Prove a force driven by a cross product between a vector and its velocity gives a spiral movement parallel to the vector I was given a problem in a Classical Mechanics course that went somehow like the following:

"Consider a particle of mass $m$ moving under the presence of a force $\vec{F} = k\hat{x}\times\vec{v}$, where $\hat{x}$ is an unit vector in the positive direction of the $x$ axis and $k$ any constant. Prove that the movement of a particle under this force is restricted to a circular motion with angular velocity $\vec{\omega}=(k/m)\hat{x}$ or, in a more general case, a spiral movement parallel to the direction of $\hat{x}$."

In an Electrodynamics elementary college course you can see and solve that a magnetic force sort of as:
$$m\ddot{\vec{r}}=\frac{q}{c}\left(\vec{v}\times\vec{B}\right)$$
with a magnetic field, say, $\vec{B}=B_0\hat{z}$ can drive a particle through a spiral movement in the precise direction of that magnetic field you customize, involving a cyclotron frequency and so, if and only if you input further initial conditions to the movement in x, y and z.
My inquiry then is, how can you prove the relation given above for the angular velocity and conclude a spiral movement, from a classical mechanics perspective? I can see there's a link between both procedures, but I cannot try solving the first one without giving a glimpse to the latter.
 A: Write $\vec{v} = \dot{x}\hat{x} + \dot{y}\hat{y} + \dot{z}\hat{z}$, where $\hat{x}, \hat{y}, \hat{z}$ denote unit vectors in the positive direction of the axis $x$, $y$, $z$ respectively. Then,
$$k \hat{x}\times\vec{v} = -k \dot{z}\hat{y} + k\dot{y}\hat{z}.$$
Equating this expression with the analogous for the force, we get a system of three differential equations, which, after dividing by $m$, are as follows:
$$\begin{align*}
\ddot{x} &= 0,\\
\ddot{y} &= -\frac{k}{m}\dot{z},\\
\ddot{z} &= \frac{k}{m}\dot{y}.
\end{align*}$$ 
The first equation is readily solved to get the description of a uniform rectilinear movement along the $x$ direction,
$$x = at + b,$$
with $a$, $b$ constants. The remaining pair of equations correspond to the movement of a particle in a central field in the $yz$ plane. To prove that it corresponds to a circular motion, just substitute $v_y = \dot{y}$ and $v_z = \dot{z}$, so that the second and third equations become 
$$\begin{align*}
\dot{v}_y &= -\frac{k}{m} v_z,\\
\dot{v}_z &= \frac{k}{m} v_y.
\end{align*}$$ 
These equations can be solved to obtain $v_y = A\cos(\frac{k}{m}t + \phi)$, $v_z = A\sin(\frac{k}{m}t + \phi)$. Since $v_y = \dot{y}$, $v_z = \dot{z}$, one last pair of integrals give the final solution for $y$ and $z$:
$$\begin{align*}
y &= A\,\frac{m}{k}\,\sin\left(\frac{k}{m} t + \phi\right),\\
z &= -A\,\frac{m}{k}\,\cos\left(\frac{k}{m} t + \phi\right).
\end{align*}$$
The equations for $x$, $y$ and $z$ are the parametric equations for a spiral movement, parallel to the direcion of $\hat{x}$. 
A: With an aid of differential geometry, velocity, acceleration and jerk can be written as:
\begin{align*}
  \mathbf{v} &= \dot{s} \, \mathbf{T} \\
  &= v \, \mathbf{T} \\
  \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\
  \mathbf{b} &=
  (\dddot{s}-\kappa^2 \dot{s}^3) \mathbf{T}+
  (3\kappa \dot{s} \ddot{s}+\dot{\kappa} \dot{s}^2) \mathbf{N}+
  \kappa \tau \dot{s}^3 \mathbf{B} \\
\end{align*}
Now $$
\mathbf{a}=\boldsymbol{\omega} \times \mathbf{v} \implies
\mathbf{a} \perp \mathbf{T} \implies
\ddot{s}=0 \implies
\dot{s}=\text{constant} \implies
v=u$$
and $$
\mathbf{b}= \dot{\mathbf{a}}=\boldsymbol{\omega} \times \mathbf{a} \implies
\mathbf{b} \perp \mathbf{a} \implies
\mathbf{b} \perp \mathbf{N} \implies
\dot{\kappa}\dot{s}^2=0 \implies
\kappa=\text{constant}$$
Also,
\begin{align*}
  \int \mathbf{a} \, dt &= \boldsymbol{\omega}  \times \int \mathbf{v} \, dt \\
  \mathbf{v} &= \mathbf{u}+\boldsymbol{\omega}  \times \mathbf{r} \\
  \mathbf{r}(0) &= \mathbf{0} \\
  \dot{\mathbf{r}}(0) &= \mathbf{u} \\
  \boldsymbol{\omega} \cdot \mathbf{v} &=
  \boldsymbol{\omega} \cdot \mathbf{u} \\
  &= \text{constant}
\end{align*}
We have
\begin{align*}
  \mathbf{v} \times \mathbf{a} &=
  \mathbf{v} \times (\boldsymbol{\omega} \times \mathbf{v}) \\
  &= v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v} \\
  \kappa &=
  \frac{|v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v}|}
       {v^3} \\
   &= \frac{|\boldsymbol{\omega} \times \mathbf{v}|}{v^2} \\
   &= \frac{|\boldsymbol{\omega} \times \mathbf{u}|}{u^2} \\
  |\mathbf{a}|
  &= |\boldsymbol{\omega} \times \mathbf{u}| \\
  &= \text{constant} \\
  \mathbf{a} \times \mathbf{b} &=
  a^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{a}) \mathbf{a} \\
  &= a^2 \boldsymbol{\omega} \\
  \tau &=
  \frac{\mathbf{v} \cdot a^2\boldsymbol{\omega}}
       {(\mathbf{v} \times \mathbf{a})^2} \\
  &= \frac{\boldsymbol{\omega} \cdot \mathbf{v}}{v^2} \\
  &= \frac{\boldsymbol{\omega} \cdot \mathbf{u}}{u^2} \\
  &= \text{constant}
\end{align*}

