Good notation for perpendicular vectors?

Question

I have a unit vector $\mathbf{\hat{a}}\in\mathbb{R^3}$. I would like to calculate two new unit vectors $\mathbf{\hat{b}}$, $\mathbf{\hat{c}}$ which are perpendicular to $\mathbf{\hat{a}}$ and to each other, i.e. $$\mathbf{\hat{a}}\cdot\mathbf{\hat{b}}=\mathbf{\hat{a}}\cdot\mathbf{\hat{c}} = \mathbf{\hat{b}}\cdot\mathbf{\hat{c}} = 0.$$

It doesn't matter which direction $\mathbf{\hat{b}}$ and $\mathbf{\hat{c}}$ point in provided the above conditions are met.

Do you know the simplest way of doing this? One way to do this is to introduce a new vector $\mathbf{\hat{x}}$ which is the unit vector in the $x$-direction. Then let $$\mathbf{\hat{b}} = \mathbf{{\hat{x}}}\times\mathbf{\hat{a}},$$ $$\mathbf{\hat{c}} = \mathbf{\hat{a}}\times(\mathbf{{\hat{x}}}\times\mathbf{\hat{a}}).$$ However, this doesn't work if $\mathbf{\hat{a}}$ is parallel to $\mathbf{\hat{x}}$.

Is there a simpler/more elegant way of expressing this?

Edit: I am interested in the case where $\mathbf{\hat{a}} = \mathbf{\hat{a}}(x,y,z)$, where we can assume $\mathbf{\hat{a}}(x,y,z)$ is smooth. It would be nice if $\mathbf{\hat{b}}(x,y,z)$ and $\mathbf{\hat{c}}(x,y,z)$ were also smooth.

Answer 1

This bears on your question, although I'm not sure that it answers it fully:

Hairy ball theorem - Wikipedia

Application to computer graphics

A common problem in computer graphics is to generate a non-zero vector in $\mathbb{R}^3$ that is orthogonal to a given non-zero vector. There is no single continuous function that can do this for all non-zero vector inputs. This is a corollary of the hairy ball theorem. To see this, consider the given vector as the radius of a sphere and note that finding a non-zero vector orthogonal to the given one is equivalent to finding a non-zero vector that is tangent to the surface of that sphere where it touches the radius. However, the hairy ball theorem says there exists no continuous function that can do this for every point on the sphere (equivalently, for every given vector).