Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=3\hat{i}-\hat{j}+2\hat{k},$ $\vec{b}=\vec{a}\times (\hat{i}-2\hat{k})$ and $\vec{c}=\vec{b}\times \hat{k}.$ Then the projection of $\vec{c}-2\hat{j}$ on $\vec{a}$:

A Vector Journey

Welcome, future engineers! Today, we are going to peel back the layers of a classic JEE Advanced vector problem. It is not just about crunching numbers; it is about understanding how vectors interact in three-dimensional space.

We are given a base vector $\vec{a}=3\hat{i}-\hat{j}+2\hat{k}$ and a series of operations that transform this vector into something entirely new. Let us embark on this journey step by step.

Constructing the Foundation

Imagine standing in a 3D coordinate system. We have our anchor, vector $\vec{a}$. The problem asks us to find the projection of a specific vector onto $\vec{a}$.

Before we can project anything, we need to know what we are projecting. We are given $\vec{b}=\vec{a}\times (\hat{i}-2\hat{k})$.

To find $\vec{b}$, we employ the determinant method. It is the most robust way to handle cross products without losing track of your components. We set up the matrix:

Expanding this, we calculate the components carefully. For the $\hat{i}$ component, we have $(-1)(-2)-(2)(0)=2$.

For the $\hat{j}$ component, we take the negative of the determinant of the remaining terms: $-[(3)(-2)-(2)(1)]=-[-6-2]=8$. Finally, for the $\hat{k}$ component, we have $(3)(0)-(-1)(1)=1$.

Thus, we arrive at $\vec{b}=2\hat{i}+8\hat{j}+\hat{k}$. Take a breath—you have successfully navigated the first hurdle.

The Transformation to $\vec{c}$

Now, the problem introduces vector $\vec{c}=\vec{b}\times \hat{k}$. Instead of setting up another determinant, let us use the distributive property of the cross product.

We have:

Distributing the cross product, we get $2(\hat{i}\times \hat{k})+8(\hat{j}\times \hat{k})+(\hat{k}\times \hat{k})$. Recalling our unit vector cross products: $\hat{i}\times \hat{k}=-\hat{j}$, $\hat{j}\times \hat{k}=\hat{i}$, and $\hat{k}\times \hat{k}=0$.

Substituting these, we find $\vec{c}=2(-\hat{j})+8(\hat{i})+0=8\hat{i}-2\hat{j}$. The complexity is melting away, isn't it?

The Final Projection

We are almost there. The question asks for the projection of $\vec{v}=\vec{c}-2\hat{j}$ onto $\vec{a}$. First, let us define $\vec{v}$ explicitly:

Now, we apply the projection formula: $\text{Projection}=\frac{\vec{v}\cdot \vec{a}}{|\vec{a}|}$.

First, the dot product $\vec{v}\cdot \vec{a}$:

Next, the magnitude $|\vec{a}|$ is calculated as:

Finally, we combine these to find the projection:

To simplify this, notice that $28=2\times 14$, and $14=(\sqrt{14}{)}^2$. Therefore:

And there it is! Through systematic steps and careful calculation, we have arrived at the final answer of $2\sqrt{14}$. Remember, in JEE Advanced, the beauty lies in the process. Keep practicing, keep visualizing, and the vectors will eventually become second nature to you.