Let a = i + 2j + 4k, b = i + λj + 4k and c = 2i + 4j + (λ² - 1)k be coplanar vectors. Then the non-z

Q: Why is the scalar triple product zero for coplanar vectors?

The scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ geometrically represents the volume of the parallelepiped formed by the three vectors. If the vectors are coplanar, they lie in the same plane, meaning the height of this parallelepiped is zero. Thus, the volume must be zero.

Q: Why do we reject $\lambda = 3$ and $\lambda = -3$?

The problem explicitly requires the vector $\vec{a} \times \vec{c}$ to be non-zero. If $\lambda = 3$ or $\lambda = -3$, then $\lambda^2 - 1 = 8$. Substituting this into $\vec{c}$, we find $\vec{c} = 2\hat{i} + 4\hat{j} + 8\hat{k} = 2(\hat{i} + 2\hat{j} + 4\hat{k}) = 2\vec{a}$. Since $\vec{c}$ is a scalar multiple of $\vec{a}$, they are parallel, and their cross product is the zero vector, which violates the condition.

Q: Is there a faster way to solve the determinant?

Absolutely! Row operations are your best friend in linear algebra. By performing $R_2 \to R_2 - R_1$, you immediately create two zeros in the second row. This reduces the complexity of the determinant expansion significantly, allowing you to solve for $\lambda$ in seconds.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} + \lambda\hat{j} + 4\hat{k}$ and $\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a} 	imes \vec{c}$ is :

Options:

(A) $-14\hat{i} - 5\hat{j}$

(B) $-10\hat{i} - 5\hat{j}$

(C) $-10\hat{i} + 5\hat{j}$

(D) $-14\hat{i} + 5\hat{j}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D space, holding three arrows representing vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. The problem states that these vectors are coplanar.

This means that there exists a single, flat sheet of paper that can contain all three of them simultaneously. In the language of linear algebra, this implies that the volume of the parallelepiped formed by these three vectors is zero.

The mathematical gatekeeper for this condition is the **Scalar Triple Product**, denoted as $[\vec{a} \ \vec{b} \ \vec{c}] = 0$. This is the foundation upon which we build our solution.

### The Algebraic Dance

To find the value of $\lambda$, we translate our geometric condition into a determinant. We place the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$ into a $3 	imes 3$ matrix:

$$
\begin{vmatrix} 1 & 2 & 4 \ 1 & \lambda & 4 \ 2 & 4 & \lambda^2 - 1 \end{vmatrix} = 0
$$

Do not rush to expand this directly. If we perform the row operation $R_2 	o R_2 - R_1$, the second row becomes $(0, \lambda - 2, 0)$.

Expanding along this row becomes trivial:

$$
(\lambda - 2) 	imes \begin{vmatrix} 1 & 4 \ 2 & \lambda^2 - 1 \end{vmatrix} = 0
$$

This simplifies to $(\lambda - 2)(\lambda^2 - 1 - 8) = 0$, which is $(\lambda - 2)(\lambda^2 - 9) = 0$. Solving this gives us three potential candidates for $\lambda$: $2$, $3$, and $-3$.

### The Hidden Trap

Many students stop here, but the JEE Advanced examiner is clever. The problem adds a crucial constraint: the vector $\vec{a} 	imes \vec{c}$ must be non-zero.

We must test our candidates. If $\lambda = 3$ or $\lambda = -3$, then $\lambda^2 - 1 = 8$. Our vector $\vec{c}$ becomes $2\hat{i} + 4\hat{j} + 8\hat{k}$, which is exactly $2\vec{a}$.

The cross product of any vector with a scalar multiple of itself is always the zero vector. Thus, $\lambda = 3$ and $\lambda = -3$ are traps and must be rejected. This leaves us with only one valid choice: **$\lambda = 2$**.

### The Final Triumph

With $\lambda = 2$ firmly in our grasp, we find $\vec{c} = 2\hat{i} + 4\hat{j} + 3\hat{k}$. Now, we calculate the final cross product $\vec{a} 	imes \vec{c}$ using the determinant method:

$$
\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 4 & 3 \end{vmatrix}
$$

Expanding this, we get:

$$
\hat{i}(2 \cdot 3 - 4 \cdot 4) - \hat{j}(1 \cdot 3 - 2 \cdot 4) + \hat{k}(1 \cdot 4 - 2 \cdot 2)
$$

This simplifies to $\hat{i}(6 - 16) - \hat{j}(3 - 8) + \hat{k}(4 - 4)$, which results in **$-10\hat{i} + 5\hat{j}$**. The zero in the $\hat{k}$ component is a beautiful confirmation of our work.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}$ , $b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then the non-zero vector $a \times c$ is :

Select Answer:

Visualized Solution

Understanding Coplanarity

The Coplanarity Condition

Setting up the Determinant

Simplifying the Determinant

Expanding the Determinant

Solving for $λ$

Checking the Non-Zero Condition

Evaluating $λ = \pm 3$

Fixing $λ = 2$

Setting up $a \times c$

Calculating the Cross Product

Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Algebraic Dance

The Hidden Trap

The Final Triumph

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+4k^, b=i^+λj^​+4k^ and c=2i^+4j^​+(λ2−1)k^ be coplanar vectors. Then the non-zero vector a×c is :