Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$ and $\vec{b}=7\hat{i}+\hat{j}-6\hat{k}$. If $\vec{r}\times \vec{a}=\vec{r}\times \vec{b}$, $\vec{r}\cdot (\hat{i}+2\hat{j}+\hat{k})=-3$, then $\vec{r}\cdot (2\hat{i}-3\hat{j}+\hat{k})$ is equal to

Welcome, my dear student. Today, we are not merely solving a problem; we are embarking on a journey into the elegant, structured world of vector algebra. In the JEE Advanced examination, the examiners do not just test your ability to calculate; they test your ability to see the 'soul' of the mathematics.

When you look at an equation like $\vec{r}\times \vec{a}=\vec{r}\times \vec{b}$, do you see just symbols, or do you see a geometric constraint waiting to be unlocked? Let us peel back the layers of this problem together.

We begin with the given condition: $\vec{r}\times \vec{a}=\vec{r}\times \vec{b}$. Many students might be tempted to immediately expand $\vec{r}$ into components $(x,y,z)$ and start calculating determinants. Stop! Take a breath.

In physics and advanced mathematics, the most powerful tool is often the simplest property. We know that the cross product is distributive. So, let us bring everything to one side of the equation:

By applying the distributive property, we can factor out the vector $\vec{r}$:

This is the 'Aha!' moment. We have transformed a complex equality into a statement about parallelism. The cross product of two vectors is zero if and only if the vectors are parallel. This tells us that our mystery vector $\vec{r}$ is constrained to lie along the direction of the vector $(\vec{a}-\vec{b})$.

Now that we know $\vec{r}$ is parallel to $(\vec{a}-\vec{b})$, we can express $\vec{r}$ using a scalar parameter, $\lambda$. This $\lambda$ is our bridge, allowing us to represent the entire family of vectors parallel to $(\vec{a}-\vec{b})$ with a single variable. First, let us calculate the difference vector:

Performing the component-wise subtraction, we get:

Thus, we can write our mystery vector as:

We have successfully reduced our unknown vector $\vec{r}$ to a single variable, $\lambda$. We look to the second piece of information provided: $\vec{r}\cdot (\hat{i}+2\hat{j}+\hat{k})=-3$. This is the anchor that fixes our vector in space.

Let us substitute our expression for $\vec{r}$ into this dot product:

Distributing the dot product, we calculate the scalar sum:

With a sigh of relief, we see that $\lambda =1$. The math has aligned perfectly. Our mystery vector $\vec{r}$ is simply $-5\hat{i}-4\hat{j}+10\hat{k}$.

The problem asks us to evaluate $\vec{r}\cdot (2\hat{i}-3\hat{j}+\hat{k})$. Now that we have fully identified $\vec{r}$, this is merely a matter of execution. Let us perform the final dot product:

And there it is. The final answer is $12$.

My dear student, notice how we did not rush into the algebra. We respected the geometry first. We used the cross product to find the direction, and the dot product to find the magnitude. This is the hallmark of a true physicist.