Question

Find the equation of the plane passing through the intresection of the planes $x-2y+z=1$ and $2x+y+z=8$ and parallel to the line with direction ratio proportional to $1,2,1,$ find also the perpendicular distance of $(1,1,1)$ from this plane.

Answer 1

The equation of the plane passing through the intersection of the given planes is

$(x-2y+z-1)+\lambda (2x+y+z-8)=0$

$\Rightarrow (1+2\lambda )x+(-2+\lambda )y+(1+\lambda )z-1-8\lambda =0$ ......$\left(1\right)$

This plane is parallel to the line whose direction ratios are proportional to $1,2,1$

So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to $1,2,1$

$\Rightarrow (1+2\lambda )1+(-2+\lambda )2+(1+\lambda )1-1-8\lambda =0$

$\Rightarrow 5\lambda -2=0$

$\Rightarrow \lambda =\frac{2}{5}$

Substituting this in $\left(1\right)$, we get

$\Rightarrow (1+2\times \frac{2}{5})x+(-2+\frac{2}{5})y+(1+\frac{2}{5})z-1-8\frac{2}{5}=0$

$\Rightarrow 9x-8y+7z-21=0$ .......$\left(2\right)$ which is the required equation of the plane.

Perpendicular distance of plane $\left(2\right)$ from $(1,1,1)$

$=\frac{|9\times 1-8\times 1+7\times 1-21|}{\sqrt{9^2+{(-8)}^2+7^2}}$

$=\frac{|-13|}{\sqrt{194}}=\frac{13}{194}$units.