Question

Find the shortest distance between the lines 

\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} .\]

 

Answer 1

\[\text{ The given equations of the lines are } \]
\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} . . . \left( 1 \right)\]
\[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} . . . \left( 2 \right)\]
\[\text{ Clearly (2) passes through the point P(3, 5, 7) } .\]
\[\text{ Let the direction ratios of the plane be proportional to a, b, c . }  \]
\[\text{ Since the plane contains line (1), it should pass through (-1, -1, -1) and is parallel to the line (1). } \]
\[\text{ Equation of the plane through (1) is } \]
\[a \left( x + 1 \right) + b \left( y + 1 \right) + c \left( z + 1 \right) = 0 . . . \left( 3 \right), \]
\[\text{ where } 7a - 6b + c = 0 . . . \left( 4 \right)\]
\[\text{ Since the plane is parallel to the line (2) } ,\]
\[a - 2b + c = 0 . . . \left( 5 \right)\]
\[\text{ Solving (4) and (5) using cross-multiplication, we get } \]
\[\frac{a}{- 4} = \frac{b}{- 6} = \frac{c}{- 8}\]
\[ \Rightarrow \frac{a}{2} = \frac{b}{3} = \frac{c}{4}\]
\[\text{ Substitutinga, b and c in (3), we get } \]
\[2 \left( x + 1 \right) + 3 \left( y + 1 \right) + 4 \left( z + 1 \right) = 0\]
\[ \Rightarrow 2x + 3y + 4z + 9 = 0 . . . \left( 6 \right)\]
\[\text{ which is the equation of the plane containing line (1) and parallel to line (2). } \]
\[\text{ Shortest distance between (1) and (2)} \]
\[ = \text{ Distance between the point P (3, 5, 7) and plane (6)} \]
\[ = \left| \frac{2 \left( 3 \right) + 3 \left( 5 \right) + 4 \left( 7 \right) + 9}{\sqrt{4 + 9 + 16}} \right|\]
\[ = \frac{58}{\sqrt{29}}\]
\[ = 2 \sqrt{29} \text{ units } \]