JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The shortest distance between the lines $x + 1 = 2y = -12z$ and $x = y + 2 = 6z - 6$ is

Choose the correct answer:

Explanation

1. Convert Lines to Standard Form

The standard symmetric form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ .

Line 1 ( $L_1$ ): $x + 1 = 2y = -12z$

Divide by $12$ : $\frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1}$
- Point $\vec{a_1} = (-1, 0, 0)$
- Direction $\vec{b_1} = (12, 6, -1)$
Line 2 ( $L_2$ ): $x = y + 2 = 6z - 6$

Write as: $\frac{x}{1} = \frac{y+2}{1} = \frac{6(z-1)}{1} \implies \frac{x}{6} = \frac{y+2}{6} = \frac{z-1}{1}$
- Point $\vec{a_2} = (0, -2, 1)$
- Direction $\vec{b_2} = (6, 6, 1)$

2. Calculate Necessary Vectors

Vector joining the points: $\vec{a_2} - \vec{a_1} = (0 - (-1), -2 - 0, 1 - 0) = (1, -2, 1)$
Cross product of directions ( $\vec{b_1} \times \vec{b_2}$ ):

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix}$

$= \hat{i}(6 - (-6)) - \hat{j}(12 - (-6)) + \hat{k}(72 - 36)$

$= 12\hat{i} - 18\hat{j} + 36\hat{k}$

3. Compute Magnitudes and Dot Product

Magnitude of cross product:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{12^2 + (-18)^2 + 36^2}$

$= \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42$
Dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ :

$(1)(12) + (-2)(-18) + (1)(36) = 12 + 36 + 36 = 84$

4. Apply Shortest Distance Formula

The shortest distance $d$ is given by:

d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|

d = \frac{84}{42} = 2

Correct Option: (2) 2

Choose the correct answer:

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