I was trying to define a function \begin{align*} f: \mathbb{R}^3 \times \mathbb{R}^3 \longrightarrow \mathbb{R}^3 \setminus \lbrace 0 \rbrace \end{align*} such that for any two vectors v and w, \begin{align*} f(\mathbf{v}, \mathbf{w}) \perp \mathbf{v}, \mathbf{w}. \end{align*} The obvious choice would be $f(\mathbf{v}, \mathbf{w}) = \mathbf{v} \times \mathbf{w}$, but of course this doesn't do the job if v and w are parallel. One solution I thought of, which doesn't work, would be to find a function $g : \mathbb{R} \longrightarrow \mathbb{R}$ satisfying \begin{align*} g(t) = 0 \Longleftrightarrow t \neq 0. \end{align*} and define \begin{align*} f(\mathbf{v},\mathbf{w}) = \mathbf{v} \times \mathbf{w} + g(|\mathbf{v} \times \mathbf{w}|)\cdot (\mathbf{v}\times R\mathbf{v}) \end{align*} where for instance \begin{align*} R = \left( \begin{matrix} 0 &-1 &0\\ 1 &0 & 0\\ 0 &0 &1 \end{matrix} \right) \end{align*} is the rotation by $90$ degrees around the $z$-axis.
The point of wanting a function like $f$ was that I didn't want to check cases all the time, ie. check whether my two vectors of the day are parallel or not. For that reason, I'm looking for some $g$ which is not defined conditionally $-$ for example, the indicator function of the singleton $\lbrace 0 \rbrace$ is not what I'm looking for. On the other hand, a closed form expression would be perfect. Obviously, $g$ can't be continuous, but that's not relevant anyway.
Depending on how you feel about assigning $0^0$ the value $1$, the function $g(t) = 0^t$ would be a possilibity, but I wonder if there are any other (preferably less controversial) choices of such a function $g$. Any thoughts?
EDIT: I'm just interested in what functions $g$, not defined conditionally, satisfy the condition I mention. The introduction about $f$ is merely what motivated the question about such functions $g$.