Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find the plane through the points $(1,1,-1)$, $(2,0,2)$ and $(0,-2,1)$.
Answer:
The general form of a plane is:
$$ Ax + By + Cz = D$$
Sometimes the following constrain is added:
$$ A^2 + B^2 + C^2 = 1$$
By inspection, we can see this plane is not parallel to the x-axis, the y-axis or the z-axis. Hence,
we can assume that the plane is of the form:
$$ Ax + By + Cz = 1 $$
Now we setup the following system of linear equations.
\begin{align*}
A + B - C &= 1 \\
2A + 2C &= 1 \\
-2B + C &= 1 \\
\end{align*}
To solve this system of equations, we get rid of $A$ and $B$ in the first equation.
\begin{align*}
2A &= 1 - 2C \\
A &= \frac{ 1 - 2C }{2} \\
-2B &= 1 - C \\
B &= \frac{ C - 1 }{2} \\
\left( \frac{ 1 - 2C }{2} \right) + \left( \frac{ C - 1 }{2} \right)  - C &= 1 \\
1 - 2C + C - 1 - 2C &= 2 \\
- 2C + C - 2C &= 2 \\
-3C &= 2 \\
C &= -\frac{2}{3} \\
B &= \frac{ -\frac{2}{3} - 1 }{2} = -\frac{2}{6} - \frac{1}{2} \\
B &= -\frac{5}{6} \\
A &= \frac{ 1 - 2\left(  -\frac{2}{3} \right)  }{2} = \dfrac{1 + \dfrac{4}{3} }{2} \\
A &= \dfrac{7}{6}
\end{align*}
Hence the equation is:
$$ \left( \dfrac{7}{6} \right) A + \left( -\frac{5}{6} \right) B + \left(  -\frac{2}{3} \right) C = 1  $$
Clearing the fraction, we get the final answer of:
$$ 7A - 5B - 4C = 6 $$
As pointed out by Paul, the correct answer is:
$$ 7x - 5y - 4z = 6 $$
 A: Yes, your method is perfectly fine, and nicely laid out, though at the very end in the original $x,y,z$ were accidentally replaced by $A,B,C$. :)
In terms of numerical computation, this is a reasonably efficient algorithm.
In terms of formulaic or abstract presentation, (conceivably what the book did), we realize that to describe a plane in 3D, we need a "normal" (=perpendicular) vector $N$ and a point $P$ on the plane. Then the plane is the set of points $X=(x,y,z)$ such that $(X-P)\cdot N=0$, where dot denotes vector "dot product" (="inner product"="scalar product").
A formulaic/conceptual trick (that hides necessary computations) is that the vector cross product $v\times w$ is orthogonal to both $v$ and $w$. So, given three points $P,Q,R$, the vectors $P-Q$ and $P-R$ (for example) are in the plane, so their cross product is orthogonal to it. So an equation for points $X=(x,y,z)$ to be in the plane is $(X-P)\cdot ((P-Q)\times(P-R))=0$.
It should probably be noted that the computation needed for the more symbolic approach is really roughly the same as the "more direct" approach: a three-by-three system of linear equations is to be solved, and Cramer's rule, with expansion-by-minors, is in-effect carried out by those vector operations. :)
A: A plane can be defined by its normal $n$ and a point $p$ of the plane. The equation of the plane  is, for any point $x$ in the plane, $x\cdot n=d$, where $d=n\cdot p$.
If we have three points in the plane, $a,b,c$ then $a.n=b.n=c.n=d$. So we can solve for $n$ by forming a matrix $$Q=\begin{pmatrix}a&b&c\end{pmatrix}^T$$
and the three equations $$Q n=\begin{pmatrix}d&d&d\end{pmatrix}^T$$ must all be true if $a,b,$ and $c$ are in the plane, then we can get $n$ by inverting the matrix, $n=A^{-1}\begin{pmatrix}d&d&d\end{pmatrix}^T$.
In the terminology of your answer, $n\equiv{1\over d}(A,B,C)^T$, so what you did was to find the normal to the plane by inverting the above matrix.
This method works except for the case that the plane goes through the origin, where $d=0$ and we cannot use the inverse.
A: You can also use standard solution which is the equation of a plane passing through a given point and a line. Suppose a plane passes through point M(x_0, y_0, z_0) and a line with following equation:
$\frac {x-x_1}{l}=\frac {y-y_1}{m}=\frac {z-z_1}{n}$
Then the equation of plane is:
$\begin {vmatrix}x-x_0&y-y_0& z-z_0\\x_1-x_0&y_1-y_0&z_1-z_0\\l&m&n\end {vmatrix}=0$
Now take for example points A an B; the components of line AB are:
$l=x_B-x_A=2-1=1$
$m=y_B-y_A=0-1=-1$
$n=z_B-z_A=2+1=3$
Take for example point $A (x_1=1, y_1=1, z_1=-1)$ and point $C(x_0=0, y_0=-2, z_0=1)$ and construct the determinant:
$\begin {vmatrix}x-0&y+2& z-1\\1-0&1+2&-1-1\\1&-1&3\end {vmatrix}=0$
Now solve this equation you finally get:
$$7x-5y-4z=6$$
