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प्रश्न
Show that the lines \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
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उत्तर
The coordinates of any point on the first line are given by
\[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} = \lambda\]
\[ \Rightarrow x = \lambda\]
\[ y = 2\lambda + 2 \]
\[ z = 3\lambda - 3\]
The coordinates of a general point on the first line are
\[\left( \lambda, 2\lambda + 2, 3\lambda - 3 \right)\]
Also, the coordinates of any point on the second line are given by
\[\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} = \mu\]
\[ \Rightarrow x = 2\mu + 2\]
\[ y = 3\mu + 6 \]
\[ z = 4\mu + 3\]
The coordinates of a general point on the second line are
\[\left( 2\mu + 2, 3\mu + 6, 4\mu + 3 \right)\]
If the lines intersect, then they have a common point. So, for some values of
\[\lambda \text{ and } \mu\] we must have
\[\lambda = 2\mu + 2, 2\lambda + 2 = 3\mu + 6, 3\lambda - 3 = 4\mu + 3\]
\[ \Rightarrow \lambda - 2\mu = 2 . . . (1)\]
\[ 2\lambda - 3\mu = 4 . . . (2)\]
\[ 3\lambda - 4\mu = 6 . . . (3)\]
\[\text{ Solving (1) and (2), we get } \]
\[\lambda = 2 \text{ and } \mu = 0\]
\[\text{ Substituting } \lambda = 2 \text{ and } \mu = 0 \text{ in (3), we get }\]
\[LHS = 3\lambda - 4\mu\]
\[ = 3\left( 2 \right) - 4\left( 0 \right)\]
\[ = 6\]
\[ = RHS\]
\[\text{ Since } \lambda = 2 \text{ and } \mu = 0 \text{ satisfy the third equation, the given lines intersect at } \left( 2, 6, 3 \right) .\]
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