Animated Solution for Mathematics - Vector Algebra: For any vector $\vec{a}$, the value of $(\vec{a}\times \hat{i}{)}^2+(\vec{a}\times \hat{j}{)}^2+(\vec{a}\times \hat{k}{)}^2$ is equal to

Unlocking the Geometry of Space

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of three-dimensional space. We are looking at the expression $(\vec{a}\times \hat{i}{)}^2+(\vec{a}\times \hat{j}{)}^2+(\vec{a}\times \hat{k}{)}^2$.

At first glance, it looks like a messy algebraic chore. But I want you to see it differently. I want you to see it as a dance of symmetry. When we deal with vectors in JEE Advanced, the goal is rarely to grind through calculations; the goal is to find the elegant path.

The Building Blocks

Imagine you are standing in the origin of a 3D Cartesian coordinate system. You have your three fundamental axes: $x,y,$ and $z$. Along these axes, we have our unit vectors: $\hat{i},\hat{j},$ and $\hat{k}$.

These are the 'atoms' of our vector space. Any vector $\vec{a}$ in this universe can be decomposed into these building blocks:

Here, $a_1,a_2,$ and $a_3$ are the scalar components. They tell us how much of $\vec{a}$ lives in each dimension. This is our starting point. Never lose sight of this decomposition; it is the key to unlocking the problem.

The Cross Product

Now, let us tackle the first term: $\vec{a}\times \hat{i}$. We substitute our component form:

When we distribute the cross product, we get three terms: $a_1(\hat{i}\times \hat{i})+a_2(\hat{j}\times \hat{i})+a_3(\hat{k}\times \hat{i})$. This is where the magic happens. Remember the rules of the cross product: a vector crossed with itself is zero because the angle between them is zero.

So, $\hat{i}\times \hat{i}=0$. For the others, we use the right-hand rule: $\hat{j}\times \hat{i}=-\hat{k}$ and $\hat{k}\times \hat{i}=\hat{j}$. Substituting these back, we get:

This is a beautiful result. Notice how the $a_1$ component vanished? That is because the cross product with $\hat{i}$ effectively 'ignores' the component along the $x$-axis.

The Power of Symmetry

Now, we need the square of the magnitude of this vector. Since $\hat{j}$ and $\hat{k}$ are orthogonal, the magnitude squared is simply the sum of the squares of the components:

Now, pause. Do you see the pattern? We did not need to calculate the other two terms from scratch. By cyclic symmetry, if crossing with $\hat{i}$ gives us $a_2^2+a_3^2$, then crossing with $\hat{j}$ must give us $a_1^2+a_3^2$, and crossing with $\hat{k}$ must give us $a_1^2+a_2^2$.

This is the 'Aha!' moment. We have turned a tedious calculation into a simple observation of structure.

The Grand Finale

Finally, we add them all together:

Grouping the terms, we see that each squared component appears exactly twice:

And what is $a_1^2+a_2^2+a_3^2$? It is the definition of the magnitude squared of our original vector, $|\vec{a}{|}^2$. Thus, the entire expression simplifies to $2|\vec{a}{|}^2$. We have arrived at the answer, not by brute force, but by understanding the geometry of the space.