JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = 11$ , $\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$ and $\vec{b} \cdot \vec{c} = -\sqrt{3} |\vec{b}|$ , then $|\vec{a} \times \vec{c}|^2$ is equal to ________.

Choose the correct answer:

Explanation

Solution

1. Vector Magnitudes and Dot Products:

$|\vec{a}|^2 = 1^2 + 2^2 + 3^2 = 14$ .
$|\vec{b}|^2 = 1^2 + 1^2 + (-1)^2 = 3 \implies |\vec{b}| = \sqrt{3}$ .
$\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (3)(-1) = 0$ (The vectors are perpendicular).
$\vec{b} \cdot \vec{c} = -\sqrt{3}(\sqrt{3}) = -3$ .

2. Scalar Triple Product Relation:

$[\vec{b} \ \vec{a} \ \vec{c}] = 27$ .

We use the determinant property for the square of the scalar triple product:

$[\vec{b} \ \vec{a} \ \vec{c}]^2 = \begin{vmatrix} \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{c} \\ \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{c} \\ \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{c} \end{vmatrix}$

$27^2 = \begin{vmatrix} 3 & 0 & -3 \\ 0 & 14 & 11 \\ -3 & 11 & |\vec{c}|^2 \end{vmatrix}$

$729 = 3(14|\vec{c}|^2 - 121) - 3(0 + 42)$ .

$243 = 14|\vec{c}|^2 - 121 - 42 \implies 14|\vec{c}|^2 = 406 \implies |\vec{c}|^2 = 29$ .

3. Finding $|\vec{a} \times \vec{c}|^2$ :

Using Lagrange's Identity:

$|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2 |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$

$|\vec{a} \times \vec{c}|^2 = 14(29) - (11)^2$

$= 406 - 121 = 285$ .

Final Answer: 285

Choose the correct answer:

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