Question

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

Answer 1

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

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

\[ \Rightarrow x = \lambda\]

\[ y = 2\lambda + 2 \]

\[ z = 3\lambda - 3\] 

The coordinates of a general point on the first line are 

\[\left( \lambda, 2\lambda + 2, 3\lambda - 3 \right)\] 
Also, the coordinates of any point on the second line are given by

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

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

\[ y = 3\mu + 6 \]

\[ z = 4\mu + 3\]

The coordinates of a general point on the second line are

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

If the lines intersect, then they have a common point. So, for some values of 

\[\lambda \text{ and  } \mu\] we must have

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

\[ \Rightarrow \lambda - 2\mu = 2 . . . (1)\]

\[ 2\lambda - 3\mu = 4 . . . (2)\]

\[ 3\lambda - 4\mu = 6 . . . (3)\]

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

\[\lambda = 2 \text{ and } \mu = 0\]

\[\text{ Substituting }  \lambda = 2 \text{ and  } \mu = 0 \text{ in (3), we get }\]

\[LHS = 3\lambda - 4\mu\]

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

\[ = 6\]

\[ = RHS\]

\[\text{ Since }  \lambda = 2 \text{ and }  \mu = 0 \text{ satisfy the third equation, the given lines intersect at }  \left( 2, 6, 3 \right) .\]