Question
Write the distance between the parallel planes $2x − y + 3z = 4$ and $2x − y + 3z = 18$.

Answer

The given equation are
$2x − y + 3z = 4 ....(1)$
The second equation of the plane is
$2x − y + 3z = 18 .....(2)$
We know that distance between two planes $ax + by + cz = d_1$  and $ax + by + cz = d_2$ is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$

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