# Minimum number of numbers needed to uniquely define a plane

As title says, what would be the minimum number of real numbers needed to uniquely define any plane in $$\mathbb{R}^3$$?

What I mean by that is, if you define a plane with 3 non-colinear points, then that would count as 9 real numbers (3 real numbers per point).

The minimum (I think) I have found so far is 5 real numbers (the angle along x-axis, the angle along y-axis and a point in the plane), but I don't know what math I could use to prove that, I do not know what to search for, what terms to use (there must be some area in mathematics about that stuff, right?) and cannot find a related question to this in here.

What I would like is pointers to resources, names of related area in mathematics, etc.

The minimum number is actually three: $$ax+by+cz=d$$ defines a plane with unit normal $$(a,b,c)$$ (which can be parametrised by two angles), whereupon $$d$$ defines how far the origin is from the plane.

• Uh, true. Did not even think of that. How would one go about proving such a statement? Jan 23, 2021 at 4:04
• @JeanClaude Because the three variables can be varied independently and they give different planes, three is the minimum. Jan 23, 2021 at 5:45

Actually three numbers suffice, since you can always multiply by a constant. You can make the length of the normal $$=1$$, so you only need two other numbers. Alternatively, as long as one of the four numbers is not zero, you can divide it out.

• I'm not sure I understand. Are you saying instead of $ax + by + cz = d$ you just have $x + \frac{b}{a}y + \frac{c}{a}z = \frac{d}{a}$? Jan 23, 2021 at 5:11
• Yes, as long as $a\ne 0$. Since $a,b,c$ cannot all be zero, any non-zero component can be used as a divisor. Jan 23, 2021 at 18:01