Question

An equation of a plane parallel to the plane $x-2y+2z-5=0$ and at a unit distance from the origin is

• A. $x–2y+2z\pm 3=0$
• B. $x–2y+2z+1=0$
• C. $x–2y+2z–1=0$
• D. $x–2y+2z+5=0$

Answer 1

The correct option is A

$x–2y+2z\pm 3=0$

Explanation for correct option:

Step1. Define the equation of the family of planes parallel to the given plane:

General equation of a plane: $a{x}_1+b{y}_1+c{z}_1+d$

Given equation of a Plane $\mathrm{P}$: $x-2y+2z-5=0$

Equation of family of a plane parallel to the given plane $\mathrm{P}$ is $Q:x-2y+2z+d=0$

Step2. Calculate the value of $\mathrm{d}$

The perpendicular distance of the plane $Q$ from the origin is $1\mathrm{unit}$

$\therefore \left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|=1$

Substitute the given values we have

$\Rightarrow \left|\frac{0-2\left(0\right)+2\left(0\right)+d}{\sqrt{1^2+2^2+2^2}}\right|=1$

$$\begin{gathered} \Rightarrow \left|\frac{d}{\sqrt{9}}\right|=1 \\ \Rightarrow \left|\frac{d}{3}\right|=1 \\ \Rightarrow d=\pm 3 \end{gathered}$$

Step3: Find the required equation:

Substitute the value of $\mathrm{d}$ in the required plane we get,

Required Plane Q: $\mathrm{x}-2\mathrm{y}+2\mathrm{z}\pm 3=0$

Hence, Option (A) is the correct answer.