Question

The distance between the planes $2x-2y+z+3=0$ and $4x-4y+2z+5=0$ is

• A. $3$
• B. $6$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{6}$

Answer 2

Since, the planes $2x-2y+z+3=0$ and $2x-2y+z+\frac{5}{2}=0$ are parallel to each other.
$\therefore$ Distance between them $=\frac{|{{c}_{2}}-{{c}_{1}}|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$
$=\frac{\left| \frac{5}{2}-3 \right|}{\sqrt{4+4+1}}$
$=\frac{\frac{1}{2}}{3}=\frac{1}{6}$