Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}-3\hat{j}+\hat{k},\vec{b}=3\hat{i}+2\hat{j}+5\hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c})\times \vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}$ and $\vec{a}\cdot \vec{c}=3$. If $\vec{b}\times \vec{c}=\vec{d}$, then $|\vec{a}\cdot \vec{d}|$ is equal to :

A Journey into Symmetry

Welcome, future engineer! Today, we are going to dive into the elegant world of vector algebra.

Often, students look at a problem involving unknown vectors like $\vec{c}$ and immediately feel the urge to find its components, $c_x,c_y,c_z$. But wait! Before you start writing down a system of equations, let's pause and look at the structure of the problem.

We are given $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}+5\hat{k}$. We are also given a relationship involving a cross product: $(\vec{a}-\vec{c})\times \vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}$.

The Distributive Expansion

Let's apply the distributive property of the cross product. Just like multiplying a scalar across a bracket, we can distribute the cross product over the subtraction.

This gives us:

This is our first major step. We have successfully separated the known vector $\vec{a}$ from the unknown $\vec{c}$.

The Anti-Commutative Twist

Now, look at the second term: $-(\vec{c}\times \vec{b})$. The problem gives us a hint: $\vec{b}\times \vec{c}=\vec{d}$.

We know that the cross product is anti-commutative, meaning $\vec{u}\times \vec{v}=-(\vec{v}\times \vec{u})$. Therefore, $-(\vec{c}\times \vec{b})$ is exactly equal to $\vec{b}\times \vec{c}$, which is $\vec{d}$.

Our equation now transforms into something much more manageable:

Isn't that beautiful? We have eliminated the need to find $\vec{c}$ entirely!

The Determinant Calculation

To find $\vec{d}$, we first need to calculate $\vec{a}\times \vec{b}$. We set up the determinant as follows:

Let's calculate the components carefully. For the $\hat{i}$ component, we have $(-3)(5)-(2)(1)=-15-2=-17$.

For the $\hat{j}$ component, we have $-[(2)(5)-(3)(1)]=-(10-3)=-7$. Finally, for the $\hat{k}$ component, we have $(2)(2)-(3)(-3)=4+9=13$.

Thus, $\vec{a}\times \vec{b}=-17\hat{i}-7\hat{j}+13\hat{k}$.

Isolating $\vec{d}$ and the Final Dot Product

Now, we substitute this back into our equation: $(-17\hat{i}-7\hat{j}+13\hat{k})+\vec{d}=-18\hat{i}-3\hat{j}+12\hat{k}$.

Subtracting the cross product from the right side, we get:

Finally, we calculate the dot product $\vec{a}\cdot \vec{d}$:

The question asks for the absolute value, $|\vec{a}\cdot \vec{d}|=|-15|=15$.

We have arrived at the solution! Remember, in JEE, the path is often hidden in the properties of the operations themselves. Keep practicing, and keep visualizing! The final answer is 15.