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Question
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
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Solution
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} \text { and } \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
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
\[3\lambda + 1 = 2\mu + 4, - \lambda + 1 = 0, - 1 = 3\mu - 1\]
\[ \Rightarrow 3\lambda - 2\mu = 3 . . . (1)\]
\[ \lambda = 1 . . . (2)\]
\[ \mu = 0 . . . (3)\]
\[\text{ From (2) and (3), we get } \]
\[\lambda = 1\]
\[\mu = 0\]
\[\text{ Substituting } \lambda = 1 \text{ and }\mu= 0 \text{ in }(1),\text{ we get } \]
\[LHS = 3\lambda - 2\mu\]
\[ = 3\left( 1 \right) - 2\left( 0 \right)\]
\[ = 3\]
\[ = RHS\]
\[\text{ Since } \lambda = 1 \text{ and } \mu = 0 \text { satisfy (1), the lines intersect }. \]
\[\text{ Substituting } \lambda = 1 \text{ and }\mu = 0\text { in the coordinates of a general point on the first line, we get} \]
\[x = 4\]
\[y = 0 \]
\[z = - 1\]
\[\text{ Hence, the given lines intersect at } \left( 4, 0, - 1 \right) .\]
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