Prove that the distance between parallel planes $\vec{n}\cdot \vec{x} = d_1 $, $\vec{n}\cdot \vec{x}=d_2$ is $|d1-d2|/||\vec{n}||$

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 ${ |d_1-d_2| \over \|n\|}$

This is a trick that will always work for finding distances between two "flat" objects.

1. First, identify the direction in which the "minimum distance" lies.
2. Second, take any two random points, one from object 1, the other from object 2.
3. 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 $\vec{x}_2\cdot{\vec{n}} = d_2$