Tetrahedron triangular section [closed]

Let us define the tetrahedron ABCD with vertices in A,B,C,D and length 1. Any point P inside it can be expressed as (a,b,c,d) where each letter represent the contribution of its corresponding vertice (Therefore a+b+c+d=1). If we cut the tetrahedron as it is shown in the previous image, it results in a triangle section with vertices in A_iB_j (random point between A and B), C and D.

I have so far the conversion to a XYZ coordinate system (B origin, BC the X axis), where $$x= \frac{c+1-b}2$$ $$y= \frac{\sqrt{3}}2a+\frac{\sqrt{3}}6d$$ $$z=\frac{\sqrt{6}}3d$$

If we define a new X'Y'Z' coordinate where A_iB_j is the origin and A_iB_j C the X' axis, my question is how to transform from one coordinate system to another. In addition to translate the x'y'z'coordinates of P from the second reference system to the triangle vertices (A_iB_j, C, D) in a similar way it was represented in the tetrahedron at the beginning of the problem.

Solution 1:

The point $P = a A + b B + c C + d D$ where $a + b + c + d=1$

You want to find the intersection of the plane $PCD$ with the segment $AB$.

Points in plane $PCD$ are spanned by two vectors: $CD$ and $CP$, and can therefore be expressed as follows:

$Q = C + \alpha CD + \beta CP $

Plugging in $CD = D - C $ and $CP = P - C$ , we get,

$Q = C + \alpha (D - C) + \beta (a A + b B + (c - 1) C + d D ) $

Since points on the segment $AB$ are of the form $ (1 - t) A + t B $ the coefficient of $C$ and the coefficient of $D$ in the above expression for $Q$ must be both zero.

Hence,

$ 1 - \alpha + \beta (c - 1) = 0 \hspace{20pt} (1) $

$ \alpha + \beta d = 0 \hspace{20pt} (2) $

Now, recall that $d = 1 - a -b - c $, therefore, the second equation becomes,

$ \alpha + \beta (1 - a - b - c ) = 0 \hspace{20pt} (3)$

Adding $(1), (3)$ we get,

$1 - \beta (a + b) = 0$

Hence, $\beta = \dfrac{1}{a + b} $

Thus the point of intersection (that you call $A_i B_j$), let's call it $E$ is given by

$E = \dfrac{a}{a + b} A + \dfrac{b}{a + b} B $

Now you want to express the point $P$ in terms of $C, D, E$

Since $P = a A + b B + c C + d D = c_1 E + c_2 C + c_3 D $

Then we have the following system of equations relating $c_1, c_2, c_3 $ to $a, b, c$:

$ \dfrac{c_1 a }{a + b} = a $

$ \dfrac{c_1 b } { a + b} = b $

$ c_2 = c$

$ c_3 = d$

The solution of this system is that $c_1 = a + b , c_2 = c , c_3 = d $

Note that $c_1 + c_2 + c_3 = 1 $ as it should be.