Question

Find the distance between the parallel lines \( \frac{x}{2} = \frac{y}{-1} = \frac{z}{2} \) and \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-4}{2} \).

Hint:

Distance between parallel lines uses the perpendicular distance formula or vector projection.

Answer 1

Lines have direction ratios \( (2, -1, 2) \), confirming they are parallel.
First line passes through \( (0, 0, 0) \), second through \( (1, -1, 4) \).
Distance between parallel lines: Find a point on one line and compute perpendicular distance to the other.
Point on first line: \( (0, 0, 0) \). Second line: \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-4}{2} = t \).
General point: \( (2t + 1, -t - 1, 2t + 4) \).
Vector from \( (0, 0, 0) \) to second line: \( (2t + 1, -t - 1, 2t + 4) \).
Direction vector: \( (2, -1, 2) \).
Perpendicularity condition: Dot product = 0.
\[ (2t + 1, -t - 1, 2t + 4) \cdot (2, -1, 2) = 2(2t + 1) + (-1)(-t - 1) + 2(2t + 4) = 4t + 2 + t + 1 + 4t + 8 = 9t + 11 = 0. \] \[ t = -\frac{11}{9}. \] Point: \( \left( 2 \cdot \frac{-11}{9} + 1, -\left( \frac{-11}{9} \right) - 1, 2 \cdot \frac{-11}{9} + 4 \right) = \left( \frac{-13}{9}, \frac{2}{9}, \frac{14}{9} \right) \).
Distance: \( \sqrt{\left( \frac{-13}{9} \right)^2 + \left( \frac{2}{9} \right)^2 + \left( \frac{14}{9} \right)^2} = \sqrt{\frac{169 + 4 + 196}{81}} = \sqrt{\frac{369}{81}} = \sqrt{\frac{41}{9}} = \frac{\sqrt{41}}{3} \).
Answer: \( \frac{\sqrt{41}}{3} \).