JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$ and $\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$ is:

Choose the correct answer:

Explanation

Solution

Shortest Distance ( $d$ ) ka formula: $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$

$\vec{a}_1 = (1, 2, 6), \vec{a}_2 = (1, -8, 4) \implies \vec{a}_2 - \vec{a}_1 = (0, -10, -2)$
$\vec{b}_1 = (2, 1, -3), \vec{b}_2 = (2, -7, 5)$

$\vec{b}_1 \times \vec{b}_2$ nikalne par:

\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = \hat{i}(5-21) - \hat{j}(10+6) + \hat{k}(-14-2) = -16\hat{i} - 16\hat{j} - 16\hat{k}

Yani hum vector $(-1, -1, -1)$ le sakte hain. Iska magnitude $|\vec{b}_1 \times \vec{b}_2| = 16\sqrt{3}$ .

d = \frac{|(0, -10, -2) \cdot (-16, -16, -16)|}{16\sqrt{3}} = \frac{|160 + 32|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}

Sahi Option: (4)

Choose the correct answer:

Explore More