We know that for a plane on the origin, its equation can be written in the form $$r\cdot n=0$$ where $r=(x,y,z)$ and $n$ is the normal to plane. It utilizes the idea that for any position vector $(x,y,z)$ on the plane, its dot product with its orthogonal vector (normal) will be 0.
However for a plane that is not on the origin its equation is $$r\cdot n=a\cdot n$$ where $a$ is a point on the plane. Now I understand $a\cdot n$ accounts for the displacement between plane and the origin but my question is how. To better frame it, how is the displacement from the origin to the plane always equal to $|a||n|\cos\theta$. I have some vague understanding for its pure mathematical reasoning but can somebody help me by explaining the graphical reason? (why is $|a||n|\cos\theta$ is the displacement?)
I'd say $\lvert a\rvert\cos\theta$ is the length of the projection of the vector $a$ onto the direction of $n$. Which is the component of $a$ that is perpendicular to the plane. This is the distance between origin and plane. The whole expression is equal to that distance only if $\lvert n\rvert=1$, otherwise you get a scaled version of the distance.