Given three points on the plane: $ A(x_1, y_1, z_1) $, $ B(x_2, y_2, z_2) $ and $ C(x_3, y_3, z_3) $.
I'm trying to obtain the equation of the plane in this format:
$ ax + by + cz + d = 0 $
I substituted given three points into the plane equation above to form this matrix equation below:
\begin{equation}
\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}
\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}
=
\begin{bmatrix} 0 \\ 0 \\ 0 \\ ? \end{bmatrix}
\end{equation}
My aim is to find the coefficients $ a $, $ b $, $ c $ and $ d $ by solving this matrix equation. However, I can't find a fourth equation to complete the equation set. Can you please write me a fourth equation to complete the set?
Note: My aim is not just finding the plane equation. My aim is to find the plane equation by this method, by means of solving a linear set of equations. I know the other more practical way of finding the plane equation, but I'm trying to find it this way on purpose. There is no reason, I just like trying different methods and playing with numbers occasionally out of interest. So, please consider this not while writing your answers and don't suggest me other methods.
Best Answer
The problem is that there isn’t a unique solution. Consider, for instance, the plane $x+y+z=1$: it can just as well be described by the equation $2x+2y+2z=2$. In short, if $d\ne 0$ you can always divide $ax+by+xz+d=0$ through by $d$ to get an equivalent equation with constant term $1$:
$$\frac{a}dx+\frac{b}dy+\frac{c}dz+1=0\;.$$
Thus, you can assume from the beginning that $d=0$ or $d=1$, depending on whether the plane passes through the origin or not. That reduces the problem to a system of three equations in three unknowns.