# Equation of a plane through 3 points in 3D space (small clarification)

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?

• It might help to think of $|A|=0$ as saying that the rows of $A$ are not linearly independent vectors.
– Karl
Sep 16, 2022 at 16:50
• @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? Sep 16, 2022 at 19:25
• 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.
– Karl
Sep 16, 2022 at 20:17
• @Karl I also think so, the point is that I want to be sure :) That's why I asked the question. Sep 17, 2022 at 11:00

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.
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 $$x,y,z$$)
• 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). Sep 17, 2022 at 12:14
• @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 Sep 17, 2022 at 13:37