JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}$ , $\vec{b} = 3\hat{i} + 5\hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ . Let $\vec{d}$ be a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$ , and $\vec{c} \cdot \vec{d} = 12$ . The $(-\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d})$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1 ( $\vec{d}$ nikaalein): $\vec{d}$ perpendicular hai $\vec{a}$ aur $\vec{b}$ ke, toh $\vec{d} = \lambda(\vec{a} \times \vec{b})$ .

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{vmatrix} = 35\hat{i} - 13\hat{j} - 21\hat{k}$

$\vec{c} \cdot \vec{d} = 12 \implies \lambda(1(35) - 1(-13) + 2(-21)) = 12 \implies \lambda(35+13-42) = 12 \implies 6\lambda = 12 \implies \lambda = 2$ .

Toh, $\vec{d} = 70\hat{i} - 26\hat{j} - 42\hat{k}$ .
Step 2 (Scalar Triple Product): Maana $\vec{v} = -\hat{i} + \hat{j} - \hat{k}$ . Humein $\vec{v} \cdot (\vec{c} \times \vec{d})$ yaani $[\vec{v} \ \vec{c} \ \vec{d}]$ nikaalna hai.

$[\vec{v} \ \vec{c} \ \vec{d}] = \begin{vmatrix} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{vmatrix}$

Row operation $R_{1} \to R_{1} + R_{2}$ karne par:

$\begin{vmatrix} 0 & 0 & 1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{vmatrix} = 1(-26 - (-70)) = 44$

Choose the correct answer:

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