Empty lattice approximation for a simple monoclinic crystal

a/b:    a/c:    $\alpha$: (degrees)   

  Symmetry points
  ($u,v,w$)  [$k_x,k_y,k_z$]
  Γ: (0, 0, 0)     [0, 0, 0]
  Y: (0, 1/2, 0)     [0, $\frac{\pi}{b \, \sin(\alpha)}$, 0]
  B: (1/2, 0, 0)     [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}$, 0]
  A: (1/2, 1/2, 0)     [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}+\frac{\pi}{b \, \sin(\alpha)}$, 0]
  Z: (0, 0, 1/2)     [0, 0, $\frac{\pi}{c}$]
  C: (0, 1/2, 1/2)     [0, $\frac{\pi}{b \, \sin(\alpha)}$, $\frac{\pi}{c}$]
  D: (1/2, 0, 1/2)     [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}$, $\frac{\pi}{c}$]
  E: (1/2, 1/2, 1/2)     [$-\frac{\pi}{a}$, $-\frac{\pi}{a \, \tan(\alpha)}+\frac{\pi}{b \, \sin(\alpha)}$, $\frac{\pi}{c}$]
Please note:
The k-values are evaluated on the straight line connecting two succeeding symmetry points.
Intermediary corners - as they occur for instance on the path between Y and A - will be neglected.
Therefore, paths like Γ-Y-C-Z or Γ-A-E-Z should be preferred over paths like Y-A-B.
\[ \begin{equation} \vec{b}_1=\frac{-2\pi}{a}(\hat{k}_x + \frac{1}{\tan(\alpha)}\hat{k}_y), \hspace{1cm} \vec{b}_2=\frac{2\pi}{b\, \sin(\alpha)}\hat{k}_y, \hspace{1cm} \vec{b}_3=\frac{2\pi}{c}\hat{k}_z. \end{equation} \]