If the vectors ai + j + k, i + bj + k and i + j + ck (a ≠ b ≠ c ≠ 1) are coplanar, then the value of

Q: Why can't I just expand the determinant directly? It seems like a standard $3 \times 3$ expansion would work.

You absolutely could, but in JEE Advanced, time is your most precious currency. Expanding directly leads to a massive polynomial that is prone to sign errors. The 'smart' way—using column operations—is designed to make the terms collapse into the form we need. Always look for the path of least resistance!

Q: Wait, why are we allowed to factor out $(1-a)$, $(1-b)$, and $(1-c)$? Is that mathematically sound?

Great catch! We are allowed to do this because the problem explicitly states $a \neq 1$, $b \neq 1$, and $c \neq 1$. Since these terms are non-zero, we are essentially dividing by a non-zero number, which is perfectly valid. If they were equal to 1, the whole problem would be undefined!

Q: Is there a physical intuition for why the scalar triple product must be zero?

Think of the scalar triple product as the volume of a parallelepiped formed by the three vectors. If the vectors are coplanar, they lie on a flat sheet. A flat sheet has no 'height' or 'thickness' in the third dimension. Therefore, the volume is zero. It's geometry masquerading as algebra!

Question

If the vectors $a\hat{i} + \hat{j} + \hat{k}$, $\hat{i} + b\hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c\hat{k}$ ($a 
eq b 
eq c 
eq 1$) are coplanar, then the value of $\frac{1}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space. You have three vectors: $\vec{u} = a\hat{i} + \hat{j} + \hat{k}$, $\vec{v} = \hat{i} + b\hat{j} + \hat{k}$, and $\vec{w} = \hat{i} + \hat{j} + c\hat{k}$.

The problem states that these vectors are coplanar. This means that if you were to draw these vectors starting from the origin, they would all lie perfectly flat on a single, two-dimensional sheet.

In the language of linear algebra, this implies they are linearly dependent and cannot span a volume. The mathematical gatekeeper of volume in 3D space is the **scalar triple product**.

If the volume of the parallelepiped formed by these vectors is zero, the vectors are coplanar. Thus, our condition is simply:
$$[\vec{u} \quad \vec{v} \quad \vec{w}] = 0$$

### The Determinant Strategy

We represent the scalar triple product as a determinant. We set the determinant of the components to zero:
$$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$$

To simplify this, we use column operations. We perform $C_1 ightarrow C_1 - C_2$ and $C_2 ightarrow C_2 - C_3$. This transforms our matrix into a much more manageable form:
$$\begin{vmatrix} a-1 & 0 & 1 \ 1-b & b-1 & 1 \ 0 & 1-c & c \end{vmatrix} = 0$$

### The Algebraic Alchemy

We now introduce the terms $\frac{1}{1-a}$, $\frac{1}{1-b}$, and $\frac{1}{1-c}$ by factoring. We factor $(1-a)$ from the first column, $(1-b)$ from the second, and $(1-c)$ from the third.

Since the problem guarantees $a, b, c 
eq 1$, we are safe from division by zero. After factoring, our determinant becomes:
$$(1-a)(1-b)(1-c) \begin{vmatrix} -1 & 0 & \frac{1}{1-a} \ 1 & -1 & \frac{1}{1-b} \ 0 & 1 & \frac{c}{1-c} \end{vmatrix} = 0$$

Since the product of our factors is non-zero, the determinant itself must be zero. Expanding along the first row, we obtain:
$$-1 \left( -\frac{c}{1-c} - \frac{1}{1-b} ight) + \frac{1}{1-a}(1 - 0) = 0$$

This simplifies to the following expression:
$$\frac{c}{1-c} + \frac{1}{1-b} + \frac{1}{1-a} = 0$$

### The Final Flourish

We rewrite the term $\frac{c}{1-c}$ as $\frac{1 - (1-c)}{1-c}$, which is equivalent to $\frac{1}{1-c} - 1$.

Substituting this back into our equation, we get:
$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} - 1 = 0$$

Moving the $-1$ to the other side, we find the final result:
$$\mathbf{\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1}$$

This journey from the geometric concept of coplanarity to the algebraic language of determinants demonstrates the essence of JEE Advanced mathematics: connecting disparate concepts to reveal a singular, elegant truth.

Enter Numerical Value:

Visualized Solution

Visualizing Coplanar Vectors

The Scalar Triple Product Condition

Setting up the Determinant

Applying Column Operations

The Transformed Determinant

Factoring Out $(1 - a)$ , $(1 - b)$ , and $(1 - c)$

The Factored Matrix Structure

Expanding Along the First Row

Simplifying the Equation

Transforming the Third Term

Finding the Final Value

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Determinant Strategy

The Algebraic Alchemy

The Final Flourish

Similar Questions

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