Find the distance $d$ between the parallel planes $$p_{1}:\ 2x+3y-z=3 \quad \hbox{and}\quad p_{2}: \ 2x+3y-z=4$$

Solution The planes $p_{1}$ and $p_{2}$ are parallel because $\mathbf{N}_{1}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}=\mathbf{N}_{2}$ and $D_{1}=3$ is not equal to $D_{2}=4$.

We find the distance between the planes by choosing a point on one plane and using the formula for distance from that point to the other plane. A point on $p_{1}$ can be found by letting $x=0$ and $y=0.$ Then $z=-3,$ and $( 0,0,-3)$ is a point on $p_{1}.$ The distance $d$ from the point $( 0,0,-3)$ to the plane $p_{2}$ is $$d=\dfrac{\left\vert \,2(0) +3(0) -1 (-3) -4\,\right\vert }{\sqrt{2^{2}+3^{2}+(-1) ^{2}}}=\dfrac{1}{\sqrt{ 14}}=\dfrac{\sqrt{14}}{14}$$