Let and be two lines. Then which of the following points lies on the line of the shortest distance between and ? [2025]

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Solution

Q.
Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_{1}$ and $L_{2}$ [2025]

1 $(- \frac{5}{3}, - 7, 1)$

2 $(2, 3, \frac{1}{3})$

3 $(\frac{14}{3}, - 3, \frac{22}{3})$

4 $(\frac{8}{3}, - 1, \frac{1}{3})$

Ans.
(3)

We have, $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$

Let $A (2 λ + 1, 3 λ + 2, 4 λ + 3)$ and $B (3 μ + 2, 4 μ + 4, 5 μ + 5)$ lies on the line $L_{1}$ and $L_{2}$ respectively.

Direction ratios of AB is

$< 3 μ - 2 λ + 1, 4 μ - 3 λ + 2, 5 μ - 4 λ + 2 >$

As $A B ⊥ L_{1}$ , we have

$2 (3 μ - 2 λ + 1) + 3 (4 μ - 3 λ + 2) + 4 (5 μ - 4 λ + 2) = 0$

$\Rightarrow 6 μ - 4 λ + 2 + 12 μ - 9 λ + 6 + 20 μ - 16 λ + 8 = 0$

$\Rightarrow 38 μ - 29 λ + 16 = 0$ ... (i)

Also, $A B ⊥ L_{2}$ , we have

$3 (3 μ - 2 λ + 1) + 4 (4 μ - 3 λ + 2) + 5 (5 μ - 4 λ + 2) = 0$

$\Rightarrow 50 μ - 38 λ + 21 = 0$ ... (ii)

Solving (i) and (ii), we get

$λ = \frac{1}{3} and μ = \frac{- 1}{6}$

$∴$ Coordinate of A is $(\frac{5}{3}, 3, \frac{13}{3})$ and $B (\frac{3}{2}, \frac{10}{3}, \frac{25}{6})$

Equation of AB is given by

$\frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{- \frac{1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}} \Rightarrow x - \frac{5}{3} = \frac{y - 3}{- 2} = z - \frac{13}{3}$

$∴$ Point $(\frac{14}{3}, - 3, \frac{22}{3})$ lies on line AB.

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