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Question
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Solution
\[\text { Let M be the foot of the perpendicular of the origin P(0, 0, 0) in the plane 2x - 3y + 4z - 6 = 0 .} \]
\[\text{ Then,PM is normal to the plane. So, the direction ratios of PM are proportional to 2, -3, 4. } \]
\[\text{ Since PM passes through P (0, 0, 0) and has direction ratios proportional to 2, -3 , 4 , the equation of PQ is } \]
\[\frac{x - 0}{2} = \frac{y - 0}{- 3} = \frac{z - 0}{4} = r (\text{ say } )\]
\[\text{ Let the coordiantes of M be } \left( 2r, - 3r, 4r \right).\]
\[\text{ Since M lies in the plane 2x - 3y + 4z - 6 = 0, } \]
\[2 \left( 2r \right) - 3 \left( - 3r \right) + 4 \left( 4r \right) - 6 = 0\]
\[ \Rightarrow 4r + 9r + 16r - 6 = 0\]
\[ \Rightarrow 29r - 6 = 0\]
\[ \Rightarrow r = \frac{6}{29}\]
\[\text{ Substituting the value of r in the coordinates of M, we get } \]
\[M = \left( 2r, - 3r, 4r \right) = \left( 2 \left( \frac{6}{29} \right), - 3 \left( \frac{6}{29} \right), 4 \left( \frac{6}{29} \right) \right) = \left( \frac{12}{29}, \frac{- 18}{29}, \frac{24}{29} \right)\]
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