JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The shortest distance between the lines $\frac{x + 2}{1} = \frac{y}{-2} = \frac{z - 5}{2}$ and $\frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0}$ is:

Choose the correct answer:

Explanation

Lines hain $\vec{r_1} = (-2, 0, 5) + \lambda(1, -2, 2)$ aur $\vec{r_2} = (4, 1, -3) + \mu(1, 2, 0)$ .

Shortest distance $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$ .

$\vec{a_2} - \vec{a_1} = (6, 1, -8)$
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -4\hat{i} + 2\hat{j} + 4\hat{k}$
$d = \frac{|(6)(-4) + (1)(2) + (-8)(4)|}{\sqrt{(-4)^2 + 2^2 + 4^2}} = \frac{|-24 + 2 - 32|}{6} = \frac{54}{6} = 9$