Suppose we want to find the distance between the parallel planes
\[ax+by+cz=d_1\]
(1)\[ax+by+cz=d_1\]
(2)Suppose a normal connects points
\[(x_1,y_1,z_1)\]
in plane (1) to point \[(x_2,y_2,z_2)\]
in plane (2). Then\[ax_1+by_1+cz_1=d_1\]
(3)\[ax_2+by_2+cz_2=d_1\]
(4)(4)-(3) gives
\[a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)=d_2-d_1\]
(5)Let the equation of the normal from plane (1) to (2) be
\[\mathbf{r}(t)= \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} + \begin{pmatrix}a\\b\\c\end{pmatrix} t\]
.Then
\[\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}= \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} + \begin{pmatrix}a\\b\\c\end{pmatrix} t_0 \rightarrow \begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}= \begin{pmatrix}a\\b\\c\end{pmatrix} t_0\]
Substitution into (5) gives
\[a^2t_0+b^2t_0+c^2t_0=d_2-d_1 \rightarrow t_0=\frac{d_2-d_1}{a^2+b^2+c^2}\]
\[\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}-= \begin{pmatrix}a\\b\\c\end{pmatrix} \frac{d_2-d_1}{a^2+b^2+c^2}\]
The distance between the planes is then
\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}=\sqrt{a^2+b^2+c^2}\frac{d_2-d_1}{a^2+b^2+c^2}= \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\]