Question

Equation of planes parallel to the plane $x - 2y + 2z + 4 = 0$ which are at a distance of one unit from the point $(1, 2, 3)$ are

• A. $x + 2y + 2z = 6$, $x + 2y + 2z = 0$
• B. $x - 2y + 2z = 0$, $x - 2y + 2z - 6 = 0$
• C. $x - 2y - 6 = 0$, $x - 2y + z = 6$
• D. $x + 2y + 2z = -6$, $x + 2y + 2z = 5$

Hint:

Parallel planes must have identical variable coefficients ($x - 2y + 2z$). Looking at the choices, options (A) and (D) change the sign of $y$ to $+2y$, and option (C) alters the $z$ term completely. This leaves option (B) as the only option that keeps the correct parallel direction coefficients!

Answer 1

The Correct Option is B

Step 1: Write a general parallel plane.
Any plane parallel to $x - 2y + 2z + 4 = 0$ keeps the same normal coefficients, so it has the form $x - 2y + 2z + \lambda = 0$. We just need the right $\lambda$.
Step 2: Recall the point-to-plane distance.
The distance from $(x_1, y_1, z_1)$ to $Ax + By + Cz + D = 0$ is $\dfrac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Step 3: Compute the denominator.
$\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.
Step 4: Substitute the point $(1, 2, 3)$ and set distance $= 1$.
$\dfrac{|1 - 2(2) + 2(3) + \lambda|}{3} = 1$, i.e. $\dfrac{|1 - 4 + 6 + \lambda|}{3} = 1$, so $|3 + \lambda| = 3$.
Step 5: Solve the absolute value.
Either $3 + \lambda = 3$ giving $\lambda = 0$, or $3 + \lambda = -3$ giving $\lambda = -6$.
Step 6: Write the two planes.
$x - 2y + 2z = 0$ and $x - 2y + 2z - 6 = 0$.
\[ \boxed{x - 2y + 2z = 0 \;\text{and}\; x - 2y + 2z - 6 = 0} \]