algebra precalculus – Confused about how we scale graph axis’ to make the axis’ dimensionless.

I am trying to understand the solution to part $mathrm{(iii)}$. But, for the question I’m asking to make sense I need to include the solutions to parts $mathrm{(i)}$ and $mathrm{(ii)}$ also:

Consider a triangular lattice where the sides of the triangles have length $d$. The figure gives a choice of unit cells (dashed lines).

Triangular Lattice

$mathrm{(i)}$ Use the sides of the unit cells as the primitive lattice vectors, $boldsymbol{a}_1$ and $boldsymbol{a}_2$. Write down these vectors in Cartesian coordinates.

$mathrm{(ii)}$ Write down a pair of reciprocal space vectors $b_{1,2}$ satisfying the condition that
$a_icdot b_j = 2pidelta_{ij}$ .
(If you want to use the explicit formula in three dimensions given in the lectures,
then you should pick as $boldsymbol{a}_3$ the unit vector in the direction out of the page.)

$mathrm{(iii)}$ The reciprocal lattice vectors G are defined by $G = h_1b_1 + h_2b_2$ where $h_{1,2}$ are integers and $boldsymbol{b}_1$ and $boldsymbol{b}_2$. Sketch the lattice that is formed by the reciprocal lattice vectors $boldsymbol{G}$ of the triangular lattice.


Solutions:

$mathrm{(i)}$ The primitive lattice vectors are $boldsymbol{a}_1 = (d, 0)$ and $boldsymbol{a}_2 = left(dfrac{d}{2},dfrac{sqrt{3}d}{2}right)$.

$mathrm{(ii)}$ A choice of the primitive lattice vectors (bold arrows in
diagram) for the reciprocal lattice is $boldsymbol{b}_1=left(dfrac{2pi}{d},-dfrac{2pi}{sqrt{3}d}right)$ and $boldsymbol{b}_2=left(0,dfrac{4pi}{sqrt{3}d}right)$. Other choices are possible, such as $−boldsymbol{b}_1$ and $−boldsymbol{b}_2$.

$mathrm{(iii)}$ $boldsymbol{G} = h_1boldsymbol{b}_1 + h_2boldsymbol{b}_2$ with integers $h_{1,2}$.
The diagram shows all the $boldsymbol{G}$ vectors plotted as points in $boldsymbol{k}$-space.
All the $boldsymbol{G}$ vectors form a periodic array in reciprocal space. This ‘reciprocal lattice’ for a triangular lattice in real space is itself a triangular lattice in $boldsymbol{k}$-space.
Reciprocal lattice

When I asked my lecturer about this scaling on the $x$ and $y$ axis he just said (something like) that it is to “avoid having factors of $dfrac{2pi}{d}$ on each increment of the $x$ and $y$ axis”. This makes sense since having a dimensionless $x$-axis looks clearer than this:

unscaled x-axis

and similarly for the $y$ axis.


So I will first factor out $dfrac{2pi}{d}$ then the reciprocal lattice vectors are $boldsymbol{b}_1=left(dfrac{2pi}{d},-dfrac{2pi}{sqrt{3}d}right)=dfrac{2pi}{d}left(1,-dfrac{1}{sqrt{3}}right)$ and $boldsymbol{b}_2=left(0,dfrac{4pi}{sqrt{3}d}right)=dfrac{2pi}{d}left(0,dfrac{2}{sqrt{3}}right)$ I thought the graph axis’ should look like this:

incorrectly scaled x-axis

and similarly for the $y$-axis.


The reason I think the graph axis should read $dfrac{2pi k_x}{d}$ and not $dfrac{k_x d}{2pi}$ (in the solution) is simply because I have factored out the $dfrac{2pi}{d}$ above so that what is plotted does not depend on $dfrac{2pi}{d}$. Math is not my strong point and I just cannot figure out why the axis reads $dfrac{k_x d}{2pi}$ instead of $dfrac{2pi k_x}{d}$ (which is what it looks like it should be). Can anyone please explain what is going on here?

Thanks in advance!