If the shortest distance between the lines $\mathrm{L}_{1}: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R}$ $L_{2}: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$ is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Three Dimensional Geometry

If the shortest distance between the lines

L_{1} : r = (2 + λ) \hat{i} + (1 - 3 λ) \hat{j} + (3 + 4 λ) \hat{k}, λ \in R

L_{2} : r = 2 (1 + μ) \hat{i} + 3 (1 + μ) \hat{j} + (5 + μ) \hat{k}, μ \in R

\frac{m}{n}

, where

gcd (m, n) = 1

, then the value of

m + n

equals.

(A)384
(B)387✓ Correct
(C)377
(D)390

Answer: (B)

Solution

Direction vectors

p = (1, - 3, 4)

q = (2, 3, 1)

;

p \times q = (- 15, 7, 9)

∣ p \times q ∣ = 355

. Taking points

A (2, 1, 3)

L_{1}

and

B (2, 3, 5)

L_{2}

A B = (0, 2, 2)

. Shortest distance

= \frac{∣ A B \cdot ( p \times q ) ∣}{∣ p \times q ∣} = \frac{∣0 + 14 + 18∣}{355} = \frac{32}{355}

. With

g cd (32, 355) = 1

m + n = 32 + 355 = 387

. Option (2).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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