Distance between parallel planes ($|(a_1-a_2)\cdot n|$)

Planes with common unit normal $n$ have distance $|(a_1-a_2)\cdot n|$. Every joining displacement has this normal component, giving a lower bound by its absolute value; a displacement parallel to the normal attains the bound.

For a common nonzero normal vector $\mathbf{n}$, not necessarily of unit Euclidean norm, the planes $\mathbf{n}\cdot \mathbf{x}=d_1$ and $\mathbf{n}\cdot \mathbf{x}=d_2$ have distance $|d_2-d_1|/\Vert \mathbf{n}\Vert$. Thus $2x-y+2z=3$ and $2x-y+2z=9$ are distance $6/3=2$ apart. First put both plane equations into the same normal-vector scale: subtracting offsets from independently rescaled equations need not give a geometric distance.