Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\hat{i}-3\hat{j}+7\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+2\vec{b})\times \vec{c}=3(\vec{c}\times \vec{a})$. If $\vec{a}\cdot \vec{c}=130$, then $\vec{b}\cdot \vec{c}$ is equal to _______

Unlocking the Hidden Parallelism

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a beautiful problem that tests not just your ability to calculate, but your ability to see the hidden geometry within algebraic expressions.

We are given two vectors, $\vec{a}=\hat{i}-3\hat{j}+7\hat{k}$ and $\vec{b}=2\hat{i}-\hat{j}+\hat{k}$, and a mysterious vector $\vec{c}$ defined by the equation $(\vec{a}+2\vec{b})\times \vec{c}=3(\vec{c}\times \vec{a})$. Our mission is to find $\vec{b}\cdot \vec{c}$ given that $\vec{a}\cdot \vec{c}=130$.

The Anti-Commutative Trap

At first glance, the equation $(\vec{a}+2\vec{b})\times \vec{c}=3(\vec{c}\times \vec{a})$ might seem intimidating. But look closer at the right side.

In the world of vectors, the cross product is anti-commutative, meaning $\vec{c}\times \vec{a}=-(\vec{a}\times \vec{c})$. This is our first key.

By substituting this into our equation, the right side transforms into $-3(\vec{a}\times \vec{c})$. Now, our equation looks like this: $(\vec{a}+2\vec{b})\times \vec{c}=-3(\vec{a}\times \vec{c})$.

The Power of Distributivity

Now, let's bring everything to one side to set the equation to the zero vector. We get $(\vec{a}+2\vec{b})\times \vec{c}+3(\vec{a}\times \vec{c})=\vec{0}$.

Here is where the magic happens. Because the cross product is distributive, we can factor out the $\times \vec{c}$ term.

This leaves us with $(\vec{a}+2\vec{b}+3\vec{a})\times \vec{c}=\vec{0}$. Simplifying the terms inside the bracket, we arrive at $(4\vec{a}+2\vec{b})\times \vec{c}=\vec{0}$.

The Geometric Insight

This is a classic JEE moment. What does it mean when the cross product of two vectors is the zero vector? It means the vectors are parallel!

Therefore, $\vec{c}$ must be parallel to the vector $4\vec{a}+2\vec{b}$. Mathematically, we express this as $\vec{c}=\lambda (4\vec{a}+2\vec{b})$, where $\lambda$ is a scalar.

Let's calculate $4\vec{a}+2\vec{b}$ explicitly:

So, $\vec{c}=\lambda (8\hat{i}-14\hat{j}+30\hat{k})$.

The Final Calculation

We are given $\vec{a}\cdot \vec{c}=130$. Let's substitute our expression for $\vec{c}$ into this dot product:

With $\lambda =\frac{1}{2}$, we find $\vec{c}=\frac{1}{2}(8\hat{i}-14\hat{j}+30\hat{k})=4\hat{i}-7\hat{j}+15\hat{k}$. Finally, we compute $\vec{b}\cdot \vec{c}$:

And there it is! The final answer is 30. You see, when you break down the problem into these logical phases, the complexity melts away, leaving only the elegance of the math.