= Solution
With constant <rest mass>, $\mathbf p=m\gamma\mathbf v$ and the <chain rule> gives $\dot\gamma=\gamma^3(\mathbf v\cdot\mathbf a)/c^2$. Hence the laboratory <relativistic force> is
$$
\mathbf F=m\gamma\mathbf a+\frac{m\gamma^3}{c^2}(\mathbf v\cdot\mathbf a)\mathbf v.
$$
Use the <vector triple product identity> $\mathbf v\times(\mathbf v\times\mathbf a)=\mathbf v(\mathbf v\cdot\mathbf a)-v^2\mathbf a$, and $\gamma^3(1-v^2/c^2)=\gamma$, to obtain
$$
\boxed{\mathbf F=m\gamma^3\left(\mathbf a+\frac{\mathbf v\times(\mathbf v\times\mathbf a)}{c^2}\right).}
$$
The <relativistic energy> is $E=m\gamma c^2$, whose <time derivative> is $m\gamma^3\mathbf v\cdot\mathbf a$. Taking the <dot product> of the first expression for $\mathbf F$ with $\mathbf v$ gives the same quantity, because $\gamma+\gamma^3v^2/c^2=\gamma^3$. Thus $dE/dt=\mathbf F\cdot\mathbf v$. Since $d/d\tau=\gamma\,d/dt$, the <four-force> is
$$
\boxed{F^\alpha=\frac{dp^\alpha}{d\tau}=\left(\frac{\gamma\mathbf F\cdot\mathbf v}{c},\gamma\mathbf F\right).}
$$
Its <tensor contraction> with the <four-velocity> vanishes: $F^\alpha U_\alpha=-\gamma^2\mathbf F\cdot\mathbf v+\gamma^2\mathbf F\cdot\mathbf v=0$. This <orthogonality> also follows by differentiating $p^\alpha p_\alpha=-m^2c^2$. Resolving the laboratory <acceleration> relative to $\mathbf v$ gives $\mathbf F_\parallel=m\gamma^3\mathbf a_\parallel$ and $\mathbf F_\perp=m\gamma\mathbf a_\perp$.
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