Question

Find the shortest distance between the lines
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$

Answer 1

The Directional ratios of first line is $2,3,4$

The Directional ratios of second line is $3,4,5$

Let the angle between them be $\theta$

We have $cos\theta =\frac{2\times 3+3\times 4+4\times 5}{\sqrt{29}\sqrt{50}}=\frac{38}{\sqrt{1450}}$

Therefore $\theta \ne 0^0$ , so the given lines are not parallel.

Since the lines intersect , the shortest distance between them will be zero