Question

(a) Find the shortest distance between the lines: \[ \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}, \quad \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}. \]

Hint:

Use the vector triple product and the cross product formula to calculate the shortest distance between skew lines.

Answer 1

For two skew lines: \[ \vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2, \] the shortest distance \( d \) is given by: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}. \] Here: \[ \vec{a}_1 = (-1, -1, -1), \quad \vec{b}_1 = (7, -6, 1), \quad \vec{a}_2 = (3, 5, 7), \quad \vec{b}_2 = (1, -2, 1). \] Compute \( \vec{b}_1 \times \vec{b}_2 \), \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \), and substitute in the formula.