NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

The number of distinct real values of $λ^{2} \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + λ^{2} \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + λ^{2} \hat{k}$ are coplanar is

Choose the correct answer:

Explanation

For three vectors to be coplanar, their scalar triple product must be zero. This is equivalent to the determinant of the matrix formed by their components being zero:

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

Applying row operation $R_1 \to R_1 + R_2 + R_3$ :

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

Taking $(\lambda^2+2)$ common from $R_1$ :

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

Applying column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ :

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

Expanding along $R_1$ :

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

Case 1: $\lambda^2 + 2 = 0 \implies \lambda^2 = -2$

(No real values for $\lambda$ )

Case 2: $(\lambda^2 - 1)^2 = 0 \implies \lambda^2 = 1$

$\lambda = \pm 1$

The distinct real values of $\lambda$ are $1$ and $-1$ .

Number of distinct real values = $2$ .

Correct Option: (B)

Choose the correct answer:

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