Both $\kappa$ and $\tau$ are constants implying the path is helical.  If $\mathbf{u} \cdot \boldsymbol{\omega}=0 \implies \tau=0$, then it'll be a circle.  While $\mathbf{u} \times \boldsymbol{\omega}=\mathbf{0} \implies \kappa=0$, that'll be a straight line.

Fitting with initial conditions:
$$\fbox{$\quad \mathbf{r}=\mathbf{u}t+
\frac{\mathbf{u} \times \boldsymbol{\omega}}{\omega^2}(\cos \omega t-1)+
\frac{\boldsymbol{\omega} \times (\mathbf{u} \times \boldsymbol{\omega})}{\omega^{3}}(\sin \omega t-\omega t) \quad \\$}$$

Some facts from differential geometry
  \begin{align*}
  s &= \int |\mathbf{v}| \, dt
  \tag{arclength} \\
  \dot{s} &= |\mathbf{v}|
  \tag{speed} \\
  &= v \\
  \mathbf{T} &= \frac{\mathbf{v}}{v}
  \tag{tangent vector}\\
  \mathbf{B} &=
  \frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|}
  \tag{binormal vector} \\
  \mathbf{N} &= \mathbf{B} \times \mathbf{T}
  \tag{normal vector} \\
  \kappa &= \frac{|\mathbf{v} \times \mathbf{a}|}{v^3}
  \tag{curvature} \\
  \tau &=
  \frac{\mathbf{v} \cdot \mathbf{a} \times \mathbf{b}}
       {(\mathbf{v} \times \mathbf{a})^2}
  \tag{torsion}
\end{align*}

A: Take the equations of motion $$ \ddot{\mathbf{r}} = \lambda \;( \hat{k} \times \dot{\mathbf{r}}) $$ and express them in cylindrical coordinates $(r,\theta,z)$
$$\begin{pmatrix} \ddot{r} - r \dot{\theta}^2 \\ r \ddot{\theta} + 2 \dot{r} \dot{\theta} \\ \ddot{z} \end{pmatrix} = \lambda \left( \begin{pmatrix}0\\0\\1 \end{pmatrix} \times \begin{pmatrix} \dot{r} \\ r \dot{\theta} \\ \dot{z} \end{pmatrix} \right) = \lambda \begin{pmatrix} -r \dot\theta \\ \dot{r} \\ 0 \end{pmatrix} $$
which is solved by
$$\begin{align} 
  \ddot{r} & = r \dot{\theta} ( \dot{\theta}-\lambda ) \\
  \ddot{\theta} & = \frac{\dot{r} }{r} ( \lambda - 2 \dot{\theta} ) \\
  \ddot{z} & =0
\end{align} $$
I see two special cases in the solutions here


*

*Set $\dot{\theta}  = \lambda$ and $\ddot{\theta}=0$ to yield 
$$ \begin{align} 
  \ddot{r} & = 0 & \dot{r} & = 0 \\
  \ddot{z} & = 0 & \dot{z} & = \mbox{const} 
\end{align} $$

*Set $\dot{\theta}  = \frac{\lambda}{2}$ and $\ddot{\theta}=0$ to yield 
$$ \begin{align} 
  \ddot{r} & = -\frac{\lambda^2}{4} \,r & r & = \mbox{harmonic} \\
  \ddot{z} & = 0 & \dot{z} & = \mbox{const} 
\end{align} $$
