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Find the Points on the Line X + 2 3 = Y + 1 2 = Z − 3 2 at a Distance of 5 Units from the Point P (1, 3, 3).

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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).

Sum
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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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Chapter 27: Straight Line in Space - Exercise 28.1 [Page 10]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.1 | Q 14 | Page 10

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