Animated Solution for Mathematics - Vector Algebra: The sum of the distinct real values of $\mu$, for which the vectors, $\mu \hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\mu \hat{j}+\hat{k}$, $\hat{i}+\hat{j}+\mu \hat{k}$ are co-planar, is :

Understanding Coplanarity

My dear student, welcome to the heart of vector algebra. Today, we are not just solving for a variable; we are exploring the geometric soul of vectors.

Imagine you are standing in a three-dimensional room. You have three arrows, $\vec{a}$, $\vec{b}$, and $\vec{c}$, all starting from the same point. Usually, these arrows would point in different directions, creating a sense of depth and volume.

But the problem gives us a constraint: these vectors are coplanar.

What does that mean? It means that despite living in a 3D world, these three vectors are trapped on a single, flat sheet of paper. They have no 'height' relative to each other. This is the physical reality we must translate into the language of mathematics.

The Volume of Nothing

How do we quantify this 'flatness'? We use the Scalar Triple Product, denoted as $[\vec{a}\vec{b}\vec{c}]$.

Geometrically, this product represents the volume of a parallelepiped formed by the three vectors. If the vectors are coplanar, they cannot form a 3D shape. The volume must be zero.

Therefore, the condition for coplanarity is simply:

This is our golden key. It transforms a geometric concept into a concrete algebraic equation.

The Determinant

To calculate this product, we arrange the components of our vectors into a $3\times 3$ matrix. Our vectors are $\vec{a}=\mu \hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+\mu \hat{j}+\hat{k}$, and $\vec{c}=\hat{i}+\hat{j}+\mu \hat{k}$.

Placing these into our determinant, we get:

I know, a determinant can look intimidating. But take a breath. Let us expand this along the first row.

We take the first element, $\mu$, and multiply it by the determinant of the remaining $2\times 2$ matrix, then subtract the second element, and so on. The expansion looks like this:

Factoring with Elegance

Now, here is where many students rush and make errors. Do not just multiply everything out! Look for patterns.

Notice that $({\mu }^2-1)$ is a difference of squares: $(\mu -1)(\mu +1)$. And notice that $(1-\mu )$ is just $-(\mu -1)$. Let us rewrite the equation with these insights:

Do you see it? The term $(\mu -1)$ is common to every single part of the expression! We can factor it out like a master conductor leading an orchestra:

Simplifying the bracket gives us ${\mu }^2+\mu -2$. We factor this quadratic as $(\mu +2)(\mu -1)$. Our final equation is:

Distinct Values

We have arrived at the roots: $\mu =1$ and $\mu =-2$. The equation $(\mu -1{)}^2=0$ gives us $\mu =1$ (a repeated root), and $\mu +2=0$ gives us $\mu =-2$.

Now, look closely at the question. It asks for the sum of the distinct real values. This is the final test of your focus.

We do not sum $1+1+(-2)$. We sum only the unique values: $1$ and $-2$.

And there it is. We have navigated the geometry, mastered the determinant, danced through the algebra, and avoided the trap. You have successfully solved the problem. Keep this clarity of thought, and you will conquer any challenge the JEE throws your way!