If $\frac{dr}{dt}=c\times r(t)$, for $c$ a constant vector, then $r(t)$ describes a particular circle 
Suppose the curve $r=r(t)$ satisfies $\frac{dr}{dt}=c\times r(t)$, where $c$ is a constant vector. Show that the curve is the circle in which the plane through $r(0)$ normal to $c$ intersects a sphere with radius $|r(0)|$ centered at the origin.

Can somebody help me with this question? 
 A: Try like this:
Let $\vec r(0) = \vec a$ (which is also a constant).
Very convenient vector forms of the equations of plane and sphere are:
$$ \vec r \cdot \vec c = const = \vec a \cdot \vec c $$
for a plane to which the point $\vec r = \vec a$ belongs; and
$$ \vec r \cdot \vec r = r^2 = const = a^2  = \vec a \cdot \vec a$$
for a sphere of radius $a = |\vec a|$. To prove that each point of the curve is in the plane, dot the curve's differential equation with $\vec c$:
$$ \vec c \cdot \frac{d\vec r}{dt} = \vec c \cdot (\vec c \times \vec r) = 0$$
but the left-hand-side is identical to $d(\vec c \cdot \vec r)/dt$ because $\vec c$ is a constant and can be taken inside the derivative. Integrating it, we see that 
 $\vec c \cdot \vec r$ is a constant on the curve. In other words, every point on the curve has the same $\vec c \cdot \vec r$ as the point $\vec r(0)$, for which it is exactly $\vec a \cdot \vec c$. Thus each point also satisfies the equation of the plane, i.e. the curve is a subset of the plane.
To show the same with the sphere, dot the equation with $\vec r$ instead:
$$ \vec r \cdot \frac{d\vec r}{dt} = \vec r \cdot (\vec c \times \vec r) = 0$$
as in the case of the plane, a vector identity transforms the left-hand side into $(1/2)d(\vec r \cdot \vec r)/dt$, so after integration,  $\vec r \cdot \vec r$ is also a constant on the curve, and for the point $\vec r(0)$ its value is exactly $\vec a \cdot \vec a$. Hence every point satisfies the equation of the sphere as well, i.e. the curve is a subset of the sphere.
The converse, that every point of the intersection of the plane and sphere belongs to the curve, can be proven with a more brute force approach, but I like the following way of thinking about it:
If we think of $t$ as time, then $d\vec r/dt$ is a velocity (of say, a particle traversing the curve). You can show that as a vector function of time, it is continuous, and its magnitude is everywhere a constant finite non-zero value. This means the particle never stops, never suddenly changes direction (therefore cannot double back), and for long enough time will travel an arbitrarily long distance. Therefore it must travel along the whole circle.
