Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and $\vec{c}=(((\vec{a}\times \vec{b})\times \hat{i})\times \hat{i})\times \hat{i}$. Then $\vec{c}\cdot (-\hat{i}+\hat{j}+\hat{k})$ is equal to

In the realm of JEE Advanced, brute force is often a time-consuming trap. When faced with a nested expression like $\vec{c}=(((\vec{a}\times \vec{b})\times \hat{i})\times \hat{i})\times \hat{i}$, the most efficient approach is to utilize the Vector Triple Product (VTP) identity.

The identity states:

This formula is powerful because it transforms a complex cross product into a simple linear combination of vectors using only scalar dot products.

Let us apply this to the innermost expression: $(\vec{a}\times \vec{b})\times \hat{i}$. Here, we set $\vec{u}=\vec{a}$, $\vec{v}=\vec{b}$, and $\vec{w}=\hat{i}$.

The identity yields:

Given $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$, the dot products are trivial: $\vec{a}\cdot \hat{i}=-5$ and $\vec{b}\cdot \hat{i}=1$.

Substituting these values, we obtain:

We are now left with the remaining operations: $((-11\hat{j}+23\hat{k})\times \hat{i})\times \hat{i}$. Using the cyclic properties of unit vectors ($\hat{j}\times \hat{i}=-\hat{k}$ and $\hat{k}\times \hat{i}=\hat{j}$), we proceed step-by-step.

First, compute the inner cross product:

Next, compute the final cross product:

Finally, we compute the dot product of the resulting vector $(11\hat{j}-23\hat{k})$ with the vector $(-\hat{i}+\hat{j}+\hat{k})$:

By choosing the right identity and leveraging symmetry, the complexity melts away. The final result is $-12$.