= Solution
Let the <Bravais lattice> have primitive vectors $a_1,a_2,a_3$, and put
$$
\Omega=a_1\cdot(a_2\times a_3).
$$
The <reciprocal lattice> has the <basis>
$$
b_1=2\pi\frac{a_2\times a_3}{\Omega},\qquad
b_2=2\pi\frac{a_3\times a_1}{\Omega},\qquad
b_3=2\pi\frac{a_1\times a_2}{\Omega}.
$$
The <scalar triple product> and <cross product> identities give
$$
a_i\cdot b_j=2\pi\delta_{ij}.
$$
Consequently every $q=\sum_jn_jb_j$ satisfies $q\cdot l\in2\pi\mathbb Z$ for every <lattice point> $l=\sum_im_ia_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be <integer>[integers]. This proves that the displayed vectors generate $\Lambda^*$.
In the <Born approximation>, suppose that the crystal potential is the <sum> of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its <Fourier transform> at the <momentum transfer> $Q=k-k'$ factors as
$$
\widetilde V(Q)=\widetilde V_0(Q)\Delta(Q),
\qquad
\Delta(Q)=\sum_{l\in S}e^{iQ\cdot l}.
$$
Thus the single-atom <scattering amplitude> is multiplied by the <crystal lattice structure factor> $\Delta(Q)$. Writing
$$
l=l_1a_1+l_2a_2+l_3a_3,
\qquad
\alpha_i=Q\cdot a_i,
$$
separates the sum into three <finite geometric series>. For $l_i=-L_i/2,\ldots,L_i/2$ this gives
$$
\boxed{
\Delta(Q)=\prod_{i=1}^3
\frac{\sin((L_i+1)\alpha_i/2)}{\sin(\alpha_i/2)}.}
$$
When the $L_i$ are large, a factor has a sharp maximum when $\alpha_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in\Lambda^*$. These are the <reciprocal-lattice peaks of a finite crystal>.
For the stated <body-centered cubic lattice> basis,
$$
a_1=\frac a2(1,1,1),\qquad
a_2=\frac a2(1,-1,1),\qquad
a_3=a(0,0,1),
$$
the reciprocal-basis formula gives
$$
b_1=\frac{2\pi}{a}(1,1,0),\qquad
b_2=\frac{2\pi}{a}(1,-1,0),\qquad
b_3=\frac{2\pi}{a}(-1,0,1).
$$
Equivalently,
$$
\Lambda^*=\frac{2\pi}{a}
\{(h,k,l)\in\mathbb Z^3:h+k+l\text{ is even}\}.
$$
The shortest nonzero reciprocal vectors have squared <Euclidean norm> $2(2\pi/a)^2$, so
$$
q_{\min}=\frac{2\sqrt2\pi}{a}.
$$
In <elastic scattering>, $|k|=|k'|=k$. If $q=k-k'\in\Lambda^*$ and $\theta$ is the angle between $k$ and $k'$, <Euclidean geometry> gives
$$
|q|=2k\sin\frac\theta2.
$$
The first possible <diffraction> peak therefore occurs at
$$
\boxed{
\theta_{\min}=2\arcsin\!\left(\frac{\sqrt2\pi}{ka}\right)
\sim\frac{2\sqrt2\pi}{ka}}
$$
as $ka\to\infty$, by the <small-angle approximation>. This is the <elastic Bragg scattering condition> for the shortest reciprocal-lattice vector.
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