Prove that the distance between parallel planes $\vec{n}\cdot \vec{x} = d_1 $, $\vec{n}\cdot \vec{x}=d_2$ is $|d1-d2|/||\vec{n}||$ Prove that the distance between parallel planes with equations $\vec{n}\cdot \vec{x}  = d_1 $ and $\vec{n}\cdot \vec{x}=d_2$ is given by $\displaystyle\frac{\left|d_1-d_2\right|}{\left\|\vec{n}\right\|}$

Not sure how to go about this. Thought after taking a discrete class I would have better intuition towards proving such things. 

I know the planes are both normal to $\vec{n}$ -- clearly the minimum distance is going to be on a line in the n direction.
Let $x_k$, $k=1,2$ be a point in each plane. Write $x_k = \alpha_k n + p_k$, where $p_k \bot n$.
Then
\begin{eqnarray}
\|x_1-x_2\|^2 &amp;=&amp; \| (\alpha_1-\alpha_2)n +(p_1-p_2)\|^2 \\
&amp;=&amp; (\alpha_1-\alpha_2)^2 \|n\|^2+\|p_1-p_2\|^2
\end{eqnarray}
Since each $x_k$ lies on each plane, we have $\langle x_k, n\rangle = d_k$, from which we get $\alpha_k = {d_k \over \|n\|^2}$, and so
$\|x_1-x_2\| = \sqrt{ ({d_1-d_2 \over \|n\|})^2+ \|p_1-p_2\|^2 }$.

It is easy to see that the distance is minimized if we choose $p_1=p_2$, in which case we get the minimum distance to be ${ |d_1-d_2| \over \|n\|}$
Same answer as "user183761" but with (perhaps) more explanation.

Given a point P(x1, y1, z1) on one plane and point Q(x2, y2, z2) on the other plane, the distance between the planes is the projection of vector PQ onto the normal vector n = (a, b, c) of the plane(s).  The vector PQ is given by (x2-x1, y2-y1, z2-z1).  By definition, for point P to lie on the plane it must satisfy the equation ax1 + by1 + cz1 = d1.  Likewise, for point Q, ax2 + by2 + cz2 = d2.  

Thus, d2 - d1 = a(x2-x1) + b(y2-y1) + c(z2-z1) = (a, b, c)· (x2-x1, y2-y1, z2-z1) = n·PQ

So |d2 – d1|/||n|| = |n·PQ|/||n|| = |projnPQ|