= Solution
Use the <Minkowski metric> $(+,-,-,-)$ and choose the incoming <electron> along $+z$, with the scattered <electron> in the $xz$ plane. In <natural units>, neglecting the <electron mass>, the <four-momenta> are
$$
p_1=(E,0,0,E),\qquad p_2=(m_p,0,0,0),\qquad
p_3=(E',E'\sin\theta,0,E'\cos\theta).
$$
The <momentum transfer> entering the <proton> is therefore
$$
\boxed{q=p_1-p_3=(E-E',-E'\sin\theta,0,E-E'\cos\theta).}
$$
The outgoing hadronic system carries $p_X=p_2+q$. The <energy transfer> is $\nu=E-E'$, while $\mathbf q$ need not point along the original beam.
\Image[/past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2021-dis-kinematics-and-detector.png]
{title=Fixed-target <deep inelastic scattering>: laboratory directions and a schematic detector}
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