Distance between two planes in space
Calculate the distance between the two planes:
$$\pi:x+3y-\sqrt{6}z-4=0$$
$$\pi':x+3y-\sqrt{6}z+1=0$$
To calculate the distance between the planes $\pi$ and $\pi'$ we can apply: $$\text{d}(\pi,\pi') = \dfrac{|D-D|}{\sqrt{A^2+B^2+C^2}}= \dfrac{|-4-1|}{\sqrt{1+9+6}}=\dfrac{5}{4}$$
$$\text{d}(\pi,\pi') = \dfrac{5}{4}$$