JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to

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Explanation

Step 1: Lines ko standard form mein likhein

Pehli line ( $L_1$ ):

\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}

Yahan point $\vec{a}_1 = (2, -1, 6)$ aur direction vector $\vec{b}_1 = (3, 2, 2)$ hai.

Doosri line ( $L_2$ ):

Dhyan dein yahan $1-y$ hai, ise standard form $y-1$ banane ke liye niche sign badalna hoga:

\frac{x - 6}{3} = \frac{y - 1}{-2} = \frac{z + 8}{0}

Yahan point $\vec{a}_2 = (6, 1, -8)$ aur direction vector $\vec{b}_2 = (3, -2, 0)$ hai.

Step 2: Vector $\vec{a}_2 - \vec{a}_1$ nikalna

\vec{a}_2 - \vec{a}_1 = (6-2, 1-(-1), -8-6) = (4, 2, -14)

Step 3: Direction vectors ka cross product ( $\vec{b}_1 \times \vec{b}_2$ ) nikalna

\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix}

= \hat{i}(0 - (-4)) - \hat{j}(0 - 6) + \hat{k}(-6 - 6)

= 4\hat{i} + 6\hat{j} - 12\hat{k}

Magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ .

Step 4: 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|}

d = \frac{|(4, 2, -14) \cdot (4, 6, -12)|}{14}

d = \frac{|(4 \times 4) + (2 \times 6) + (-14 \times -12)|}{14}

d = \frac{|16 + 12 + 168|}{14} = \frac{196}{14} = 14

Final Answer:

Dono lines ke beech ki shortest distance 14 unit hai.

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