If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{\ell}=\frac{y-2}{3}=\frac{z-b}{4}, \ell \neq 0$ is (3, 5, 7) then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Three Dimensional Geometry

If the foot of the perpendicular from point

(4, 3, 8)

on the line

L_{1} : \frac{x - a}{ℓ} = \frac{y - 2}{3} = \frac{z - b}{4}, ℓ \neq = 0

is (3, 5, 7) then the shortest distance between the line

L_{1}

and line

L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

is equal to :

Answer: (B)

The foot

(3, 5, 7)

lies on

L_{1}

, where

\frac{5 - 2}{3} = 1

, so

3 = a + ℓ

and

7 = b + 4

, giving

b = 3

. The segment from

(3, 5, 7)

(4, 3, 8)

has direction

(1, - 2, 1)

, which is perpendicular to

(ℓ, 3, 4)

ℓ - 6 + 4 = 0 \Rightarrow ℓ = 2, a = 1

L_{1}

passes through

(1, 2, 3)

along

(2, 3, 4)

, while

L_{2}

passes through

(2, 4, 5)

along

(3, 4, 5)

. Then

(2, 3, 4) \times (3, 4, 5) = (- 1, 2, - 1), (2, 4, 5) - (1, 2, 3) = (1, 2, 2)

and the shortest distance is

\frac{∣ ( 1 , 2 , 2 ) \cdot ( - 1 , 2 , - 1 ) ∣}{∣ ( - 1 , 2 , - 1 ) ∣} = \frac{∣ - 1 + 4 - 2∣}{6} = \frac{1}{6}

The shortest distance between the two lines is

\frac{1}{6}

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option