Distance Between Two Parallel Planes

Theorem 6.21. The distance between the two PARALLEL planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ (same normal direction ratios $a,b,c$) is

Proof. Pick any point $A(x_1,y_1,z_1)$ on the SECOND plane, so $ax_1+by_1+cz_1+d_2=0$, i.e. $ax_1+by_1+cz_1=-d_2$. Now apply §6.8.12's point-to-plane distance formula, using $A$ and the FIRST plane:

$\delta=\frac{|ax_1+by_1+cz_1+d_1|}{\sqrt{a^2+b^2+c^2}}=\frac{|-d_2+d_1|}{\sqrt{a^2+b^2+c^2}}=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}.$ …