JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}| = \sqrt{31}, 4|\vec{b}| = |\vec{c}| = 2$ and $2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a})$ . If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$ , then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2$ is equal to ________.

Choose the correct answer:

Explanation

Solution:

Diye gaye magnitudes: $|\vec{a}| = \sqrt{31}, |\vec{b}| = \frac{1}{2}, |\vec{c}| = 2$ .

Condition: $2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a}) \implies \vec{a} \times (2\vec{b} + 3\vec{c}) = 0$ .

Iska matlab $\vec{a} = \lambda(2\vec{b} + 3\vec{c})$ .

Dono side ka square karne par:

|\vec{a}|^2 = \lambda^2 |2\vec{b} + 3\vec{c}|^2 = \lambda^2 (4|\vec{b}|^2 + 9|\vec{c}|^2 + 12\vec{b} \cdot \vec{c})

31 = \lambda^2 (4(\frac{1}{4}) + 9(4) + 12(\frac{1}{2})(2)\cos\frac{2\pi}{3}) = \lambda^2 (1 + 36 - 6) = 31\lambda^2

$\lambda^2 = 1 \implies \lambda = 1$ . Toh $\vec{a} = 2\vec{b} + 3\vec{c}$ .

Ab, $\vec{a} \times \vec{c} = (2\vec{b} + 3\vec{c}) \times \vec{c} = 2(\vec{b} \times \vec{c})$ .

Aur $\vec{a} \cdot \vec{b} = (2\vec{b} + 3\vec{c}) \cdot \vec{b} = 2|\vec{b}|^2 + 3\vec{b} \cdot \vec{c} = 2(\frac{1}{4}) + 3(\frac{1}{2})(2)(-\frac{1}{2}) = \frac{1}{2} - \frac{3}{2} = -1$ .

Magnitude $|\vec{a} \times \vec{c}| = 2|\vec{b}||\vec{c}|\sin\frac{2\pi}{3} = 2(\frac{1}{2})(2)(\frac{\sqrt{3}}{2}) = \sqrt{3}$ .

Final value:

\left(\frac{\sqrt{3}}{-1}\right)^2 = 3

Answer: 3

Choose the correct answer:

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