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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.
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Solution
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 ,
\[3\lambda - 1 = \mu + 2, 5\lambda - 3 = 3\mu + 4, 7\lambda - 5 = 5\mu + 6\]
\[ \Rightarrow 3\lambda - \mu = 3 . . . (1)\]
\[ 5\lambda - 3\mu = 7 . . . (2)\]
\[ 7\lambda - 5\mu = 11 . . . (3)\]
\[\text{ Solving (1) and (2), we get } \]
\[\lambda = \frac{1}{2} \]
\[\mu = - \frac{3}{2}\]
\[\text { Substituting } \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text { in (3), we get } \]
\[LHS = 7\lambda - 5\mu\]
\[ = 7\left( \frac{1}{2} \right) - 5\left( - \frac{3}{2} \right)\]
\[ = 11\]
\[ = RHS\]
\[\text{ Since } \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text{ satisfy (3), the given lines intersect .} \]
\[\text{ Substituting the value of } \lambda \text{ in the general coordinates of the first line, we get } \]
\[x = \frac{1}{2}\]
\[y = - \frac{1}{2}\]
\[z = - \frac{3}{2}\]
\[\text{ Hence, the given lines intersect at point } \left( \frac{1}{2}, - \frac{1}{2}, - \frac{3}{2} \right) .\]
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