Sec1Sess8
A way to find if a point is in a plane region is finding if <math>det( \langle P_1,P_{new} \rangle \ \langle P_1,P_2 \rangle \ \langle P_1,P_3 \rangle) = 0</math> where <math>P_{new}</math> is the point being tested
This tests if a volume of 0 is made if the new point is connected to the plane's points
Different way to check if point is in plane:
Normal vector is one that is perpendicular to a plane (points out at a 90* angle from a plane)
if vector <math>\langle P_1, P_{new} \rangle </math> is _|_ (perpendicular) to the vector <math> \vec{N} </math> (the normal vector) than the point in question is on the plane
This tests if the new point is perpendicular to the plane.
To find if vector are perpendicular it can be looked for if their dot products are <math>0:
(\vec{P_1,P_{new}})\ .\ \vec{N} = 0</math> where "." is dot product
How can <math> \vec{N} </math> be found? The cross product of points can be taken: <math> \vec{N} = (\vec{P_1,P_2})\ \ </math> x <math>\ \ (\vec{P_1,P_3}) </math> .
It seems the cross product can be taken <math>(\vec{P_1,P_3})</math> x <math>(\vec{P_1,P_2})</math> and this also creates N
"Triple Product":
Since N can be replaced than as seen above:
<math>0 = (\vec{P_1,P_{new}})\ .\ ((\vec{P_1,P_3})</math> x <math>(\vec{P_1,P_2}))</math>
shows <math>P_{new}</math> in the plane
This is the "triple product" and is equivalent to the determinant
"equation of the plane"
use <math>P_{new} = </math><x,y,z> symbolically
"equation of the plane" = N . (<math>\vec{P_1,P_{new}}</math>) = 0 and create symbolic solution
Example:
if N = <-2,1,1> and <math>p_1</math> = <1,3,1>
N . (<math>P_1</math>,<math>P_{new}</math>) = <-2,1,1> . <x-1,y-3,z-1> = 0
-2(x-1) + (y - 3) + (z-1) = 0
-2(x) + y + z = 2
equation of the plane = -2(x) + y + z = 2
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/session-8-equations-of-planes/MIT18_02SC_we_7_comb.pdf