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Table of Contents
Remark: the distance from the origin to a plane
If we write $0=(0,0,0)$ for the origin in $\rt$ and apply the formula above to the plane $\Pi:ax+by+cz=d$ with $B=(d/a,0,0)$ (assuming that $a\ne 0$) then we obtain \[ \dist(0,\Pi)=\frac{|d|}{\|\nn\|}\] where $\nn$ is the normal vector $\nn=\c abc$.
So as $d$ varies (with the normal vector $\nn$ fixed), we obtain parallel planes at different distances to the origin $0$; the larger $d$ is, the further the plane is from $0$.
The distance between parallel planes
If $\Pi_1$ and $\Pi_2$ are parallel planes, then the shortest distance between them is given by \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)\] for any point $A$ is $\Pi_1$. The reason is that for parallel planes, changing $A$ to a different point in $\Pi_1$ does not change $\dist(A,\Pi_2)$.
Of course, if the planes $\Pi_1$ and $\Pi_2$ are not parallel, then they intersect (in many points: in a whole line). So for non-parallel planes we always have $\dist(\Pi_1,\Pi_2)=0$.
Example
The distance between the planes $3x+4y-2z=5$ and $3x+4y-3z=1$ is $0$, since the normal vectors $\c34{-2}$ and $\c34{-3}$ are not scalar multiples of one another, so they are in different directions, so the planes are not parallel.
Example
The planes $\Pi_1:3x+4y-2z=5$ and $\Pi_2:3x+4y-2z=1$ have the same normal vector $\c34{-2}$, so they are parallel. Their distance is given by $\dist(A,\Pi_2)$ where $A$ is any point in $\Pi_1$, and to find this we also need a point $B$ in $\Pi_2$.
We can choose $A=(1,0,-1)\in \Pi_1$ and $B=(1,0,1)\in \Pi_2$. (Of course, there are lots of different possible choices here, but they should all give the same answer!) Then $\vec {AB}=\c002$ and \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2) = \frac{|\nn\cdot \vec{AB}|}{\|n\|}=\frac{|0+0+(-2)2|}{\sqrt{3^2+4^2+(-2)^2}} = \frac4{\sqrt{29}}.\]
Exercise: a formula for the distance between parallel planes
Show that the distance between the parallel planes $\Pi_1:ax+by+cz=d_1$ and $\Pi_2:ax+by+cz=d_2$ is \[\dist(\Pi_1,\Pi_2)=\frac{|d_2-d_1|}{\|\nn\|},\] where $\nn=\c abc$.
Example
To find the distance between $x+3y-5z=4$ and $2x+6y-10z=11$ we can rewrite the second equation as $x+3y-5z=11/2$ to see that this is a parallel plane to the first, with common normal vector $\nn=\c13{-5}$. By the formula in the exercise the distance between these planes is \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]