Question

If the distance between parallel planes $2x-y+3z-4=0$ and $6x-3y+9z+13=0$ is $D$. Then the value of $126D^2=$

Answer 1

The given equations are, $2x-y+3z-4=0\Rightarrow 2x-y+3z=4$
Multiplying equation by $3$ we get,
$\Rightarrow 6x-3y+9z=12\cdots \left(i\right)$
$6x-3y+9z+13=0\Rightarrow 6x-3y+9z=-13\cdots \left(ii\right)$
Comparing the given equations with $ax+by+cz=d_1$ and $ax+by+cz=d_2$​ we get,
$a=6,b=-3,c=9,d_1=12,d_2=-13$
Distance between plane $1$ and $2$ $=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$
$D=\frac{|12+13|}{\sqrt{36+9+81}}=\frac{25}{\sqrt{126}}$
$\therefore 126D^2=625$