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

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