Given a vector normal to a plane, how to find two vectors parallel to the plane? I understand that I can find the normal vector by using cross product on the two vectors parallel to the plane, but I can't seem to find anywhere that tells me how to do it in reverse, and using the cross product formulas and methods are extremely messy, slow, and prone to error.
How would I find the two parallel vectors quickly and efficiently?
 A: Let's assume we're working in $\mathbb R^3$ and the plane is $$ax+by+cz+d=0.$$ A normal vector to the plane is $[a,b,c].$ If $c \ne 0$ then $[0,-c,b]$ and $[-c,0,a]$ are two linearly independent vectors perpendicular to $[a,b,c]$ and thus parallel to the plane $ax+by+cz+d=0.$
A: Given the normal vector $n=(a,b,c)$, a vector $v=(p,q,r)$ parallel to the plane is perpendicular to the normal vector and thus satisfies $n\cdot v=0$. If you put arbitrary values for $p,q$ and $c\ne0$ then $r=-\frac{pa+qb}c$ ensures $v$ is parallel to the plane; it should not take long to repeat this process (arbitrarily assuming two of $p,q,r$, then finding the third) to find two linearly independent parallel vectors. Just don't divide by zero.

However, in a sense, any formula for deriving vectors perpendicular to the normal vector, and hence parallel to the plane, must be messy. The hairy ball theorem guarantees that any continuous function $f:\mathbb R^3\to\mathbb R^3$ such that the input $v$ and output $w$ satisfy $v\cdot w=0$ must have some $v\ne\mathbf0$ for which $f(v)=\mathbf 0$, so you don't get a nontrivial parallel vector.
A: Assuming you are in dimension 3 (though it would also work in higher dimensions by changing "plane" to "hyperplane").
Let’s call your vector $v$, and your plane $P$.
You know that $v \perp P$.
Let’s call $v^\perp$ the space of vectors that are orthogonal to $v$. That space is of dimension 2. and $$P \subseteq v^\perp$$
But P is also of dimension 2. Thus $P = v^\perp$.
Now we need to find any two vectors that are in $v^\perp$. Ie we need any two vectors that are orthogonal to $v$. Ie we need $a$ and $b$ that verify
$$v^T a = 0$$
$$v^T b = 0$$
Those are linear equations (linear systems in the coordinates of a and b). You know how to solve them. In this case they have an infinity of solutions, you just need to pick any 2 that are not colinear
