The number of distinct real values of λ, for which the vectors -λ²i + j + k, i - λ²j + k and i + j -

Q: Why do we set the determinant to zero? Can't we just use cross products?

You absolutely can use cross products, but it is a trap! Calculating $\vec{a} \cdot (\vec{b} \times \vec{c})$ is the definition of the volume of a parallelepiped. If the volume is zero, the vectors are coplanar. Using the determinant is the most elegant and efficient path in JEE Advanced, as it avoids the tedious expansion of cross products.

Q: Wait, why do we reject the solution from $\lambda^2 + 1 = 0$?

Great catch! The question explicitly asks for *real* values of $\lambda$. Since $\lambda^2 = -1$ implies $\lambda = \pm i$, these are complex numbers. Always check the domain constraints in the problem statement; in JEE, ignoring the 'real' constraint is a classic way to lose marks.

Q: Is there a faster way to solve the determinant without row operations?

You could expand it directly, but you will end up with a cubic in $\lambda^2$ that is much harder to factor. Row operations reveal the symmetry and common factors immediately. In JEE, always look for symmetry before brute-forcing!

Question

The number of distinct real values of $\lambda$, for which the vectors $-\lambda^2\hat{i} + \hat{j} + \hat{k}$, $\hat{i} - \lambda^2\hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - \lambda^2\hat{k}$ are coplanar, is

Options:

(A) zero

(B) one

(C) two

(D) three

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Three vectors are given as $\vec{a} = -\lambda^2\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \lambda^2\hat{j} + \hat{k}$, and $\vec{c} = \hat{i} + \hat{j} - \lambda^2\hat{k}$. These vectors span a parallelepiped in 3D space.

Our objective is to find the values of $\lambda$ such that these vectors become coplanar. Coplanarity occurs when the volume of the parallelepiped they span is zero.

### The Scalar Triple Product

The condition for coplanarity is defined by the Scalar Triple Product being equal to zero, expressed as $[\vec{a}\ \vec{b}\ \vec{c}] = 0$. This is equivalent to the determinant of the matrix formed by the components of the vectors:

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

### The Elegance of Row Operations

To simplify the determinant, we apply row operations $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$. This transformation yields:

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

We observe that $(\lambda^2 + 1)$ is a common factor in both the first and second rows. Factoring this out twice, we obtain:

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

### Final Calculation

Expanding the remaining $3 	imes 3$ determinant, we calculate:

$$
-1[(-1)(-\lambda^2) - (1)(1)] - 1[(0)(-\lambda^2) - (1)(1)] = 0
$$

This simplifies to $-1(\lambda^2 - 1) + 1 = 0$, which further reduces to $2 - \lambda^2 = 0$. Combining this with our earlier factor, the master equation becomes:

$$
(\lambda^2 + 1)^2 (2 - \lambda^2) = 0
$$

### The Final Verdict

We analyze the two cases resulting from the product:

1.  $\lambda^2 + 1 = 0 \Rightarrow \lambda^2 = -1$. Since we are restricted to real numbers, this yields no real solutions.
2.  $2 - \lambda^2 = 0 \Rightarrow \lambda^2 = 2$.

The real values of $\lambda$ that force the vectors into a flat plane are **$\lambda = \sqrt{2}$** and **$\lambda = -\sqrt{2}$**.

Animated Solution for Mathematics - Vector Algebra: The number of distinct real values of $λ$ , for which the vectors $- λ^{2} \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} - λ^{2} \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - λ^{2} \hat{k}$ are coplanar, is

Select Answer:

Visualized Solution

Visualizing the Vectors

The Condition for Coplanarity

Setting up the Determinant

Simplifying the Determinant

Factoring Out Common Terms

Expanding the Simplified Determinant

The Complete Factored Equation

Analyzing Case 1: $λ^{2} + 1 = 0$

Analyzing Case 2: $2 - λ^{2} = 0$

Final Count of Real Values

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Scalar Triple Product

The Elegance of Row Operations

Final Calculation

The Final Verdict

Similar Questions

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