Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}-7\hat{j}+5\hat{k},\vec{b}=\hat{i}+\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}-3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times \vec{a}=\vec{c}\times \vec{a}$ and $\vec{r}\cdot \vec{b}=0$, then $|\vec{r}|$ is equal to:

Welcome, fellow traveler in the world of JEE Advanced mathematics. Today, we are not just solving a problem; we are peeling back the layers of vector geometry to reveal a hidden structure.

We are given three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, and we are tasked with finding a vector $\vec{r}$ that satisfies two distinct conditions. This is a classic scenario where algebra meets geometry.

The first condition, $\vec{r}\times \vec{a}=\vec{c}\times \vec{a}$, is the heartbeat of this problem. Many students look at this and instinctively want to cancel $\vec{a}$ from both sides. But wait! In vector algebra, we cannot simply divide by a vector.

Instead, we bring everything to one side:

By the distributive property of the cross product, this becomes:

Now, pause and visualize. What does it mean for the cross product of two vectors to be the zero vector? It means they are collinear!

Therefore, the vector $(\vec{r}-\vec{c})$ must be parallel to $\vec{a}$. This leads us to the elegant parametric form:

Geometrically, this equation describes a straight line in 3D space passing through the tip of $\vec{c}$ and extending in the direction of $\vec{a}$. Every point on this line is a potential candidate for $\vec{r}$.

We have a line of candidates, but we need the specific one that satisfies the second condition: $\vec{r}\cdot \vec{b}=0$. This condition tells us that $\vec{r}$ must be perpendicular to $\vec{b}$.

We substitute our parametric expression for $\vec{r}$ into this condition:

Expanding this, we get:

This is the moment where the abstract geometry turns into concrete arithmetic. We calculate the dot products:

Substituting these back, we find $-2+7\lambda =0$, which gives us the scalar:

With $\lambda =\frac{2}{7}$ in hand, we can finally define our vector $\vec{r}$ explicitly. Substituting $\lambda$ back into $\vec{r}=\vec{c}+\lambda \vec{a}$, we get:

Grouping the components, we arrive at:

The final step is to find the magnitude $|\vec{r}|$. Using the formula:

We factor out the common term to get:

We have successfully navigated the line, applied the perpendicular constraint, and arrived at the solution. Remember, in JEE Advanced, the math is the language, but the geometry is the story.