Question

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection. 

Answer 1

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

\[\frac{x - 0}{4 - 0} = \frac{y + 1}{5 + 1} = \frac{z + 1}{1 + 1} = \lambda\]

\[ \Rightarrow x = 4\lambda\]

\[ y = 6\lambda - 1 \]

\[ z = 2\lambda - 1\] 

The coordinates of a general point on AB are 

\[\left( 4\lambda, 6\lambda - 1, 2\lambda - 1 \right)\]

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

\[\frac{x - 3}{3 + 4} = \frac{y - 9}{9 - 4} = \frac{z - 4}{4 - 4} = \mu\]

\[ \Rightarrow x = 7\mu + 3\]

\[ y = 5\mu + 9 \]

\[ z = 4\]

The coordinates of a general point on CD are 

\[\left( 7\mu + 3, 5\mu + 9, 4 \right)\] 

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

\[\lambda \text{ and  } \mu\]

we must have

\[4\lambda = 7\mu + 3, 6\lambda - 1 = 5\mu + 9, 2\lambda - 1 = 4\]

\[ \Rightarrow 4\lambda - 7\mu = 3 . . . (1)\]

\[ 6\lambda - 5\mu = 10 . . . (2) \]

\[ \lambda = \frac{5}{2} . . . (3)\]

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

\[\lambda = \frac{5}{2} \]

\[\mu = 1\]

\[\text { Substituting } \lambda = \frac{5}{2} \text{ and }  \mu = 1 \text{ in (1), we get } \]

\[LHS = 4\lambda - 7\mu\]

\[ = 4\left( \frac{5}{2} \right) - 7\left( 1 \right)\]

\[ = 3\]

\[ = RHS\]

\[\text{ Since }  \lambda = \frac{5}{2} \text{ and } \mu = 1 \text{ satisfy (3), the given lines intersect .}  \]

\[\text{ Substituting the value of  } \lambda \text{ in the coordinates of a general point on the line AB, we get } \]

\[x = 10\]

\[y = 14 \]

\[z = 4\]

\[\text{ Hence, AB and CD intersect at point }  \left( 10, 14, 4 \right) .\]