Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}+\hat{j}-\hat{k},\vec{b}=((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to tackle a problem that, at first glance, might seem like a chaotic mess of cross products. You see an expression like $\vec{b}=((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i}$ and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for a series of simple, elegant steps. We are not going to be intimidated by this nested structure; instead, we are going to peel it back, layer by layer, like an onion.

Imagine you are standing in a 3D space, and you have a vector $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$. This is our starting point. We are then asked to perform a sequence of operations.

The key to solving any nested expression is to work from the inside out. We don't need to see the whole path at once; we just need to see the next step. Our first task is to compute the innermost part: $\vec{c}=\vec{a}\times (\hat{i}+\hat{j})$.

We use the determinant method, which is our most reliable friend in vector algebra. We set up the determinant with $\hat{i},\hat{j},\hat{k}$ in the first row, the components of $\vec{a}$ in the second, and the components of our target vector in the third:

Expanding this, we get $\hat{i}(0-(-1))-\hat{j}(0-(-1))+\hat{k}(2-1)$, which simplifies beautifully to $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. See? The beast is already starting to shrink.

Now that we have $\vec{c}$, the next layer of the onion is $\vec{d}=\vec{c}\times \hat{i}$. Again, we use the determinant method. This is a mechanical process, but it requires focus.

Don't rush. A single sign error here could derail the entire calculation.

Expanding this gives us $\hat{i}(0)-\hat{j}(0-1)+\hat{k}(0-(-1))$, which results in $\vec{d}=\hat{j}+\hat{k}$. We are making progress! Finally, we reach the outermost layer: $\vec{b}=\vec{d}\times \hat{i}$.

Expanding this gives us $\hat{i}(0)-\hat{j}(0-1)+\hat{k}(0-1)$, which simplifies to $\vec{b}=\hat{j}-\hat{k}$. We have successfully tamed the beast. Our vector $\vec{b}$ is simply $\hat{j}-\hat{k}$.

Now, the problem asks for the square of the projection of $\vec{a}$ onto $\vec{b}$. Geometrically, the projection of $\vec{a}$ onto $\vec{b}$ is like shining a light perpendicular to $\vec{b}$ and looking at the shadow that $\vec{a}$ casts on it. The formula for this projection $P$ is:

We have $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=0\hat{i}+1\hat{j}-1\hat{k}$. Let's calculate the dot product $\vec{a}\cdot \vec{b}$:

Now, the magnitude of $\vec{b}$ is:

So, the projection $P$ is:

We are almost there! The question asks for the square of the projection. This is a classic JEE twist—always check if they want the projection or its square.

We have $P=\sqrt{2}$, so:

And there you have it! We started with a complex, nested expression, broke it down into manageable pieces, and arrived at a clean, elegant result. This is the essence of physics and mathematics: taking the complex and making it simple through systematic, logical steps.