Show that the motion is a circular path Question:
The position $\mathbf r$ of a particle satisfies
$$ \ddot {\mathbf r} = f\dot {\mathbf r} \times \mathbf k$$
where $f$ is a known constant and $\mathbf k$ is the unit vector in the $z$ direction.
Given that $\dot {\mathbf r} \cdot \mathbf k \equiv 0$, show that the path of the particle is a circle.

Attempt:
Firstly, there are a few quantities that I know are constant:
$$\frac{d}{dt}\big(|\dot{\mathbf r}|^2 \big) = 2\ddot {\mathbf r} \cdot \dot {\mathbf r} = 2f(\dot{\mathbf r} \times \mathbf k) \cdot \dot{\mathbf r} = 0 \implies |\dot{\mathbf r}|^2 \equiv \text{constant}$$
$$\frac{d}{dt}\big(\mathbf r \cdot \mathbf k\big) = \dot{\mathbf r} \cdot \mathbf k =0 \implies {\mathbf r} \cdot \mathbf k \equiv \text{constant}$$
Moreover, I noticed that
$$\frac{d^2\dot{\mathbf r}}{dt^2} = f^2\dot{\mathbf r}$$
and solving this gives
$$\dot{\mathbf r}(t) = \mathbf c_1 \cos(ft)+\mathbf c_2 \sin (ft)$$
but I am unsure how to deduce from this that $\mathbf r$ traces out a circle.
 A: Let $\mathbf u = \mathbf r + \frac1f (\dot{\mathbf r} \times \mathbf k)$, we have
$$\dot{\mathbf u} = \dot{\mathbf r} + \frac1f(\ddot{\mathbf r} \times \mathbf k)
= \dot{\mathbf r} + ((\dot{\mathbf r} \times \mathbf k) \times \mathbf k)
= \dot{\mathbf r} + ((\dot{\mathbf r}\cdot \mathbf k) \mathbf k - \dot{\mathbf r}) = \mathbf 0$$
This means $\mathbf u$ is a constant. Notice
$$\frac{d}{dt}|\mathbf r - \mathbf u|^2
= 2(\mathbf r-\mathbf u)\cdot \dot{\mathbf r} = -\frac{2}{f} (\dot{\mathbf r} \times \mathbf k)\cdot \dot{\mathbf r} = \mathbf 0$$
The expression $|\mathbf r-\mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $\mathbf u$ with radius $R$. Together with what you know $\mathbf r \cdot \mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
A: Insert the solution you found of the derived equations into the original equations.
\begin{multline}
-f\vec c_1\sin(ft)+f\vec c_2\cos(ft)=\frac{d}{dt}\dot{\vec r}
\\
=\ddot{\vec r}=f(\vec c_1×\vec k)\cos(ft)+f(\vec c_2×\vec k)\sin(ft).
\end{multline}
Comparing coefficients for the same trigonometric functions gives
$$
\vec c_2=~~\vec c_1×\vec k,\\
\vec c_1=-\vec c_2×\vec k.
$$
This is only possible if both vectors are orthogonal to $\vec k$, of the same length and orthogonal to each other. With $\vec k=(0,0,1)$ you get $\vec c_1=(a,b,0)$ and $\vec c_2=(b,-a,0)$. This means that $\dot{\vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $\vec r$ circulates.
A: One way is to show that the curvature is constant. Recall that
$$\kappa = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} = \frac{||\dot{\mathbf{r}}(t) \times \ddot{\mathbf{r}}(t)||}{||\dot{\mathbf{r}}(t)||^3} $$
Which should be straightforward since you have $\dot{\mathbf{r}}(t)$. 
It might help to write: 
\begin{align}
\dot{\mathbf{r}}(t) &= \mathbf{c_1}\cos(ft) + \mathbf{c_2}\sin(ft)\\
&= \left[\begin{array}{c}c_{11}\\c_{12}\\c_{13}\end{array}\right]\cos(ft) + \left[\begin{array}{c}c_{21}\\c_{22}\\c_{23}\end{array}\right]\sin(ft)\\
&=(c_1\cos(ft) + c_2\sin(ft))\hat{\mathbf{i}} + (c_3\cos(ft) + c_4\sin(ft))\hat{\mathbf{j}} + (c_5\cos(ft) + c_6\sin(ft))\hat{\mathbf{k}} 
\end{align}
Then just do the grinding to get the final result.
EDIT: I just noticed that it might be easier to replace quantities like $\ddot{\mathbf{r}}(t) = f\dot{\mathbf{r}}(t) \times \mathbf{k}$ into the curvature expression as well.
EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.
