Construct a vector orthogonal to a given vector in $\mathbb{R}^3$ without singularities

Question

I have a unit vector $v = (x_v, y_v, z_v)$ in $\mathbb{R}^3$ and I want to construct another vector $u$ which is orthogonal to $v$. The construction process should be a straight-forward formula without singularities (divisions by zero, roots of negative number, etc.) and special cases.

For example, we can take the plane equation

$$ A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 $$

for a plane which contains $(x_0, y_0, z_0)$ point and $(A,B,C)$ is a normal vector for it. In our case it boils down to

$$ x_vx + y_vy + z_vz = 0 $$

and finally

$$ x = \frac{- y_vy - z_vz}{x_v} $$

so we can chose any $y$ and $z$ and we'll get $x$ such that $(x,y,z)$ is orthogonal to $v$. But what if $x_v$ is zero? In this case the plane contains $Ox$ axis, thus the choice of $x$ does not matter for any $y$ and $z$ from the plane. But I want to implement the construction of such vector in a GPU shader program, so processing special cases (branching) is unwanted because:

• It breaks performance of the program
• $x_v$ may be arbitrary close but not equal to zero
• $x_v$ may smoothly vary with time, so special case may introduce some animation artifacts.

Another approach is to take another predefined (i.e. hard-coded) vector $a$ and make $u = v \times a$, but the only restriction on $v$ is it's unitness, so this approach faces same problems if $a$ and $v$ are collinear or close to that.

I also tried to represent $v$ as some rotation of $(1,0,0)$, but I ended up with

$$ \beta = \arcsin(-z), $$ $$ \cos{\gamma} = \frac{x}{\cos{\beta}}. $$

For $z = \pm{1}$ it faces the same problem as above and it corresponds to a case when we rotate $(0,0,1)$ about $Oz$ axis, which does not make sense and always equals to identity rotation.

Could you give me a clue how to construct an orthogonal vector without singularities or an outline of a proof if it's impossible?

Answer 1

There is no continuous map $f\colon S^2\to S^2$ such that $f(x)\perp x$ for all $x$.

This is a consequence of the hairy ball theorem.

Answer 2

Hagen von Eitzen's answer excludes a simple expression producing such a vector ${\bf u}$, given ${\bf v}$. Nevertheless the practical problem remains. I propose the following:

Let ${\bf e}_k$ $(1\leq k\leq 3)$ be the three standard basis vectors of ${\mathbb R}^3$. Given any ${\bf v}\in{\mathbb R}^3$ choose $$j\in{\rm argmin}\bigl\{|{\bf v}\cdot{\bf e}_k|\>\bigm|\>1\leq k\leq 3\bigr\}\ ,$$ and put ${\bf u}:={\bf v}\times{\bf e}_j\>$.