Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$. Let $\vec{c}$ be the vector such that $\vec{a}\times \vec{c}=\vec{b}$ and $\vec{a}\cdot \vec{c}=3$. Then $\vec{a}\cdot ((\vec{c}\times \vec{b})-\vec{b}-\vec{c})$ is equal to :

Imagine you are standing in a 3D coordinate space. You have two solid, known vectors, $\vec{a}$ and $\vec{b}$, and a third, elusive vector $\vec{c}$.

The problem presents you with two constraints: $\vec{a}\times \vec{c}=\vec{b}$ and $\vec{a}\cdot \vec{c}=3$.

Your goal is to evaluate the expression $E=\vec{a}\cdot ((\vec{c}\times \vec{b})-\vec{b}-\vec{c})$.

At first glance, this looks like a nightmare of vector algebra. You might be tempted to start solving for the components of $\vec{c}$.

Stop! Take a deep breath. In JEE Advanced physics and mathematics, the most elegant path is rarely the one that requires brute force. Let us embark on a journey to simplify this expression using the sheer beauty of vector properties.

Our first step is to demystify the expression. We have a dot product interacting with a bracketed sum of vectors.

Just like in basic algebra, the dot product is distributive over vector addition and subtraction. We can expand our expression $E$ as follows:

Suddenly, the "nightmare" has transformed into three distinct, manageable terms. We have successfully broken the problem into smaller, bite-sized pieces.

Look at the first term: $\vec{a}\cdot (\vec{c}\times \vec{b})$. This is a classic Scalar Triple Product.

One of the most powerful tools in your vector toolkit is the ability to interchange the dot and cross operations within an STP. Specifically:

Why is this useful? Because the problem explicitly gives us $\vec{a}\times \vec{c}=\vec{b}$!

By performing this swap, we transform the first term into $\vec{b}\cdot \vec{b}$, which is simply the square of the magnitude of $\vec{b}$, denoted as $|\vec{b}{|}^2$.

Given $\vec{b}=3\hat{i}-3\hat{j}+3\hat{k}$, we calculate:

The first term is conquered.

Now, consider the second term: $\vec{a}\cdot \vec{b}$. You could calculate this using components, but let's use our geometric intuition.

We know $\vec{b}=\vec{a}\times \vec{c}$. By the very definition of the cross product, the resulting vector $\vec{b}$ must be perpendicular to both $\vec{a}$ and $\vec{c}$.

And what is the dot product of any two perpendicular vectors? It is zero. Thus, $\vec{a}\cdot \vec{b}=0$. This term vanishes, simplifying our life significantly.

Finally, we have the third term: $\vec{a}\cdot \vec{c}$. Do we need to find $\vec{c}$? Absolutely not!

The problem statement graciously provides this value: $\vec{a}\cdot \vec{c}=3$. We simply plug it in. Now, we assemble our findings back into the expanded expression:

Performing the final arithmetic, we get $E=24$.

We have arrived at the solution without ever needing to solve for the mysterious vector $\vec{c}$. This is the essence of JEE Advanced problem solving: identifying the structure, applying the right theorems, and letting the math simplify itself.