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 \[ d=\dfrac{\left\vert \,2(0) +3(0) -1 (-3) -4\,\right\vert }{\sqrt{2^{2}+3^{2}+(-1) ^{2}}}=\dfrac{1}{\sqrt{ 14}}=\dfrac{\sqrt{14}}{14} \]