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. 
 A: 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 &=& \| (\alpha_1-\alpha_2)n +(p_1-p_2)\|^2 \\
&=& (\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 ${ |d_1-d_2| \over \|n\|}$
A: This is a trick that will always work for finding distances between two "flat" objects.


*

*First, identify the direction in which the "minimum distance" lies.

*Second, take any two random points, one from object 1, the other from object 2.

*Finally, find the length of the projection of your random vector to your direction vector.


I'll put my solution in a spoiler tag so that you can work it out on your own first.
Let $\vec{x}_1$ be a position vector in line $1$, and $\vec{x}_2$ a position vector in line $2$. 
Then, the length of the projection is:

 $|Proj_{\vec{n}}(\vec{x}_2-\vec{x}_1)| = \dfrac{|(\vec{x}_2 - \vec{x}_1)\cdot\vec{n}|}{||\vec{n}||}$

Now use the fact that

 $\vec{x}_1\cdot{\vec{n}} = d_1$ and $\vec{x}_2\cdot{\vec{n}} = d_2$

A: 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 $\dfrac{\left|d_1-d_2\right|}{\left\|\vec{n}\right\|}$.
p1 = r . n = D1
p2 = r . n = D2
Point on P1: r1
Point on P2: r2
|(r1 - r2) . n |/ |n|    = |[ (r1 . n) - (r2. n) ]| / |n|  = |d1 - d2| / |n|
A: Same answer as "user183761" but with (perhaps) more explanation.
Given a point $P(x_1, y_1, z_1)$ on one plane and point $Q(x_2, y_2, z_2)$ 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 $(x_2-x_1, y_2-y_1, z_2-z_1)$.  By definition, for point $P$ to lie on the plane it must satisfy the equation $ax_1 + by_1 + cz_1 = d_1$.  Likewise, for point $Q$, $ax_2 + by_2 + cz_2 = d_2$.
Thus,
$$d_2 - d_1 = a(x_2-x_1) + b(y_2-y_1) + c(z_2-z_1)\\ = (a, b, c)^T(x_2-x_1, y_2-y_1, z_2-z_1) \\ = n·PQ$$
So $|d2 – d1|/||n|| = |n·PQ|/||n|| = |projnPQ| $
