NIMCET 2022 — Mathematics PYQ

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Question

NIMCET 2022 — Mathematics PYQ

NIMCET | Mathematics | 2022

If $a = λ \hat{i} + \hat{j} - 2 \hat{k}$ , $b = \hat{i} + λ \hat{j} - 2 \hat{k}$ , and $c = \hat{i} + \hat{j} + \hat{k}$ such that the scalar triple product $[a b c] = 7$ , then the values of $λ$ are:

Choose the correct answer:

Explanation

Solution

The scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ is given by the determinant of the coefficients of the vectors:

[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} \lambda & 1 & -2 \\ 1 & \lambda & -2 \\ 1 & 1 & 1 \end{vmatrix}

Given that $[\vec{a} \ \vec{b} \ \vec{c}] = 7$ , we set up the equation:

\begin{vmatrix} \lambda & 1 & -2 \\ 1 & \lambda & -2 \\ 1 & 1 & 1 \end{vmatrix} = 7

Expanding the determinant along the first row:

\lambda(\lambda(1) - (-2)(1)) - 1(1(1) - (-2)(1)) + (-2)(1(1) - \lambda(1)) = 7

\lambda(\lambda + 2) - 1(1 + 2) - 2(1 - \lambda) = 7

\lambda^2 + 2\lambda - 3 - 2 + 2\lambda = 7

Simplify the equation:

\lambda^2 + 4\lambda - 5 = 7

\lambda^2 + 4\lambda - 12 = 0

Factoring the quadratic equation:

\lambda^2 + 6\lambda - 2\lambda - 12 = 0

\lambda(\lambda + 6) - 2(\lambda + 6) = 0

(\lambda + 6)(\lambda - 2) = 0

Solving for $\lambda$ :

\lambda = 2, \ -6