Question

Find the shortest distance between the lines

\[\frac{x - 2}{- 1} = \frac{y - 5}{2} = \frac{z - 0}{3} \text{ and }  \frac{x - 0}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{2} .\]

 

Answer 1

\[\text{ The given equations of the lines are } \]
\[\frac{x - 2}{- 1} = \frac{y - 5}{2} = \frac{z - 0}{3} . . . \left( 1 \right)\]
\[\frac{x - 0}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{2} . . . \left( 2 \right)\]
\[\text{ Clearly (2) passes through the pointP(0, -1, 1).} \]
\[\text{ Let the direction ratios of the plane be proportional to a, b, c . } \]
\[\text{ Since the plane containing line (1) should pass through (2, 5, 0) and is parallel to the line (1) } ,\]
\[\text{ equation of the plane passing through (1) is } \]
\[a \left( x - 2 \right) + b \left( y - 5 \right) + c \left( z - 0 \right) = 0 . . . \left( 3 \right), \]
\[\text{ where}  -a + 2b + 3c = 0 . . . \left( 4 \right)\]
\[\text{ Since the plane is parallel to line (2), } \]
\[2a - b + 2c = 0 . . . \left( 5 \right)\]
\[\text{ Solving (4) and (5) using cross-multiplication, we get } \]
\[\frac{a}{7} = \frac{b}{8} = \frac{c}{- 3}\]
\[\text{ Substitutinga, b and c in (3), we get} \]
\[7 \left( x - 2 \right) + 8 \left( y - 5 \right) - 3 \left( z - 0 \right) = 0\]
\[ \Rightarrow 7x + 8y - 3z - 54 = 0 . . . \left( 6 \right), \]
\[\text{ which is the equation of the plane containing line (1) and parallel to line (2).} \]
\[\text{ Shortest distance between (1) and (2) } \]
\[ = \text{  Distance between the point P(0, -1, 1) and plane (6) } \]
\[ = \left| \frac{7 \left( 0 \right) + 8 \left( - 1 \right) - 3 \left( 1 \right) - 54}{\sqrt{49 + 64 + 9}} \right|\]
\[ = \frac{65}{\sqrt{122}} \text{ units } \]