What's the slickest way to show $[x,v,x\times v]\in SO(3)$? Given $x,v$ are orthonormal in $\Bbb R^3$ i.e. $\|x\|=\|v\|=1,\,\langle x,v\rangle = 0$, what's the slickest way to show that $[x,v,x\times v]\in SO(3)$? In particular, how to show that $\det [x,v,x\times v] = 1$ instead of $-1$?
This is of course intuitively obvious and easy to wrap one's mind around, but I just couldn't find a rigorous proof. For example, it is tempting to say "well we can always find some rotation $R$ which sends $[e_1,e_2,e_3]$ to $[x,v,x\times v]$" but this ends up circular reasoning because a rotation is exactly an element in $SO(3)$.
Another possible approach I think is to show the tangent bundle $TS^2$ is a connected topological space. Once we show it, we can proceed to show $T_1S^2$, the unit tangent bundle, being obviously a continuous image of $TS^2$ by a normalisation map must be connected too, and again its image under the determinant map $\det$ must also be connected and hence can only be the singleton $\{1\}$. But how to show $TS^2$ is connected in the first place which might be much harder?
 A: Consider using the formula $$\det[\mathbf{a},\mathbf{b},\mathbf{c}]=(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}$$ which is a direct consequence of expanding the determinant along the third column.
A: Depends on how do you define the cross product.
Using the coordinatewise definition, Rene provided you with a strong hint.
If, though, we define the cross product through the Hodge star
$$a \times b = \star (a \wedge b),$$
the determinant of three 3-dimentional vectors as
$$\det(a,b,c) = \star(a \wedge b \wedge c),$$
and use the identity
$$a \wedge (\star b) = \langle a, b \rangle \omega,$$
where $\omega \in \Lambda^3 \mathbb{R}^3$ is the preffered volume element (usually $e_1 \wedge e_2 \wedge e_3$ in terms of a basis), and the definition of scalar product of $k$-vectors, then the answer is a straightforward alrebraic manipulation:
$$\det(x,v,x \times v)=\star(x \wedge v \wedge (x \times v)) = \star(x \wedge v \wedge \star(x \wedge v)) = \star(\|x \wedge v\|^2 \omega) = \|x \wedge v\|^2 = \\ = \begin{vmatrix}\langle x,x \rangle & \langle x,v \rangle \\ \langle v,x \rangle & \langle v,v \rangle \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$$
