Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=-\hat{i}-\hat{k}$, $\vec{b}=-\hat{i}+\hat{j}$ and $\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times \vec{b}=\vec{c}\times \vec{b}$ and $\vec{r}\cdot \vec{a}=0$, then the value of $\vec{r}\cdot \vec{b}$ is .........

Unlocking the Unknown

Welcome, future IITians! Today, we are going to embark on a journey through the elegant world of vector algebra. Often, when we see a problem involving cross products and dot products, our instinct is to immediately jump into component calculations.

But wait! Before we start crunching numbers, let's pause and visualize the geometric soul of the problem. We are given three vectors: $\vec{a}=-\hat{i}-\hat{k}$, $\vec{b}=-\hat{i}+\hat{j}$, and $\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$. Our mission is to find the value of $\vec{r}\cdot \vec{b}$, given two constraints on $\vec{r}$.

Decoding the Cross Product

The first condition is $\vec{r}\times \vec{b}=\vec{c}\times \vec{b}$. Instead of treating this as a system of equations, let's look at the structure. If we bring everything to one side, we get $\vec{r}\times \vec{b}-\vec{c}\times \vec{b}=\vec{0}$.

Because the cross product is distributive, this simplifies beautifully to:

Now, ask yourself: when is the cross product of two vectors zero? Only when they are parallel! This means the vector $(\vec{r}-\vec{c})$ must be parallel to $\vec{b}$.

In the language of vectors, this is our 'Aha!' moment. We can express this relationship using a scalar $\lambda$:

The Parametric Line of Possibilities

Rearranging our equation, we get $\vec{r}=\vec{c}+\lambda \vec{b}$. Geometrically, this is the equation of a line in 3D space. It tells us that $\vec{r}$ is not just a single point; it is a collection of points lying on a line that passes through the tip of $\vec{c}$ and runs parallel to $\vec{b}$.

As $\lambda$ changes, we slide along this line. But which point on this line is our $\vec{r}$? That is where the second condition comes in.

The Orthogonality Constraint

We are told that $\vec{r}\cdot \vec{a}=0$. This is the 'pin' that locks our vector $\vec{r}$ into place. It tells us that $\vec{r}$ must be perpendicular to $\vec{a}$.

Let's substitute our parametric form $\vec{r}=\vec{c}+\lambda \vec{b}$ into this condition:

Expanding this using the distributive property of the dot product, we get:

This is a simple linear equation in $\lambda$. To solve it, we just need the values of $\vec{c}\cdot \vec{a}$ and $\vec{b}\cdot \vec{a}$.

The Final Calculation

Let's calculate these dot products with precision. For $\vec{c}\cdot \vec{a}$:

And for $\vec{b}\cdot \vec{a}$:

Substituting these back into our equation, we have $-4+\lambda (1)=0$, which gives us $\lambda =4$. Now that we have our scalar, we can find the exact vector $\vec{r}$:

Finally, the problem asks for $\vec{r}\cdot \vec{b}$:

And there it is! The answer is 9. Through visualization and systematic algebra, we have navigated the constraints to find the solution.