Question

The distance between the planes $x+2y+3z+7=0$ and $2x+4y+6z+7=0$ is

• A. $\frac{\sqrt{7}}{2\sqrt{2}}$
• B. $\frac{7}{2}$
• C. $\frac{\sqrt{7}}{2}$
• D. $\frac{7}{2\sqrt{2}}$

Answer 1

The correct option is A $\frac{\sqrt{7}}{2\sqrt{2}}$

The equation of second plane can be rearrange as $x+2y+3z+\frac{7}{2}=0$.

The distance between parallel planes $Ax+By+cZ+D_1=0$ and $Ax+By+Cz+D_2$ is given by

$\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$.

Hence, for the given problem distance between planes is given by:

$\frac{|7-\frac{7}{2}|}{\sqrt{1+4+9}}=\frac{\sqrt{7}}{2\sqrt{2}}$.