Curve in plane intersecting a sphere I'm given a curve $r = r(t)$, with a derivative $\dfrac{dr}{dt}$ = $cr(t)$. (Where $c$ is a constant vector)
I'm trying to show that the curve is the circle which the plane through $r(0)$, normal to $c$, intersects a sphere (with radius $|r(0)|$).
I know that if $(r'(t) - r'(0))$ is perpendicular to $c$, then no component of $r'(t)$  is pointing away from the plane, and thus must lie in it. 
So thus,  $\dfrac{dr}{dt} [(r(t) - r(0)) · c]$ will give me $0$. And that therefore $[(r(t) - r(0)) · c] = A$, a constant. Which, if taken at $t=0$, implies that the constant is $0$ and that the curve lies in the plane.
But, I don't think I'm really understanding intuitively what I'm trying to solve. Any explanations?
 A: You have a fixed $c\in\mathbb{R}^3$ and a vector field given by $F(x)=c\times x$. You are looking for the trajectories of this vector field.
Without loss of generality, assume for convenience $c=(0,0,1)$. It follows that the vector field $F$ is every where parallel to the $xy$ plane. Thus, all the trajectories have constant $z$. 
Consider the point $p=(x,y,z)$. The vector field at $p$ is perpendicular to $p$ and to $c$, so it is also perpendicular to the projection of $p$ on the $xy$ plane. Hence, the trajectories are circles, with centers on the $z$ axis.
A: To show it lies in that plane: 
$$c \cdot (r'(t) - r'(0)) = c \cdot (c \times r(t) - c \times r(0))\\
 = c \cdot (c \times (r(t) - r(0))) = (r(t) - r(0))\cdot (c \times c) = 0$$
(the cross product of $c$ with anything else is going to be orthogonal to $c$).
To show the path lies in a sphere, show $\| r(t)\|$ is constant. That is, show
$$
\frac{d}{dt} r(t) \cdot r(t) \rangle = 2 r'(t)\cdot r(t)
$$
is zero. (you can use the same basic idea as the first part)
You have now shown that the image of this curve lies in the intersection of a plane and a sphere, i.e. it lies in the circle you have described.
