Question

The distance (in units) between the parallel planes $x+2y-3z=2$ and $2x+4y-6z+7=0$ is:

• A. $\frac{2}{\sqrt{14}}$
• B. $\frac{11}{\sqrt{56}}$
• C. $\frac{7}{\sqrt{56}}$
• D. $\frac{9}{\sqrt{14}}$

Answer 1

The correct option is B $\frac{11}{\sqrt{56}}$

Given plane equations are $x+2y-3z=2\Rightarrow 2x+4y-6z-4=0$ and $2x+4y-6z+7=0$
Distance between parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$ Here $d_1=-4,d_2=7$
$\therefore$ distance between given parallel planes is $\frac{11}{\sqrt{56}}$