Question

Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{           and                  } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\]   intersect. Find their point of intersection.

Answer 1

The coordinates of any point on the first line are given by 

\[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda\]

\[ \Rightarrow x = 3\lambda - 1\]

\[ y = 5\lambda - 3\]

\[ z = 7\lambda - 5\] 

The coordinates of a general point on the first line are 

\[\left( 3\lambda - 1, 5\lambda - 3, 7\lambda - 5 \right)\] 

The coordinates of any point on the second line are given by 

\[\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} = \mu\]

\[ \Rightarrow x = \mu + 2\]

\[ y = 3\mu + 4 \]

\[ z = 5\mu + 6\]

The coordinates of a general point on the second line are

\[\left( \mu + 2, 3\mu + 4, 5\mu + 6 \right)\] 

If the lines intersect, then they have a common point. So, for some values of  \[\lambda \text{ and } \mu\] we must have ,

\[3\lambda - 1 = \mu + 2, 5\lambda - 3 = 3\mu + 4, 7\lambda - 5 = 5\mu + 6\]

\[ \Rightarrow 3\lambda - \mu = 3 . . . (1)\]

\[ 5\lambda - 3\mu = 7 . . . (2)\]

\[ 7\lambda - 5\mu = 11 . . . (3)\]

\[\text{ Solving (1) and (2), we get } \]

\[\lambda = \frac{1}{2} \]

\[\mu = - \frac{3}{2}\]

\[\text { Substituting }  \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text { in (3), we get } \]

\[LHS = 7\lambda - 5\mu\]

\[ = 7\left( \frac{1}{2} \right) - 5\left( - \frac{3}{2} \right)\]

\[ = 11\]

\[ = RHS\]

\[\text{ Since } \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text{ satisfy (3), the given lines intersect .}  \]

\[\text{ Substituting the value of       } \lambda \text{ in the general coordinates of the first line, we get } \]

\[x = \frac{1}{2}\]

\[y = - \frac{1}{2}\]

\[z = - \frac{3}{2}\]

\[\text{ Hence, the given lines intersect at point }  \left( \frac{1}{2}, - \frac{1}{2}, - \frac{3}{2} \right) .\]