2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Uh, true. Did not even think of that. How would one go about proving such a statement? $\endgroup$ – Jean-Claude Jan 23 at 4:04
  • $\begingroup$ @JeanClaude Because the three variables can be varied independently and they give different planes, three is the minimum. $\endgroup$ – Parcly Taxel Jan 23 at 5:45
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I'm not sure I understand. Are you saying instead of $ax + by + cz = d$ you just have $x + \frac{b}{a}y + \frac{c}{a}z = \frac{d}{a}$? $\endgroup$ – Jean-Claude Jan 23 at 5:11
  • $\begingroup$ 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. $\endgroup$ – herb steinberg Jan 23 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.