The shortest distance between lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, where $L_{1}: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_{2}$ is the line passing through the points $\mathrm{A}(-4,4,3) \cdot \mathrm{B}(-1,6,3)$ and perpendicular to the line $\frac{\mathrm{x}-3}{-2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}-1}{1}$, is

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Three Dimensional Geometry

The shortest distance between lines

L_{1}

and

L_{2}

, where

L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}

and

L_{2}

is the line passing through the points

A (- 4, 4, 3) \cdot B (- 1, 6, 3)

and perpendicular to the line

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

, is

Answer: (C)

L_{1}

has direction

d_{1} = (2, - 3, 2)

and passes through

P_{1} = (1, - 1, - 4)

.

For

L_{2}

: direction

A B = B - A = (- 1 - (- 4), 6 - 4, 3 - 3) = (3, 2, 0)

. Checking against the given perpendicularity condition,

d_{1}^{'} \cdot (3, 2, 0) = (- 2) (3) + 3 (2) + 1 (0) = 0

, so this direction is indeed perpendicular to

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

, confirming

L_{2}

has direction

d_{2} = (3, 2, 0)

through

P_{2} = (- 4, 4, 3)

P_{1} P_{2} = P_{2} - P_{1} = (- 5, 5, 7)

d_{1} \times d_{2} = \hat{i} 23 \hat{j} - 3 2 \hat{k} 20 = \hat{i} (0 - 4) - \hat{j} (0 - 6) + \hat{k} (4 + 9) = (- 4, 6, 13)

∣ d_{1} \times d_{2} ∣ = 16 + 36 + 169 = 221

P_{1} P_{2} \cdot (d_{1} \times d_{2}) = (- 5) (- 4) + (5) (6) + (7) (13) = 20 + 30 + 91 = 141

.

Shortest distance

= \frac{141}{221}

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option