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Question
Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{ and } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\] do not intersect.
Solution
The coordinates of any point on the first line are given by
\[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} = \lambda\]
\[ \Rightarrow x = 3\lambda + 1\]
\[ y = 2\lambda - 1 \]
\[ z = 5\lambda + 1\]
The coordinates of a general point on the first line are
\[\left( 3\lambda + 1, 2\lambda - 1, 5\lambda + 1 \right)\]
The coordinates of any point on the second line are given by
\[\frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2} = \mu\]
\[ \Rightarrow x = 4\mu - 2\]
\[ y = 3\mu + 1 \]
\[ z = - 2\mu - 1\]
The coordinates of a general point on the second line are
\[\left( 4\mu - 2, 3\mu + 1, - 2\mu - 1 \right)\]
If the lines intersect, then they have a common point. So, for some values of \[\lambda \text{ and } \mu\]
we must have
\[3\lambda + 1 = 4\mu - 2, 2\lambda - 1 = 3\mu + 1, 5\lambda + 1 = - 2\mu - 1\]
\[ \Rightarrow 3\lambda - 4\mu = - 3 . . . (1)\]
\[ 2\lambda - 3\mu = 2 . . . (2)\]
\[ 5\lambda + 2\mu = - 2 . . . (3)\]
\[\text{ Solving (1) and (2), we get } \]
\[\lambda = - 17 \]
\[\mu = - 12\]
\[\text{ Substituting } \lambda = - 17 \text{ and } \mu = - 12 \text{ in (3), we get } \]
\[LHS = 3\lambda + 2\mu\]
\[ = 3\left( - 17 \right) + 2\left( - 12 \right)\]
\[ = - 75\]
\[ \neq - 2\]
\[ \Rightarrow LHS \neq RHS\]
\[\text{ Hence, the given lines do not intersect } .\]
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