Question

Determine whether the following pair of lines intersect or not: 

\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]

Answer 1

\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} \text {  and  } \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]

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

\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} = \lambda\]

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

\[ y = - \lambda + 1 \]

\[ z = - 1\]

The coordinates of a general point on the first line are

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

Also, the coordinates of any point on the second line are given by

\[\frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3} = \mu\]

\[ \Rightarrow x = 2\mu + 4\]

\[ y = 0\]

\[ z = 3\mu - 1\]

The coordinates of a general point on the second line are

\[\left( 2\mu + 4, 0, 3\mu - 1 \right)\]

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

\[3\lambda + 1 = 2\mu + 4, - \lambda + 1 = 0, - 1 = 3\mu - 1\]

\[ \Rightarrow 3\lambda - 2\mu = 3 . . . (1)\]

\[ \lambda = 1 . . . (2)\]

\[ \mu = 0 . . . (3)\]

\[\text{ From (2) and (3), we get } \]

\[\lambda = 1\]

\[\mu = 0\]

\[\text{ Substituting }  \lambda = 1 \text{ and }\mu= 0 \text{ in }(1),\text{ we get } \]

\[LHS = 3\lambda - 2\mu\]

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

\[ = 3\]

\[ = RHS\]

\[\text{ Since } \lambda = 1 \text{ and } \mu = 0 \text { satisfy (1), the lines intersect }. \]

\[\text{ Substituting } \lambda = 1 \text{ and }\mu = 0\text { in the coordinates of a general point on the first line, we get} \]

\[x = 4\]

\[y = 0 \]

\[z = - 1\]

\[\text{ Hence, the given lines intersect at } \left( 4, 0, - 1 \right) .\]