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Question
Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]
Hence, or otherwise, deduce the length of the perpendicular.
Solution
\[\text{ Let M be the foot of the perpendicular of the point P(5, 4, 2) on the line } \frac{x + 1}{2}=\frac{y - 3}{3}=\frac{z - 1}{- 1}\]
\[\text{ Therefore, its equation is } \]
\[\frac{x + 1}{2}=\frac{y - 3}{3}=\frac{z - 1}{- 1}= r\]
\[\text{ Then,M is in the form } \left( 2r - 1, 3r + 3, - r + 1 \right)\]
\[\text{ Direction ratios of MP are } 2r - 1 - 5, 3r + 3 - 4, - r + 1 - 2 \text{ or } 2r - 6, 3r - 1, - r - 1 . \]
\[\text{ Since MP is perpendicular to the given line } (2, 3, -1),\]
\[2 \left( 2r - 6 \right) + 3 \left( 3r - 1 \right) - 1 \left( - r - 1 \right) = 0 (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 4r - 12 + 9r - 3 + r + 1 = 0\]
\[ \Rightarrow 14r - 14 = 0\]
\[ \Rightarrow r = 1\]
\[\text{ So } ,M = \left( 2r - 1, 3r + 3, - r + 1 \right) = \left( 2 \left( 1 \right) - 1, 3 \left( 1 \right) + 3, - 1 + 1 \right) = \left( 1, 6, 0 \right)\]
\[\text{ Length of the perperndicular,MP } = \sqrt{\left( 1 - 5 \right)^2 + \left( 6 - 4 \right)^2 + \left( 0 - 2 \right)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2 \sqrt{6} \text{ units } \]
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