Show that $\sigma(t)$ lies on a sphere of radius 1 and center at the origin. Let $A$ and $B$ two unitary vectors in $\mathbb{R}³$, such that $A \cdot B = 0$, if $\sigma(t)=A\cos{t}+B \sin{t}$, show that $\sigma(t)$ lies on a sphere of radius 1 and center at the origin.
 A: Hint: follow your nose and compute $\langle \sigma(t),\sigma(t)\rangle$, using that $\|A\|=\|B\|=1$ and $\langle A,B\rangle=0$.
A: A very simple problem with some very engaging consequences.
It is easy to see that
$\langle \sigma(t), \sigma(t) \rangle = \langle A \cos t + B \sin t, A \cos t + B\sin t \rangle$$
= \langle A, A \rangle \cos^2 t + 2\langle A, B \rangle \cos t \sin t + \langle B, B \rangle \sin^2 t = \cos^2 t + \sin^2 t = 1, \tag 1$
which shows that $\sigma(t)$ lies in the unit sphere.  
A few other engaging facts concerning $\sigma(t)$:  the map
$\sigma:[0, 2\pi] \to S^2 \subset \Bbb R^3 \tag 2$
is manifestly periodic of period $2\pi$; $\sigma$ is injective on $[0, 2\pi)$; this easily follows from the linear independence of $A$ and $B$; $\sigma(t)$ in fact lies in the plane spanned by $A$ and $B$, which is normal to the vector $A \times B$ and contains the origin; any such plane intersects $S^2$ in a great circle, so in fact $\sigma(t)$ traces out that unique great circle containing $A$ and $B$.  This gives us a sweet way to parametrize geodesics (great circles) in the sphere $S^2$, which is an issue which arises in other problems.  Finally, it appears we may relax the condition $\langle A, B \rangle = 0$ to mere linear independence of $A$ and $B$ and obtain conclusions similar to those at which we have here arrived.  It would be fun and useful to work these things out in greater detail, and maybe I will--later.
Nota Bene: If
$\langle A, B \rangle \ne 0, \tag 3$
but
$A \times B = 0, \tag 4$
that is, $A$ and $B$ are linearly independent, we may still find our great circle by replacing $B$ by $B - \langle A, B \rangle A$, which is easily see to be normal to $A$, and then simply take
$\sigma(t) = A \cos t + (B - \langle A, B \rangle A)\sin t, \tag 5$
and we have a map
$\sigma:[0, 2\pi) \to S^2 \tag 6$
which expresses a great circle in terms of two non-orthogonal vectors.  End of Note.
