One of the properties of the reciprocal lattice vector is shown in the following
Theorem. For any family of lattice planes separated by a distance $d$ there are reciprocal
lattice vectors $\mathbf{G}_m$ perpendicular to the planes, the shortest of which $\mathbf{g}_{hkl}$have a length
$$
|\mathbf{g}_{hkl}|=\frac {2\pi}{d}
$$
If $\mathbf {G} _{m}=m_{1}\mathbf {b} _{1}+m_{2}\mathbf {b} _{2}+m_{3}\mathbf {b} _{3}$ is a reciprocal lattice vector pointing to the fixed direction identified by $m=(m_1, m_2, m_3) \in \mathbb{Z}^3$ and $\mathbf {R} _{n}=n_{1}\mathbf {a} _{1}+n_{2}\mathbf {a} _{2}+n_{3}\mathbf {a}_{3}$ is the generic direct lattice vector, the equation
$$
\mathbf{R}_n \cdot \mathbf{G}_m = 2\pi Z, \qquad Z\in\mathbb{Z}
$$
represents a parallel sheaf of planes $\pi^{(m)}_Z$ all perpendicular to the line set by $\mathbf{G}_m$.
I sketched a method
that seems to derive the interplanar distance, namely $\displaystyle d=(\mathbf{R}_{n'}-\mathbf{R}_n)\cdot\frac{\mathbf{G}_m}{|\mathbf{G}_m|}$, where $\mathbf{R}_{n'}$ and $\mathbf{R}_n$ are two arbitrary direct lattice vectors satisfying $\mathbf{R}_{n'} \cdot \mathbf{G}_m = 2\pi (Z+1)$ and $\mathbf{R}_{n} \cdot \mathbf{G}_m = 2\pi Z$.
This can't be right, since the theorem requires the shortest vector in the direction of $\mathbf{G}_m$, and not $\mathbf{G}_m$ itself. Where is the mistake?
Best Answer
Each reciprocal lattice vector $\mathbf{G}$ defines a set of equally-spaced, parallel planes via $$ \mathbf{G}\cdot\mathbf{R}=2\pi m\,, $$ one for every integer $m$. Nearest-neighbor planes have $|\Delta m| =1$. Your calculation indeed shows that the distance between consecutive planes is exactly $2\pi/G$. For completeness, I'll sketch the proof here.
Let $\mathbf{r}_1$ and $\mathbf{r}_2$ be points in consecutive planes, i.e. they satisfy $$ \mathbf{G}\cdot\mathbf{r}_1=2\pi m\,,~~~~~~~ \mathbf{G}\cdot\mathbf{r}_2=2\pi (m+1)\,. $$ Then, $\mathbf{G}\cdot(\mathbf{r}_2 - \mathbf{r}_1)=2\pi$. Dividing by G, we get $\hat{\mathbf{G}}\cdot(\mathbf{r}_2 - \mathbf{r}_1)=2\pi/G$, where $\hat{\mathbf{G}}$ is a unit vector in the direction of ${\mathbf{G}}$. We can clearly interpret the left-hand side as the component of the vector $\mathbf{r}_2 - \mathbf{r}_1$ that is perpendicular to the planes, and hence its length must be the distance between the planes.
The issue is that this set of planes is not a family of lattice planes except in the special case where $\mathbf{G}$ is the smallest reciprocal lattice vector in its direction. The problem is that if $\mathbf{G}$ is not the shortest, then there are too many planes; that is, some of the planes don't have any lattice points at all. (In fact, one of these planes is either a lattice plane$-$it has a two-dimensional lattice of points of the 3D lattice$-$or is it has exactly zero lattice points.)
We only get a family of lattice planes$-$that is, a set of equally-spaced, parallel planes, each of which contains an infinite number of points of the lattice$-$if $\mathbf{G}$ is the shortest reciprocal lattice vector in the direction $\hat{\mathbf{G}}$.
This is a claim. To see this, we proceed by way of contradiction. First form the set of planes corresponding to $\mathbf{G}'=2\mathbf{G}$, where $\mathbf{G}$ is a reciprocal lattice vector. Then, $\mathbf{G}'$ is not the shortest reciprocal lattice vector along this direction. The $m=0$ plane is clearly a lattice plane (because $\mathbf{R}=0$ is in this plane), so consider the plane $m=1$. If there is lattice point $\mathbf{R}$ in this plane, then $$ \mathbf{G}'\cdot\mathbf{R} = 2\pi\,, $$ which means that $$ \mathbf{G}\cdot\mathbf{R} = \pi\,. $$ At the same time, since $\mathbf{G}$ is a reciprocal lattice vector, it must satisfy (by definition of the reciprocal lattice!) $$ e^{i\mathbf{G}\cdot\mathbf{R}}=1 $$ which further implies that $$ \mathbf{G}\cdot\mathbf{R} = 2\pi m $$ for some integer $m$. This contradicts the statement above that $\mathbf{G}\cdot\mathbf{R} = \pi$. Hence, our assumption that there was a lattice point $\mathbf{R}$ in this plane must be false. Hence that plane is empty! There are too many planes if we don't use the shortest reciprocal lattice vector along that direction!
Finally, we should probably show that the set of planes formed using a shortest reciprocal lattice vector really is a family of lattice planes, but the proof is essentially dual to the one that we just did. We pretty much just have to reverse the role of the lattice vectors and reciprocal lattice vectors.