Find the shortest distance between the lines (x + 1)/7 = (y + 1)/-6 = (z + 1)/1 and (x - 3)/1 = (y - 5)/-2 = (z - 7)/1

Question

Find the shortest distance between the lines \(\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}\) and \(\frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1}\)

(x + 1)/7 = (y + 1)/-6 = (z + 1)/1 and (x - 3)/1 = (y - 5)/-2 = (z - 7)/1

Answer 1

The shortest distance between the lines \(\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1}\) and \(\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2}\) is given by

The equations of the given lines are

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6) 

= -16 – 36 – 64 = -116

Hence, the required shortest distance between the given lines \(=\left|\frac{-116}{\sqrt{116}}\right| = \sqrt{116}\) = 2√29 units.