Hp 50g Math question: Intersection of 2 triangulars

[peacecalc]

Hello Han,

thank you for lucid answer!

Quote: The most difficult step is to determine the position of a point the the plane H relative to triangle one.

To determine their positions relative to triangle one, you can use a linear system. Since we already have the three vertices from triangle one being coplanar, any other point in H must be a linear combination of the vertices of triangle one. The solution can even be normalized. That is, if A is the matrix whose entries are the coordinates of the vertices (in column form) of triangle one, and b⃗ is a point on H, then the solution x⃗ =λ⃗ to A⋅x⃗ =b⃗ can also be made to satisfy
∑j=13λj=1

where λj is the j-th component of λ⃗ . In a sense, λj is a weight on the j-th vertex of triangle one and is a measure of how close b⃗ is to that vertex. Moreover, the sign of λj determines the relative position of b⃗ to vertex j. (Look up Radon partitions; there are also lots of theorems regarding d+2 points in a d-dimensional space.) That is, if λj is positive, then b⃗ and vertex j are on the same side of the line passing through the remaining two vertices of triangle one. Thus, if λj are all positive, then b⃗ lies in the relative interior of triangle one. If λj=0 for a specific value of j then b⃗ actually lies one an edge of triangle one.

That is maybe the solution to avoid a lot of "IF THEN" structures, which often slow down the execution speed of programs.