Question

Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane

Answer 1

\[\text{ The equation of the plane passing through the intersection of the given planes is } \]
\[\left( x - 2y + z - 1 \right) + \lambda \left( 2x + y + z - 8 \right) = 0\]
\[ \Rightarrow \left( 1 + 2\lambda \right) x + \left( - 2 + \lambda \right) y + \left( 1 + \lambda \right) z - 1 - 8\lambda = 0 . . . \left( 1 \right)\]
\[\text{ This plane is parallel to the line whose direction ratios are proportional to 1,2,1 } .\]
\[\text{ So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to 1, 2, 1 }  . \]
\[ \Rightarrow \left( 1 + 2\lambda \right) 1 + \left( - 2 + \lambda \right) 2 + \left( 1 + \lambda \right) 1 = 0\]
\[ \Rightarrow 1 + 2\lambda - 4 + 2\lambda + 1 + \lambda = 0\]
\[ \Rightarrow 5\lambda - 2 = 0\]
\[ \Rightarrow \lambda = \left( \frac{2}{5} \right)\]
\[\text{ Substituting this in (1), we get} \]
\[\left( 1 + 2 \left( \frac{2}{5} \right) \right) x + \left( - 2 + \left( \frac{2}{5} \right) \right) y + \left( 1 + \left( \frac{2}{5} \right) \right) z - 1 - 8 \left( \frac{2}{5} \right) = 0\]
\[ \Rightarrow 9x - 8y + 7z - 21 = 0 . . . \left( 2 \right), \text{ which is the required equation of the plane.} \]
\[\text{ Perpendicular distance of plane (2) from (1, 1, 1) } \]
\[ = \frac{\left| 9 \left( 1 \right) - 8 \left( 1 \right) + 7 \left( 1 \right) - 21 \right|}{\sqrt{9^2 + \left( - 8 \right)^2 + 7^2}}\]
\[ = \frac{\left| -13 \right|}{\sqrt{194}}\]
\[ = \frac{13}{\sqrt{194}} \text{ units } \]