Question

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 \( L_2 \)?

• A. \( \left( \frac{-5}{3}, -7, 1 \right) \)
• B. \( (2, 3, \frac{1}{3}) \)
• C. \( \left( \frac{8}{3}, -1, \frac{1}{3} \right) \)
• D. \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \)

Hint:

The shortest distance between two skew lines is given by the line joining the points of intersection on each line.

Answer 1

The Correct Option is D

Let the line PQ be the line of shortest distance between the lines \( L_1 \) and \( L_2 \). 

The points on these lines are: Point \( P \) lies on \( L_1 \) and is of the form \( P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \) 

Point \( Q \) lies on \( L_2 \) and is of the form \( Q(3\mu + 2, 4\mu + 4, 5\mu + 5) \). 
 

Dr's of PQ are \( 3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2 \).
 

$(3\mu - 2\lambda + 1)2 + (4\mu - 3\lambda + 2)3 + (5\mu - 4\lambda + 2)4 = 0$ 
 

$38\mu - 29\lambda + 16 = 0 \quad \text{...(1)}$ $PQ \perp L_2$ 
 

$(3\mu - 2\lambda + 1)3 + (4\mu - 3\lambda + 2)4 + (5\mu - 4\lambda + 2)5 = 0$ 
 

$50\mu - 38\lambda + 21 = 0 \quad \text{...(2)}$ By (1) \& (2) $\lambda = \frac{1}{3}, \quad \mu = \frac{-1}{6}$ 
 

$\therefore P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad \& \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$ 
 

Line PQ $\frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{\frac{-1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}}$ $\frac{x - \frac{5}{3}}{1} = \frac{y - 3}{-2} =\frac{z - \frac{13}{3}}{1}$ Point $\left(\frac{14}{3}, -3, \frac{22}{3}\right)$ lies on the line PQ.