# Confused about Miller indices and parallel planes

I am doing the classic exercise of finding the distance between interplanar planes in a cubic lattice. So i need to demonstrate that the distance d of the (h,k,l) planes is $$d = a/(h²+k²+l²)^{1/2}$$.

So, using elementary geometry, the distance between parallel planes is $$|c-c'|/\sqrt{m²+n²+p²}$$, where the plane equations are $$0 = xm + y n + z p - c, 0 = x m + y n +z p - c'$$.

Now, since (h,k,l) are miller indices, i thought that the right equation for the planes would be $$0 = x/h + y/k + z/l - c$$, since the miller indices are the reciprocal of these "geometric indices".

Unfortunately, using this equation and the formula, we get $$d = a/(1/h²+1/k²+1/l²)^{1/2}$$.

I think the main problem here is that i am interpreting the miller indices wrong, even so I don't know where my error is... Could you help me?

$$\mathbf{The\;General \;Method}$$
Let $$\boldsymbol{a_1}$$, $$\boldsymbol{a_2}$$ and $$\boldsymbol{a_3}$$ be the crystallographic axes. Then the planes with Miller indices $$(hkl)$$ intercept the crystallographic axes at $$n\boldsymbol{a_1}/h,n\boldsymbol{a_2}/k$$ and $$n\boldsymbol{a_3}/l$$, where $$n$$ is an integer. In general, the vectors $$h\boldsymbol{a_1}+k\boldsymbol{a_2}+l\boldsymbol{a_3}$$ is not perpendicular to the $$(hkl)$$ planes. This is why the general method of determination of the interplanar spacing requires reciprocal lattice basis $$\boldsymbol{b_1},\boldsymbol{b_2}$$ and $$\boldsymbol{b_3}$$, as the reciprocal lattice vectors $$\boldsymbol{G}_{hkl}=h\boldsymbol{b_1}+k\boldsymbol{b_2}+l\boldsymbol{b_3}$$ is perpendicular to the family of $$(hkl)$$ planes. Expoliting this property, the $$(hkl)$$ interplanar spacing $$d_{hkl}$$ can be found by projecting the vector joining the separation of two adjacent planes at the $$\boldsymbol{a_1}$$ axis onto the unit normal, that is $$d_{hkl}=\frac{\boldsymbol{a_1}}{h}\cdot \frac{\boldsymbol{G}_{hkl}}{|\boldsymbol{G}_{hkl}|}. \tag{1}\label{eq1}$$ Since $$\boldsymbol{b_i}\cdot \boldsymbol{a_j}=2\pi\delta_{ij},$$ Eq.\eqref{eq1} can be expressed as $$d_{hkl}=\frac{2\pi}{|\boldsymbol{G}_{hkl}|}. \tag{2}\label{eq2}$$
$$\mathit{Simple\, Cubic \,Lattice}$$
For simple cubic lattice, $$|\boldsymbol{a_1}|=|\boldsymbol{a_2}|=|\boldsymbol{a_3}|=a.$$ The reciprocal lattice vector is $$\boldsymbol{G}_{hkl}=\frac{2\pi}{a}(h\hat x+k\hat y+l\hat z),$$ where $$\hat x, \hat y$$ and $$\hat z$$ are orthogonal unit vectors and by Eq.\eqref{eq2}, the interplanar spacing is $$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}. \tag{3}\label{eq3}$$
$$\mathbf{Your\;Method}$$
Having presented the general formalism, now I attempt to use your reasoning to find the interplanar spacing for the simple cubic lattice. The nearest $$(hkl)$$ plane from the origin is found by setting $$n=1,$$ that is the plane which intercepts the crystallographic axes at $$\boldsymbol{a_1}/h,\boldsymbol{a_2}/k$$ and $$\boldsymbol{a_3}/l$$. Let the plane equation for this plane has the form $$Ax/a+By/a+Cz/a=D.$$ The x-interception is $$a/h$$, so we must have $$D/A=1/h$$. By similar reasoning, we get $$D/B=1/k$$ and $$D/C=1/l$$. So the plane equation has the form $$\frac{hx}{a}+\frac{ky}{a}+\frac{lz}{a}=1. \tag{4}\label{eq4}$$ From Eq.\eqref{eq4}, we see that the normal vector to the plane is $$\frac{h\hat x}{a}+\frac{k\hat y}{a}+\frac{l\hat z}{a}.$$ It can be shown that the adjacent plane to this plane should be $$hx/a+ky/a+lz/a=0$$ or $$hx/a+ky/a+lz/a=2.$$ Using the formula for distance $$D$$ between parallel planes $$ax+by+cz=d_0$$ and $$ax+by+cz=d_1$$, $$D=\frac{|d_0-d_1|}{\sqrt{a^2+b^2+c^2}}\tag{5}\label{eq5}$$ and $$|d_0-d_1|=1$$, the interplanar spacing is found to be $$d_{hkl}=\frac{1}{\sqrt{(h/a)^2+(k/a)^2+(l/a)^2}}=\frac{a}{\sqrt{h^2+k^2+l^2}},\tag{6}\label{eq6}$$ which is identical to Eq.\eqref{eq3}.