Question

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.

Answer 1

\[\text{ Let M be the foot of the perpendicular of the point P(3, 2, 1) in the plane 2x - y + z + 1 = 0 }. \]
\[\text{ Then,PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, -1, 1. } \]
\[\text{ Since PM passes through P (3, 2, 1) and has direction ratios proportional to 2, -1, 1,} \]
\[\text{ equation of PQ is } \]
\[\frac{x - 3}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1} = r (\text{ say })\]
\[\text{ Let the coordinates of M be } \left( 2r + 3, - r + 2, r + 1 \right).\]
\[\text{ Since M lies in the plane } 2x - y + z + 1 = 0, \]
\[2 \left( 2r + 3 \right) - \left( - r + 2 \right) + r + 1 + 1 = 0\]
\[ \Rightarrow 4r + 6 + r - 2 + r + 2 = 0\]
\[ \Rightarrow 6r = - 6\]
\[ \Rightarrow r = - 1\]
\[\text{ Substituting the value of r in the coordinates of M, we get } \]
\[M = \left( 2r + 3, - r + 2, r + 1 \right) = \left( 2 \left( - 1 \right) + 3, - \left( - 1 \right) + 2, - 1 + 1 \right) = \left( 1, 3, 0 \right)\]
\[\text{ Now, the length of the perpendicular fromPonto the given plane } \]
\[ = \frac{\left| 2 \left( 3 \right) - \left( 2 \right) + \left( 1 \right) + 1 \right|}{\sqrt{4 + 1 + 1}}\]
\[ = \frac{6}{\sqrt{6}}\]
\[ = \sqrt{6} \text{ units }\]