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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.
Solution
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
\[4\lambda = 7\mu + 3, 6\lambda - 1 = 5\mu + 9, 2\lambda - 1 = 4\]
\[ \Rightarrow 4\lambda - 7\mu = 3 . . . (1)\]
\[ 6\lambda - 5\mu = 10 . . . (2) \]
\[ \lambda = \frac{5}{2} . . . (3)\]
\[\text { Solving (2) and (3), we get } \]
\[\lambda = \frac{5}{2} \]
\[\mu = 1\]
\[\text { Substituting } \lambda = \frac{5}{2} \text{ and } \mu = 1 \text{ in (1), we get } \]
\[LHS = 4\lambda - 7\mu\]
\[ = 4\left( \frac{5}{2} \right) - 7\left( 1 \right)\]
\[ = 3\]
\[ = RHS\]
\[\text{ Since } \lambda = \frac{5}{2} \text{ and } \mu = 1 \text{ satisfy (3), the given lines intersect .} \]
\[\text{ Substituting the value of } \lambda \text{ in the coordinates of a general point on the line AB, we get } \]
\[x = 10\]
\[y = 14 \]
\[z = 4\]
\[\text{ Hence, AB and CD intersect at point } \left( 10, 14, 4 \right) .\]
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