Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Well, this is a homewrok question (which I know I should not be asking, but I cannot find an answer to this anywhere):

The exercise is as follows: i) Find the equation of the plane of $\mathbb{R}^3$ that contains the following three points: (0,0,1),(-1,3,5) and (1,-1,0) ii) Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points

So I have managed to solve (i) using a lot of Google and a third party tool (my notes are not all that good), but I cannot find (ii) anywhere. Can someone prove this providing an example and explaining it properly so that I can understand the reason this is impossible?

Thanks in advance!

P.S: This is an exercise from a course on Matrixes and how to solve Systems using matrices, in case anyone wants to know.

share|cite|improve this question
up vote 0 down vote accepted

From the phrasing of the question, it sounds like you understand that there is only one plane through those three points in part i). Knowing that, then we can say this:

Hint: If you find a point $P$ that does not satisfy the equation you found in i), then that point, along with the original three points, is an example of a set of four points that doesn't fit in a single plane.

share|cite|improve this answer
You would also need to show that there is only one plane containing the three points that you were given. – Kris Williams May 14 '13 at 13:35
@KrisWilliams Already in the answer content. If he found "the" plane, I guess he realizes there's only one. – rschwieb May 14 '13 at 13:36
Well, I understand that there is only one plane that contains the three points. So to give an example I shall just write a set of points that is not part of the plane? :O – Angelos Chalaris May 14 '13 at 13:36
@Spiritios No, because you might accidentally pick four points that lie in a different plane. While you have these three points in a plane you are already familiar with, and you know all the points that lie in that plane and that lie outside of it, you should pick a fourth point that is guaranteed to be outside of the plane. You will be able to argue "these four points are not contained in a plane" I hope to spur you to fill out that argument, though :P – rschwieb May 14 '13 at 13:39
Yeah, that's what I meant to say, wirte the original three points plus a point that does not work with the equation. Gotcha! ;) – Angelos Chalaris May 14 '13 at 13:40

Let $ax+by+cz=d$ be the equation of a plane (where $\{a,b,c\}\not=\{0\})$. Show that $(0,0,0),(0,0,1),(0,1,0),(0,0,1)$ cant satisfy the equation $ax+by+cz=d$ simultaneously.

share|cite|improve this answer
This is actually a good way to prove it, I think! And a vey simple example, probably was overthinking it... – Angelos Chalaris May 14 '13 at 13:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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