<s>[INST] 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:
$$ 7x - 5y - 4z = 6 $$
 [/INST] 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. :)
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