Let the vectors (2 + a + b)i + (a + 2b + c)j - (b + c)k, (1 + b)i + 2j - bk and (2 + b)i + 2j + (1 -

Q: Why do we use the scalar triple product instead of just checking for linear dependence?

The scalar triple product is the most elegant way to measure the volume spanned by three vectors. If they are coplanar, they don't span any 3D volume, hence the volume is zero. It is the direct, mathematical definition of coplanarity in 3D space.

Q: Can I just expand the determinant directly without row operations?

You could, but it is a trap! JEE examiners love to test your ability to simplify. Direct expansion of a 3x3 determinant with variables is highly prone to calculation errors. Always simplify first to save time and ensure accuracy.

Q: What if I get a non-zero value after my calculation?

If you get a non-zero value, it means the vectors are linearly independent and span a 3D volume. In this problem, we are given they are coplanar, so the result must be zero. If you don't get zero, re-check your row operations or your expansion signs!

Question

Let the vectors $(2 + a + b)\hat{i} + (a + 2b + c)\hat{j} - (b + c)\hat{k}, (1 + b)\hat{i} + 2\hat{j} - b\hat{k}$ and $(2 + b)\hat{i} + 2\hat{j} + (1 - b)\hat{k}, a, b, c \in \mathbb{R}$ be co-planar. Then which of the following is true?

Options:

(A) $2b = a + c$

(B) $3c = a + b$

(C) $a = b + 2c$

(D) $2a = b + c$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we talk about coplanarity, we are really talking about the absence of volume. The scalar triple product, denoted as $[\vec{v}_1 \vec{v}_2 \vec{v}_3]$, is the mathematical tool that calculates the volume of the parallelepiped formed by these vectors.

If the vectors are coplanar, the volume is zero. Therefore, our condition is simply $[\vec{v}_1 \vec{v}_2 \vec{v}_3] = 0$.

This is our gateway. We set up the determinant:

$$\Delta = \begin{vmatrix} 2+a+b & a+2b+c & -(b+c) \ 1+b & 2 & -b \ 2+b & 2 & 1-b \end{vmatrix} = 0$$

### The Art of Simplification

In JEE Advanced, brute force is the enemy of elegance. Look at the second and third rows; notice how similar they are.

If we perform the operation $R_3 	o R_3 - R_2$, we get:

$$(2+b) - (1+b) = 1$$

$$2 - 2 = 0$$

$$(1-b) - (-b) = 1$$

Suddenly, our third row is just $[1, 0, 1]$. This is the beauty of linear algebra—a little bit of manipulation clears the fog.

We can do the same for the first row using $R_1 	o R_1 - R_2$. The $b$ terms vanish, and the expression becomes manageable.

### The Final Execution

With our simplified determinant, we have:

$$\begin{vmatrix} a+1 & a+2b+c-2 & -c \ 1+b & 2 & -b \ 1 & 0 & 1 \end{vmatrix} = 0$$

Expanding along the third row is now a breeze. We multiply the elements by their cofactors, noting that the zero makes our life much easier.

After careful expansion, we arrive at the beautiful, simple relation:

$$a - 2b + c = 0$$

Rearranging this, we get **$2b = a + c$**.

### The Hidden Harmony

Look at that result: $2b = a + c$. It is the classic condition for three numbers to be in an **Arithmetic Progression (A.P.)**.

We started with complex vector components, and through the elegance of row operations and the logic of coplanarity, we uncovered a fundamental sequence. This is why we study mathematics—to find the hidden order in the chaos.

Animated Solution for Mathematics - Vector Algebra: Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}, (1 + b) \hat{i} + 2 \hat{j} - b \hat{k}$ and $(2 + b) \hat{i} + 2 \hat{j} + (1 - b) \hat{k}, a, b, c \in R$ be co-planar. Then which of the following is true?

Select Answer:

Visualized Solution

Visualizing Coplanarity

The Determinant Condition

Simplifying Row 3

Simplifying Row 1

The New Determinant

Expansion & Simplification

The Final Relation

Summary & Takeaway

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Art of Simplification

The Final Execution

The Hidden Harmony

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