Question

Find the shortest distance between the lines

\[\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}\] and

\[3x - y - 2z + 4 = 0 = 2x + y + z + 1\]

 

Answer 1

The equation of the plane containing the line

\[3x - y - 2z + 4 = 0 = 2x + y + z + 1\] is 

\[\left( 3x - y - 2z + 4 \right) + \lambda\left( 2x + y + z + 1 \right) = 0\]

\[\text{ Or } \left( 3 + 2\lambda \right)x + \left( \lambda - 1 \right)y + \left( \lambda - 2 \right)z + \left( \lambda + 4 \right) = 0 . . . . . \left( 1 \right)\]

If it is parallel to the line

\[\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}\] , then 

\[2\left( 3 + 2\lambda \right) + 4\left( \lambda - 1 \right) + \left( \lambda - 2 \right) = 0\]

\[ \Rightarrow 9\lambda = 0\]

\[ \Rightarrow \lambda = 0\]

Putting

\[\lambda = 0\]   in (1), we get

\[3x - y - 2z + 4 = 0 . . . . . \left( 2 \right)\]

This is the equation of the plane containing the second line and parallel to the first line.
Now, the line

\[\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}\]

passes through (1, 3, −2).

∴ Shortest distance between the given lines

= Length of the perpendicular from (1, 3, −2) to the plane

\[3x - y - 2z + 4 = 0\]

\[= \left| \frac{3 \times 1 - 3 - 2 \times \left( - 2 \right) + 4}{\sqrt{3^2 + \left( - 1 \right)^2 + \left( - 2 \right)^2}} \right|\]
\[ = \left| \frac{3 - 3 + 4 + 4}{\sqrt{9 + 1 + 4}} \right|\]
\[ = \frac{8}{\sqrt{14}} \text{ units } \]