This horribly unreadable webpage describes a method to find the Cartesian equation for a plane given its parametric equations. I'll try to type the method out here in a neater fashion:
The parametric equations for a plane are given, for example:
\begin{cases}
x = 1 + 2r + 3s \\
y = 4 + 5r + 6s \\
z = 8 + 8r + 9s
\end{cases}We want to find an equivalent condition for that a point $(x, y, z)$ is on this plane, in the form of a Cartesian equation $ax+by+cz+d=0$. This is precisely when the above system of equations has a solution in $r$ and $s$. We can write this in matrix form:
$$
\begin{bmatrix}2&3 \\ 5&6 \\8&9\end{bmatrix}
\begin{bmatrix}r \\ s \end{bmatrix}
=\begin{bmatrix}x-1 \\ y-4 \\ z-8 \end{bmatrix}
$$Missing step here. The webpage says:
Since the direction vectors $\mathbf{AB}$ and $\mathbf{AC}$ give a different direction, the rank of the matrix of coefficients is 2.
The system has a solution for $r$ and $s$ if and only if the characteristic determinant is zero.Apparently, this determinant is:
$$
\begin{vmatrix}2&3&x-1\\5&6&y-4\\8&9&z-8\end{vmatrix}=0
$$Which simplifies to
$$-3x +6y -3z +3 = 0$$
This is the correct Cartesian equation for the plane.
Why does this work? What "missing step" is happening – why does looking at this determinant immediately yield the solution? The webpage I linked describes this as the characteristic determinant, which is a term that's used this way nowhere else on the Internet, as far as I can tell.
Note: All the other sources on this mention that you can take the cross product of the two known vectors $(2, 5, 8)$ and $(3, 6, 9)$ to find a normal vector to the plane, giving you $(a,b,c)$, and then you can easily find $d$. I already understand this approach. I just want to know why this one works.
Best Answer
Think about it in terms of vectors. The vector equation of the plane is $$ (x,y,z) = (1,4,8) + r(2,5,8) + s(3,6,9). $$ In other words, the plane passes through $(1,4,8)$, and is spanned by $(2,5,8)$ and $(3,6,9)$. Therefore, $$ (x,y,z)-(1,4,8) = (x-1,y-4,z-8) = r(2,5,8) + s(3,6,9), $$ so it is a linear combination of $(2,5,8)$ and $(3,6,9)$. Or if you prefer, the parallelepiped with edges given by $(x-1,y-4,z-8)$, $(2,5,8)$ and $(3,6,9)$ has zero volume. Either one of these means that the determinant $$\begin{vmatrix}2&3&x-1\\5&6&y-4\\8&9&z-8\end{vmatrix}$$ vanishes (the vectors don't form a basis of $\mathbb{R}^3$ if and only if they are all in the same plane, so this also means that you can't always solve $$ a(x-1,y-4,z-8) + b(2,5,8) + c(3,6,9) = (e,f,g), $$ or $$ \begin{pmatrix}2&3&x-1\\5&6&y-4\\8&9&z-8\end{pmatrix} (b,c,a) = (e,f,g), $$ so the matrix on the left must be singular.
Hopefully at least one of those explanations makes sense to you.
The idea is that the determinant is alternating multilinear: what this actually means is that if one column is made of linear combinations of the other columns, the determinant must be zero.
More explicitly, think of it as a map $\det$ that takes $n$ vectors in $\mathbb{R}^n$ and spits out a real number. It is also linear in each variable (vector), and the sign of this real number is changed if two of the vectors swap places: e.g. in $\mathbb{R}^3$, $$ \det{(\mathbf{a},\mathbf{b},\mathbf{c})} = -\det{(\mathbf{b},\mathbf{a},\mathbf{c})} $$ and so on. This means firstly that $$ \det{(\mathbf{a},\mathbf{a},\mathbf{c})} = -\det{(\mathbf{a},\mathbf{a},\mathbf{c})} $$ (swapping $\mathbf{a}$ with $\mathbf{a}$), and so $$ \det{(\mathbf{a},\mathbf{a},\mathbf{c})} = 0 $$ for any $\mathbf{a}$ and $\mathbf{c}$. Secondly, the linearity gives $$ \det{(r\mathbf{b}+s\mathbf{c},\mathbf{b},\mathbf{c})} = r\det{(\mathbf{b},\mathbf{b},\mathbf{c})}+s\det{(\mathbf{c},\mathbf{b},\mathbf{c})} = 0, $$ which is why if one column is a linear combination of the others, the determinant is zero.