= Solution
Neglecting gravity, the <equation of motion> is $m\ddot{\mathbf x}=a\dot{\mathbf x}\times\mathbf x/r^3$, where $r=|\mathbf x|>0$. Dotting with the <velocity> gives $d(m|\dot{\mathbf x}|^2/2)/dt=0$, because a <cross product> is perpendicular to its factors. Thus the <speed> is constant. The <vector triple product> gives
$$
\frac d{dt}(m\mathbf x\times\dot{\mathbf x})=\frac a{r^3}\mathbf x\times(\dot{\mathbf x}\times\mathbf x)=a\left(\frac{\dot{\mathbf x}}r-\frac{\mathbf x(\mathbf x\cdot\dot{\mathbf x})}{r^3}\right)=a\frac d{dt}\frac{\mathbf x}r.
$$
Hence the <generalized angular momentum in a radial monopole field> is
$$
\boxed{\mathbf L=m\mathbf x\times\dot{\mathbf x}-a\frac{\mathbf x}r=\text{constant}}.
$$
Dotting with $\mathbf x$ gives $\mathbf L\cdot\mathbf x=-ar$, or $\widehat{\mathbf L}\cdot\widehat{\mathbf x}=-a/|\mathbf L|$. For $\mathbf L\ne0$, the direction of $\mathbf x$ therefore makes the fixed angle $\boxed{\cos\beta=-a/|\mathbf L|}$ with the axis $\mathbf L$, describing a <right circular conical surface> with vertex at the North Pole. Orthogonality gives $|\mathbf L|^2=m^2|\mathbf x\times\dot{\mathbf x}|^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary <angular momentum> is nonzero the cone is the usual orbital plane. If $\mathbf L=0$, the norm identity forces $a=0$ and $\mathbf x\times\dot{\mathbf x}=0$, giving a radial straight line.
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