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Given line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ Which of the following statements is/are correct?

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Shortest distance between $L_{1}$ and $L_{2}$ is $\frac{1}{\sqrt{6}}$

The shortest distance between $L_{1}$ and $L_{2}$ is $\frac{1}{\sqrt{3}}$

The equation of the line of the shortest distance is given by $\frac{x - (\frac{5}{3})}{1} = \frac{y - 3}{- 2} = \frac{z - (\frac{13}{3})}{1}$

The equation of the line of the shortest distance is given by $\frac{x + (\frac{5}{3})}{1} = \frac{y - 3}{- 2} = \frac{z - (\frac{13}{3})}{1}$

answer is A, C.

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Detailed Solution

(i) The two given lines are

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = r_{1} (say)$

$and \frac{x - 2}{2} = \frac{y - 4}{4} = \frac{z - 5}{5} = r_{2} (say)$

Any point on (i) is given by $P (2 r_{1} + 1, 3 r_{1} + 2, 4 r_{1} + 3)$ and

any point on (ii) is given by $Q (3 r_{2} + 2, 4 r_{2} + 4, 5 r_{2} + 5)$

direction rations of PQ are given by

$3 r_{2} - 2 r_{1} + 1, 4 r_{2} - 3 r_{1} + 2 and 5 r_{2} - 4 r_{1} + 2$

Since PQ is perpendicular to (i) we get

$2 (3 r_{2} - 2 r_{1} + 1) + 3 (4 r_{2} - 3 r_{1} + 2) + 4 (5 r_{2} - 4 r_{1} + 2) = 0$

$or 38 r_{2} - 29 r_{1} + 16 = 0$

Also PQ is perpendicular to (ii), we get

$3 (3 r_{2} - 2 r_{1} + 1) + 4 (4 r_{2} - 3 r_{1} + 2) + 5 (5 r_{2} - 4 r_{1} + 2) = 0 or 50 r_{2} - 38 r_{1} + 16 = 0$

Solving (iii) and (iv) we obtain $r_{2} = (\frac{1}{6}), r_{1} = (\frac{1}{3})$

Therefore, coordinates of P and Q are $(\frac{5}{3}, 3, \frac{13}{3})$ and $(\frac{3}{3}, \frac{10}{3}, \frac{25}{6})$ respectively Thus

$P Q^{2} = {(\frac{3}{2} - \frac{5}{3})}^{2} + {(\frac{10}{3} - 3)}^{2} + {(\frac{25}{6} - \frac{13}{3})}^{2}$

$= {(- \frac{1}{6})}^{2} + {(\frac{1}{3})}^{2} + {(- \frac{1}{6})}^{2} = \frac{1}{6} or$

The equation of the line of the shortest distance is given by

$\frac{x - (\frac{5}{3})}{(\frac{3}{2}) - (\frac{5}{3})} = \frac{y - 3}{(\frac{10}{3}) - 3} = \frac{z - (\frac{13}{3})}{(\frac{25}{6}) - (\frac{13}{3})}$

$\frac{x - (\frac{5}{3})}{- (\frac{1}{6})} = \frac{y - 3}{(\frac{1}{3})} = \frac{z - (\frac{13}{3})}{- (\frac{1}{6})}$

$\frac{x - (\frac{5}{3})}{1} = \frac{y - 3}{- 2} = \frac{z - (\frac{13}{3})}{1}$

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