How many parameters does an equation of a plane have?

Question

The traditional way I learned to write the equation of a plane is:

$$Ax + By + Cz + D = 0$$ where $\langle A,B,C\rangle$ is a normal vector to the plane and $D$ is the dot product between the normal and another point on the plane.

However, with some simple rearrangements, we can make the equation for the plane look like this:

$$-\frac{A}{C}x - \frac{B}{C}y - \frac{D}{C} = z = a'x + b'y + c'$$

As you can see this equation seems to only have 3 parameters, $a', b', c'$. And in terms of the physical intuition for what they mean, the new normal to the plane is $\langle a', b', -1\rangle$ and $c'$ is the dot product between the new normal and another point on the plane.

Why is the 4 parameter formulation of the equation of a plane more popular? Are there any problems with the 3 parameter formulation?

In fact, one cool benefit I see from the 3 parameter formulation is the ability to fit a plane to a cloud of points using a system of equations (not sure I see a way of doing it using the 4 parameter way):

$$ \begin{bmatrix} x_0 & y_0 & 1 \\ x_1 & y_1 & 1 \\ &... \\ x_n & y_n & 1 \\ \end{bmatrix} \begin{bmatrix} a' \\ b' \\ c' \end{bmatrix} = \begin{bmatrix} z_0 \\ z_1 \\ ... \\ z_n \end{bmatrix} $$

Answer 1

Any plane where $C=0$, i.e. perpendicular to the $xy$-plane, cannot be converted into the form $z=ax+by+d$ because of a divide-by-zero. But then again there is no inherent reason to favour normalising any of the coefficients to $1$, so to avoid having to grapple with divisions by zero (and to preserve symmetry to enable simpler calculations) the plane equation is left with four parameters.

Answer 2

Any plane in $\Bbb R^3$ can be written in the form $x=d$ or $ax+y=d$ or $ax+by+z=d$. So with this interpretation, the parameter set of all planes in $\Bbb R^3$ is $(\Bbb R^0\cup\Bbb R^1\cup\Bbb R^2)\times\Bbb R$.

Can we give more structure to the parameter set? The Grassmann manifold of $2$-planes in $\Bbb R^3$ denoted by $G_2(\Bbb R^3)$ is the (parameter) space of the planes in $\Bbb R^3$ which pass through the origin. By translating each element of $G_2(\Bbb R^3)$ along its normal vector, we can obtain all planes in $\Bbb R^3$. Therefore, the parameter space of all planes is $G_2(\Bbb R^3)\times\Bbb R^1$. We know that $G_2(\Bbb R^3)\cong G_1(\Bbb R^3)=\Bbb R P^2$, the real projective plane and as a set it is the disjoint union $\Bbb R^0\cup\Bbb R^1\cup\Bbb R^2$. Hence, we reach the same parameter set with a topological meaning.