We can derive the Bragg equation from the von Laue equations derived in the previous problem. First show that $\mathbf{S}=\mathbf{s}-\mathbf{s}_{0}$ bisects the angle between $\mathbf{s}_{0}$ and $\mathbf{s}$ and is normal to the plane from which the X-radiation would be specularly reflected (the angle of incidence equals the angle of reflection). Now show that the distance from the origin of the $\mathbf{a}, \mathbf{b}$, and c axes to the $h k l$ plane is given by
$$
d=\frac{\mathbf{a}}{h} \cdot \frac{\mathbf{S}}{|\mathbf{S}|}=\frac{\mathbf{b}}{k} \cdot \frac{\mathbf{S}}{|\mathbf{S}|}=\frac{\mathbf{c}}{l} \cdot \frac{\mathbf{S}}{|\mathbf{S}|}=\frac{\lambda}{|\mathbf{S}|}
$$
Last, show that $|\mathbf{S}|=\left[\left(\mathbf{s}-\mathbf{s}_{0}\right) \cdot\left(\mathbf{s}-\mathbf{s}_{0}\right)\right]^{1 / 2}=[2-2 \cos 2 \theta]^{1 / 2}=2 \sin \theta$, which leads to the Bragg equation $d=\lambda / 2 \sin \theta$