If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Three Dimensional Geometry

If the shortest distance between the lines

\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}

and

\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}

\frac{13}{29}

, then a value of

λ

is :

Answer: (C)

Lines:

L_{1}

through

(λ, 4, 3)

direction

(2, 3, 4)

;

L_{2}

through

(2, 4, 7)

direction

(4, 6, 8) = 2 (2, 3, 4)

— the two lines are parallel (same direction

(2, 3, 4)

).
For parallel lines, shortest distance

= \frac{∣ d \times ( r _{2} - r _{1} ) ∣}{∣ d ∣}

with

r_{2} - r_{1} = (2 - λ, 0, 4)

d = (2, 3, 4)

d \times (r_{2} - r_{1}) = (12 - 0, 4 (2 - λ) - 8, 0 - 3 (2 - λ)) = (12, - 4 λ, 3 λ - 6)

∣ d \times (\cdot) ∣^{2} = 144 + 16 λ^{2} + (3 λ - 6)^{2}, ∣ d ∣ = 29

Setting distance

= \frac{13}{29}

144 + 16 λ^{2} + (3 λ - 6)^{2} = 169 ⟹ 16 λ^{2} + 9 λ^{2} - 36 λ + 36 = 25 ⟹ 25 λ^{2} - 36 λ + 11 = 0

.
Factoring:

(25 λ - 11) (λ - 1) = 0 ⟹ λ = 1

λ = 11/25

; the listed option is

λ = 1

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option