Question

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

 

Answer 1

\[\text{ Let M be the foot of the perpendicular of the point P (1, 1, 2) in the plane 2x - 2y + 4z + 5 = 0 } \]
\[\text{ Then,PM is normal to the plane. So, the direction ratios of PM are proportional to 2, -2, 4.} \]
\[\text{ Since PM passes through P (1, 1, 2) and has direction ratios proportional to 2, -2 and 4 , equation of PQ is } \]
\[\frac{x - 1}{2} = \frac{y - 1}{- 2} = \frac{z - 2}{4} = r (\text{ say } )\]
\[\text{ Let the coordiantes of M be } \left( 2r + 1, - 2r + 1, 4r + 2 \right).\]
\[\text{ Since M lies in the plane } 2x - 2y + 4z + 5 = 0, \]
\[2 \left( 2r + 1 \right) - 2 \left( - 2r + 1 \right) + 4 \left( 4r + 2 \right) + 5 = 0\]
\[ \Rightarrow 4r + 2 + 4r - 2 + 16r + 8 + 5 = 0\]
\[ \Rightarrow 24r + 13 = 0\]
\[ \Rightarrow r = \frac{- 13}{24}\]
\[\text{ Substituting this in the coordinates of M, we get } \]
\[M = \left( 2r + 1, - 2r + 1, 4r + 2 \right) = \left( 2 \left( \frac{- 13}{24} \right) + 1, - 2 \left( \frac{- 13}{24} \right) + 1, 4 \left( \frac{- 13}{24} \right) + 2 \right) = \left( \frac{- 1}{12}, \frac{25}{12}, \frac{- 1}{6} \right)\]
\[\text{ Now, the length of the perpendicular from P onto the given plane } \]
\[ = \frac{\left| 2 \left( 1 \right) - 2 \left( 1 \right) + 4 \left( 2 \right) + 5 \right|}{\sqrt{4 + 4 + 16}}\]
\[ = \frac{13}{\sqrt{24}} \text{ units } \]