Question

Find the equation of a plane which is parallel to the plane $x-2y+2z=5$ and whose distance from the point $(1,2,3)$ is $1$.

Answer 1

plane $x-2y+2z=5$ (1,2,3)

distance = 1 unit

normal vector is $\overrightarrow{n}=<1,-2,2>$

Equation of th plane parallel to the

plane given passing through $(1,2,3)$ is

$\overrightarrow{n}.<x-x_0,y-y_0,z-z_0>=0$

$\overrightarrow{n}<x-1,y-2,z-3>=0$

$<1,-2,2>$ $<x-1,y-2,z-3>=0$

$x-1+(-2)(y-2)+2(z-3)=0$

$x-1-2y+4+2z-6=0$

$x-2y+2z-3=0$

Let $x=0,y=0$

$0-0+2z=5\Rightarrow z=5/2$

$\to P_1(0,0,5/2)$

Let D = 1, then

$D=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}=\frac{|1\times 0-2\times 0+2\times \frac{5}{2}+d|}{\sqrt{1+4+4}}$

$1=\frac{|5+d|}{3}\Rightarrow |5+d|=3||5+d=-3$

$d=-8$

$d=3-5=-2$

first of $e{q}^n$, $x-2y+2z-2=0$

second $e{q}^n$, $x-2y+2z-8=0$