Forming equation of a plane by solving linear equation set

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:

$$\begin{bmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ ? & ? & ? & ? \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ ? \end{bmatrix}$$

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.

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If you really want to solve a system of four equations (which isn't really necessary, as Brian Scott pointed out), then the fourth one that you're missing could be almost anything. For example $a+b+c = 1$ would work. The only purpose of this fourth equation is to remove the scaling indeterminacy in $a,b,c,d$ that was explained in Brian's answer.

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I got it. I couldn't understand Brian's message then, but know you made it very clear to me. Thank you. –  hkBattousai Aug 19 '12 at 13:40

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 $ax+by+xz+d=0$ 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.

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To get an equation of the plane you're interested in ...

(1) Put $x$, $y$, $z$, and 1 in place of your four ? symbols

(2) Take the determinant of the resulting $4 \times 4$ matrix and set it equal to 0.

You were sooooo close :-)

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Very interesting. Why should the determinant be equal to 1? –  hkBattousai Jun 3 '12 at 15:07
If any of the three points (A, B, or C) is substituted for the fourth row of the matrix, then there will be two equal rows, so the detrminant will be zero. This shows that the three points A, B, C satisfy the equation. The equation is linear in x, y, z, so it must be a plane. –  bubba Jun 4 '12 at 2:47
OK, I understand the fact that determinant of a matrix will be zero if all rows of it are not linearly independent. But, what you said in your answer was a different thing; you claimed that the determinant must be equal to 1. I am asking for a proof or explanation for this. –  hkBattousai Jun 4 '12 at 6:45
Sorry. My mistake. I meant =0, not =1. I corrected the mistake. –  bubba Jun 5 '12 at 1:02
Then, what is the purpose of making the determinant zero? I'm trying to find a fourth equation which is linearly independent of the other three. –  hkBattousai Jun 5 '12 at 10:09

In the equation $ax + by + cz + d = 0$, the vector $(a,b,c)^T$ is the normal to the plane. The parameter d defines the distance from the origin and may be determined given a point on the plane.

The plane equation may be written as follows: $$\vec{n}^T(\vec{x} - \vec{p_0}) = 0$$

The point $\vec{p_0}$ is any point on the plane (you can choose any one of the three points you have). Essentially the equation says that the dot product of the normal with any vector on the plane is zero (if $\vec{x}$ is a point on the plane, then $\vec{x} - \vec{p_0}$ is a vector parallel to the plane).

So the problem is now to find the normal to the plane given the three points you have. Then determine $d$ using one of the points. The solution is simple: Use the three points to derive two vectors on the plane ($\vec{v_1} = \vec{B} - \vec{A}$ and $\vec{v_2} = \vec{C} - \vec{A}$). Now compute the cross product of these two vectors. The cross product is perpendicular to the two vectors and therefore the normal to the plane.

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