polygon. Every slice of this fan is identical, and this uniformity is the secret weapon that will collapse our complex-looking summation into a simple algebraic expression.

The Cross Product as an Area Vector

Let us recall the definition of the vector cross product. For any two vectors $\vec{A}$ and $\vec{B}$, the cross product is defined as $\vec{A}\times \vec{B}=|\vec{A}||\vec{B}|\sin (\theta )\hat{n}$.

Physically, the magnitude of this cross product represents twice the area of the triangle formed by the two vectors. In our case, for any consecutive pair ${\overrightarrow{OA}}_i$ and ${\overrightarrow{OA}}_{i+1}$, the cross product is:

Here, $\hat{n}$ is the unit normal vector perpendicular to the plane of the polygon. Notice that the magnitude $a^2\sin (\frac{2\pi }{n})$ depends only on the radius $a$ and the number of sides $n$.

Since the entire polygon lies on a flat plane, the direction $\hat{n}$ is identical for every single pair. Every term in our summation is, quite literally, the same vector.

The Power of Uniformity

We are tasked with evaluating the sum ${\sum }_{i=1}^{n-1}({\overrightarrow{OA}}_i\times {\overrightarrow{OA}}_{i+1})$. Because we have established that each term is identical, we simply count the terms.

There are exactly $n-1$ terms in this sequence. Therefore, the sum becomes:

This is the moment where many students stop, thinking they are done. However, we must match the target expression: $(1-n)({\overrightarrow{OA}}_2\times {\overrightarrow{OA}}_1)$.

The Final Twist

This is where the anti-commutative property of the cross product saves the day. Remember that $\vec{A}\times \vec{B}=-(\vec{B}\times \vec{A})$.

We know that:

By the anti-commutative property, we can write:

Now, substitute this back into our summation result. We replace the constant term with $-({\overrightarrow{OA}}_2\times {\overrightarrow{OA}}_1)$:

Distributing the negative sign, we arrive at the final result:

The Elegance of the Result

We started with a geometric shape, applied the symmetry of the polygon, utilized the definition of the cross product, and finished with a simple algebraic manipulation.

The result is not just a proof; it is a testament to the power of looking for patterns. When you face a JEE Advanced problem, do not rush to calculate. Pause, visualize the symmetry, and let the geometry guide your hand.