Animated Solution for Mathematics - Vector Algebra: If the vectors $\vec{a},\vec{b}$ and $\vec{c}$ form the sides $BC,CA$ and $AB$ respectively of a triangle $ABC$, then

Imagine standing at vertex $A$ of a triangle. You walk along side $AB$, then $BC$, then $CA$. You are back at $A$. This is the essence of the vector sum $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.

It is not just math; it is a physical reality. When we cross this with $\vec{a}$, we are essentially projecting the area of the triangle. The beauty of the cross product is its anti-commutative nature, which allows us to flip the vectors and find the hidden symmetry.

By crossing with $\vec{a}$ and then $\vec{b}$, we reveal that $\vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$. This is the hallmark of cyclic symmetry in geometry.

Let us embark on this journey. First, we visualize the triangle $ABC$. The vectors $\vec{a},\vec{b},\vec{c}$ are arranged in a continuous, cyclic order, forming a closed loop.

According to the Triangle Law of Vector Addition, if you start at a point and return to it, your net displacement is zero. Applying this law, we can write our foundational equation:

This simple equation holds the key to finding the relationship between their cross products. This is a favorite concept of JEE.

To find a relation involving cross products, let us manipulate this equation. We will take the cross product of vector $\vec{a}$ with the entire equation on both sides:

Next, we use the distributive property of the cross product over vector addition. We distribute vector $\vec{a}$ to each term inside the bracket:

Here is a crucial property to remember: the cross product of a vector with itself is always zero, because the angle between them is zero. So, $\vec{a}\times \vec{a}$ becomes the zero vector. Our equation simplifies to:

We can move the term $\vec{a}\times \vec{c}$ to the right side of the equals sign. When we do this, its sign changes:

Now, recall the anti-commutative property of cross products: $\vec{u}\times \vec{v}=-(\vec{v}\times \vec{u})$. Substituting this back, we find that:

We can repeat this exact same logical process. This time, let us take our original vector addition equation and take the cross product with vector $\vec{b}$ on both sides:

Distributing vector $\vec{b}$, we get:

Just like before, the self-cross product, $\vec{b}\times \vec{b}$, is zero. So the equation reduces to:

Rearranging this, we move $\vec{b}\times \vec{a}$ to the right side, giving us $\vec{b}\times \vec{c}=-(\vec{b}\times \vec{a})$. Applying the anti-commutative property again, $-(\vec{b}\times \vec{a})$ becomes $\vec{a}\times \vec{b}$. Therefore:

Finally, let us look at the two results we have derived. Combining these, we get the beautiful symmetric relation:

The elegance of this result lies in the cyclic symmetry of the triangle. You have successfully navigated the vector landscape!