True or false? The three points $(1,2,0), (1,1,1)$ and $(0,0,-1)$ lie in the same plane in $\mathbb{R}^{3}$

I think an equivalent question would be if these vectors are linearly dependent?

So I made this in matrix form and used sarrus rule to calculate the determinant.

$$\begin{vmatrix} 1 & 1 & 0\\ 2 & 1 & 0\\ 0 & 1 & -1 \end{vmatrix}\left.\begin{matrix} 1 & 1 & \\ 2 & 1 & \\ 0 & 1 & \end{matrix}\right|$$

And the determinant is $-1$ and thus the statement is false.

Did I do it correctly? Please tell me, I'm preparing for an exam.

  • 5
    $\begingroup$ Three (different) points always lie in a common plane. $\endgroup$
    – Martin R
    Commented Feb 28, 2017 at 14:28
  • $\begingroup$ @MartinR even stronger, we can of course say that three distinct points define a place in $\mathbb{R}^3$... OP, What is your definition of a plane? $\endgroup$ Commented Feb 28, 2017 at 14:30
  • 1
    $\begingroup$ @MartinR: they don't even need to be different. $\endgroup$
    – user65203
    Commented Feb 28, 2017 at 14:44
  • $\begingroup$ @BrevanEllefsen Being distinct is not sufficient. Consider three points on a line. They don't define a unique plane. $\endgroup$
    – Bobbie D
    Commented Feb 28, 2017 at 15:16
  • $\begingroup$ @BobbieD ah, you're right. Forgot to explicitly write that condition. OK, so three distinct and not-colinear points in $\mathbb{R}^3$ $\endgroup$ Commented Feb 28, 2017 at 15:24

1 Answer 1


The thing you did wrong is that, considering these points as vectors you add the starting point $(0,0,0)$, so you try to see that these points and the point $(0,0,0)$ is on the same plane or not. And your result shows that they are not. You may choose one point as a starting point, then you have 2 vectors, which must lie in a plane. In general, if you have more than 3 points, you may choose one of them, say A, as a starting point, then calculate the vectors starting with A and ending with each of other points, then see if these vectors belong to the same plane or not.


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