TITLE: Parametric equation for a plane perpendicular to a vector
QUESTION [5 upvotes]: The implicit equation for a plane perpendicular to a given vector at the origin is $ax + by + cz = 0$.
I can write this in parametric form as $x = t, y = u, z = -\frac{at + bu}{c}$.
The only problem is that I can't  use this equation when $c = 0$, and $a$, $b$ and $c$ are each going to be zero at different times so I can't avoid division by zero by shuffling around the coordinates.  Is there a way to formulate this plane parametrically without restricting values of $a$, $b$ and $c$?

REPLY [4 votes]: A solution is to parametrize the plane by $\mathbb R^3$.
This may seem a strange idea since a plane has dimension $2$ and $\mathbb R^3$ has dimension $3$, but, for the plane of equation $ax+by+cz=0$, such a parametrization is
$$
x=\gamma b-\beta c,\quad
y=\alpha c-\gamma a,\quad
z=\beta a-\alpha b,\quad
(\alpha,\beta,\gamma)\in\mathbb R^3.
$$
Then, as was to be expected, the parameter $(\alpha,\beta,\gamma)$ corresponding to a given point $(x,y,z)$ in the plane is not uniquely defined. The set of parameters corresponding to $(x,y,z)$ is exactly the affine line $L(x,y,z)$ in $\mathbb R^3$ defined as
$$
L(x,y,z)=(\alpha,\beta,\gamma)+\mathbb R\cdot(a,b,c)=\left\{(\alpha+\lambda a,\beta+\lambda b,\gamma+\lambda c)\mid\lambda\in\mathbb R\right\},
$$
where $(\alpha,\beta,\gamma)$ denotes any parameter corresponding to $(x,y,z)$. Note that $L(x,y,z)$ depends on $(x,y,z)$ but not on the choice of $(\alpha,\beta,\gamma)$.