Question

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular.

Answer 1

\[\text{ Let M be the foot of the perpendicular of the point P(2, 3, 7) in the plane 3x - y - z = 7 } . \]
\[\text{ Then,PM is normal to the plane. So, the direction ratios of PM are proportional to 3, -1, -1.} \]
\[\text{ Since PM passes through P (2, 3, 7) and has direction ratios proportional to 3, -1 and -1 , equation of PQ is } \]
\[\frac{x - 2}{3} = \frac{y - 3}{- 1} = \frac{z - 7}{- 1} = r (\text{ say} )\]
\[\text{ Let the coordinates of M be} \left( 3r + 2, - r + 3, - r + 7 \right).\]
\[\text{ Since M lies in the plane } 3x - y - z = 7, \]
\[3 \left( 3r + 2 \right) - \left( - r + 3 \right) - \left( - r + 7 \right) = 7\]
\[ \Rightarrow 9r + 6 + r - 3 + r - 7 = 7\]
\[ \Rightarrow 11r = 11\]
\[ \Rightarrow r = 1\]
\[\text{ Substituting this in the coordinates of M, we get } \]
\[M = \left( 3r + 2, - r + 3, - r + 7 \right) = \left( 3 \left( 1 \right) + 2, - 1 + 3, - 1 + 7 \right) = \left( 5, 2, 6 \right)\]
\[\text{ Now, the length of the perpendicular from P onto the given plane } \]
\[ = \frac{\left| 3 \left( 2 \right) - 3 - 7 - 7 \right|}{\sqrt{9 + 1 + 1}}\]
\[ = \frac{11}{\sqrt{11}}\]
\[ = \sqrt{11} \text{ units } \]