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

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of 3D vector space.

When you look at vectors like $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, don't just see numbers. Visualize them as arrows piercing through space, defining directions and magnitudes.

Our goal is to evaluate the expression $(\vec{a}-\vec{b})\cdot \vec{d}$, where $\vec{d}$ is built from a chain of cross products. Let's break this down.

The standard approach is to follow the instructions step-by-step. First, we define $\vec{c}=\vec{a}\times \vec{b}$. Using the determinant method, we set up our matrix:

Expanding this, we find $\vec{c}=-\hat{i}-\hat{j}+0\hat{k}$. Next, we define $\vec{d}=\vec{c}\times \vec{a}$. Again, we set up the determinant:

Calculating this, we get $\vec{d}=-2\hat{i}+2\hat{j}-2\hat{k}$. Finally, we find $(\vec{a}-\vec{b})=-2\hat{i}+2\hat{j}+5\hat{k}$ and take the dot product with $\vec{d}$.

The result is $(-2)(-2)+(2)(2)+(5)(-2)=4+4-10=-2$. It works, but it feels like a lot of heavy lifting, doesn't it?

Now, let's put on our 'JEE Advanced' hats. Is there a more beautiful way? Yes! Let's look at the expression $(\vec{a}-\vec{b})\cdot \vec{d}$.

We can distribute the dot product:

Recall that $\vec{d}=\vec{c}\times \vec{a}$. By the very definition of the cross product, $\vec{d}$ is perpendicular to $\vec{a}$. Therefore, the dot product $\vec{a}\cdot \vec{d}$ is exactly $0$.

Our expression simplifies instantly to $-\vec{b}\cdot \vec{d}$. Now, substitute $\vec{d}=\vec{c}\times \vec{a}$ back into the expression:

This is the scalar triple product! Using the cyclic property, we know that $\vec{b}\cdot (\vec{c}\times \vec{a})=\vec{c}\cdot (\vec{a}\times \vec{b})$.

Since we defined $\vec{c}=\vec{a}\times \vec{b}$, the expression becomes:

Since $\vec{c}=-\hat{i}-\hat{j}$, the magnitude squared is $|\vec{c}{|}^2=(-1{)}^2+(-1{)}^2=2$. Thus, our final answer is $-2$.

Look at the difference between the two methods. The first was a mechanical grind; the second was a dance of properties.

In your journey toward becoming an engineer, always look for the underlying structure of the problem. When you see cross products, think of orthogonality.

When you see triple products, think of volumes and cyclic symmetry. Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of physics and mathematics!