Proof: Determinant of 3 Non-Colinear Points is 0

[Dethrone]

View attachment 4291

I understand the intuition behind it, but I'm unable to prove it. Essentially, three non-colinear points define a plane, and so by adding one more point on the plane, it becomes dependent. This means that the determinant is 0, since there is probably a dependent row lying around. Also, how is that determinant even set up in the first place?

Any ideas?

 

[Euge]

Hi Rido12,

Here are some things to keep in mind. Let $x,y,z$ range over $\Bbb R$, so we may consider the determinantal condition as an equation in $\Bbb R^3$. By expansion of the determinant along the first row, you can see that it gives the equation of a plane. Note that if you substitute any of the points $(x_i,y_i,z_i)$ for $(x,y,z)$, the determinant is zero. Therefore, all three points lie in the plane. On the other hand, let $Ax + By + Cz + D = 0$ be the equation of plane containing the points $(x_i,y_i,z_i)$. If $(x_0,y_0,z_0)$ lies on this plane, then $\mathbf{X}\mathbf{v} = \mathbf{0}$, where $\mathbf{X}$ is the matrix

$$\begin{bmatrix}x_0&y_0&z_0&1\\
x_1&y_1&z_1&1\\
x_2&y_2&z_2&1\\
x_3&y_3&z_3&1\\
\end{bmatrix}$$

and $\mathbf{v}$ is the matrix

$$\begin{bmatrix}A\\B\\C\\D\end{bmatrix}$$

Since $\mathbf{v} \neq \mathbf{0}$, it is a nontrivial solution of the homogenous system $\mathbf{X}\mathbf{u} = \mathbf{0}$, and hence $\operatorname{det}(X) = 0$.

 

[Dethrone]

Hi Euge!

That all makes sense, but how can we show that $\mathbf{v}$ is a non-trivial solution? I guess that would be equivalent to showing it is not invertible, or has dependent rows/columns?

 

[Euge]

Consider that the coefficients $A, B, C$ form a vector normal to the plane. The unit normal to the plane exists because the vector from $(x_1, y_1, z_1)$ to $(x_2, y_2, z_2)$ is linearly independent from the vector from $(x_1, y_1, z_1)$ to $(x_3, y_3, z_3)$.

 

[Dethrone]

That makes sense, but I'm thinking I'm not grasping some other aspects related to this question. Suppose we only had three points and wanted to determine the coefficients of the plane Ax+By+Cx+D=0 that passed through those three points. That is equivalent to solving $Xv=0$, where $X$ is:

$\begin{bmatrix}
x_1&y_1&z_1&1\\
x_2&y_2&z_2&1\\
x_3&y_3&z_3&1\\
\end{bmatrix}$

and $v$:

$\begin{bmatrix}A\\B\\C\\D\end{bmatrix}$

But seeing as the system is independent, then only the trivial solution exists ($A=B=C=D=0$). Shouldn't there also be a unique solution (i.e $A=4,B=-1,C=5,D=-4 \implies 4x-1y+5z=4$ whereby $(x_i,y_i,z_i)$ are satisfied for all three points?)

 

[Euge]

The coefficients $ A, B, C, D $ were constant from the start. The goal is not to find the formula for the coefficients. The point is $(A, B, C)$ is nonzero, and therefore $(A, B, C, D)$ is nonzero.