= Solution
The three <primitive lattice translation vectors> join a cube corner to the three adjacent face centres. Thus the <Bravais lattice> is a <face-centered cubic lattice>, with conventional cubic side $a$.
Write the given <primitive lattice translation vectors> as $\mathbf a_1,\mathbf a_2,\mathbf a_3$. Their <primitive unit cell> <volume> is
$$
\Omega=\mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3)=\frac{a^3}{4},\qquad
\mathbf a_2\times\mathbf a_3=\frac{a^2}{4}(-1,1,1).
$$
The <reciprocal lattice> convention $\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}$ then gives
$$
\boxed{\begin{aligned}
\mathbf b_1&=\frac{2\pi}{a}(-1,1,1),\\
\mathbf b_2&=\frac{2\pi}{a}(1,-1,1),\\
\mathbf b_3&=\frac{2\pi}{a}(1,1,-1).
\end{aligned}}
$$
Each <reciprocal lattice> vector has units $\mathrm{m^{-1}}$ when $a$ is in metres. Integer combinations generate a <body-centered cubic lattice> of conventional cubic side $4\pi/a$; the point $(2\pi/a)(1,1,1)$ is its body centre. Directly taking the nine <dot products> checks the signs and the $2\pi$ normalization.
Back to article page
