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Let's assume we're given 3 points in the 3-dimensional space which are not collinear and which have the following coordinates in some coordinate system A $ = \{O,\overrightarrow{e_1},\overrightarrow{e_2},\overrightarrow{e_3}\}$ (of the 3-d space).

$P_1 = (x_1, y_1, z_1)$

$P_2 = (x_2, y_2, z_2)$

$P_3 = (x_3, y_3, z_3)$

Then (I recall that) the equation of the plane passing through these points is as follows.

$$\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix} = 0$$

This here is a 3x3 determinant. But for this to hold true, the coordinate system A which we have, should it be orthogonal or not?

I think not, but I am not quite sure. I mean, I think even if the system A is not orthogonal, this is still the equation of the unique plane passing through the 3 given points.

Could someone more knowledgeable confirm if this is true?

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  • $\begingroup$ It might help to think of $|A|=0$ as saying that the rows of $A$ are not linearly independent vectors. $\endgroup$
    – Karl
    Sep 16, 2022 at 16:50
  • $\begingroup$ @Karl Yes, that's exactly how I remember this equation for myself. The question is: does A need to be orthogonal, or it can be any coordinate system in the 3-d space? $\endgroup$ Sep 16, 2022 at 19:25
  • $\begingroup$ Right. The equation is certainly satisfied at your three points, and any full-rank linear transformation (such as a change of coordinates, orthogonal or not) takes a plane to a plane, so I think the answer is no. $\endgroup$
    – Karl
    Sep 16, 2022 at 20:17
  • $\begingroup$ @Karl I also think so, the point is that I want to be sure :) That's why I asked the question. $\endgroup$ Sep 17, 2022 at 11:00

2 Answers 2

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It doesn't need to be orthogonal.

If the coordinates are not orthogonal, then you can transform them to any orthogonal system by multiplying by some non-singular $3\times 3$ matrix $T$. By linearity, $T(a-b)=Ta-Tb$ so differences between vector coordinates are transformed by the same matrix as the vector coordinates themselves. And determinants are multiplicative: $|TM|=|T||M|$, so given $|T|\neq 0$ (because the transformation was stated to be non-singular), we find $|TM|=0$ if and only if $|M|=0$. If the determinant is zero in a non-orthogonal coordinate system, then it must be zero in any orthogonal one too.

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In Affine Geometry the equation of a plane containing three independent points$A,B,C$ is given by the fact that a fourth point $P (x,y,z)$ shall be such that $$ \mathop {AP}\limits^ \to = \lambda \mathop {AB}\limits^ \to + \mu \mathop {AC}\limits^ \to \quad \Rightarrow \quad \left| {\begin{array}{*{20}c} {x - x_A } & {x_B - x_A } & {x_C - x_A } \\ {y - y_A } & {y_B - y_A } & {y_C - y_A } \\ {z - z_A } & {z_B - z_A } & {z_C - z_A } \\ \end{array}} \right| = \left| {\begin{array}{*{20}c} x & {x_A } & {x_B } & {x_C } \\ y & {y_A } & {y_B } & {y_C } \\ z & {z_A } & {z_B } & {z_C } \\ 1 & 1 & 1 & 1 \\ \end{array}} \right| = 0 $$

(your version of the determinant will be a cubic in $x,y,z$)

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  • $\begingroup$ Oh, cubic. Right. Yes, I understand. Btw, I think you should write an equivalence sign there instead of $\Rightarrow$ (because it's if and only if). $\endgroup$ Sep 17, 2022 at 12:14
  • $\begingroup$ I corrected my question. The real question is: does the coordinate system need to be orthogonal or it can be any affine coordinate system? I think you're saying it does not need to be orthogonal. $\endgroup$ Sep 17, 2022 at 12:16
  • $\begingroup$ @peter.petrov: actually the two determinants are equal (apart from the sign I did not check, maybe the A column shall be put at the end) $\endgroup$
    – G Cab
    Sep 17, 2022 at 13:31
  • $\begingroup$ @peter.petrov: $A,B,C$ just need to be independent (i.e. not aligned) and if they are, they are so in an orthogonal as well as non-orthogonal system $\endgroup$
    – G Cab
    Sep 17, 2022 at 13:37

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