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Question
Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] at a distance of 5 units from the point P (1, 3, 3).
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Solution
The coordinates of any point on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] are given by
\[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = \lambda\]
\[ \Rightarrow x = 3\lambda - 2, y = 2\lambda - 1, z = 2\lambda + 3 . . . (1)\]
Let the coordinates of the desired point be \[\left( 3\lambda - 2, 2\lambda - 1, 2\lambda + 3 \right)\]
The distance between this point and (1, 3, 3) is 5 units.
\[\therefore \sqrt{\left( 3\lambda - 2 - 1 \right)^2 + \left( 2\lambda - 1 - 3 \right)^2 + \left( 2\lambda + 3 - 3 \right)^2} = 5\]
\[ \Rightarrow \left( 3\lambda - 3 \right)^2 + \left( 2\lambda - 4 \right)^2 + \left( 2\lambda \right)^2 = 25\]
\[ \Rightarrow 17 \lambda^2 - 34\lambda = 0\]
\[ \Rightarrow \lambda\left( \lambda - 2 \right) = 0\]
\[ \Rightarrow \lambda = 0 \text{ or } 2\]
Substituting the values of \[\lambda\] in (1) we get the coordinates of the desired point as (-2,-1,3) and (4, 3 , 7) .
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