Given three rays and a right triangle, draw the pyramid having those rays as edges and that triangle as base

Question

I am looking for suggestions for solving the following geometrical problem algebraically.

I have three rays in the 3D space emanating from an origin $O$, say $\alpha$, $\beta$, $\gamma$. I wanted to find a plane $\pi$ intersecting these three lines in the points $A$, $B$, $C$ respectively, so that the triangle $ABC$ is right, say in $B$, and that its sides have known lengths $\overline{AB}$ and $\overline{BC}$. Obviously, $\overline{AC}^2$ = $\overline{AB}^2$ + $\overline{BC}^2$.

Basically, given three rays out of an origin and a right triangle, the plane $\pi$ draws a pyramid having those rays as the edges and that right triangle as the base. It does not matter whether the pyramid is acute or obtuse.

I also understand that the problem may have no solution in particular circumstances. When a solution exists, though, I would like to determine the position of the intersection points $A$, $B$ and $C$ between thosee rays $\alpha, \beta, \gamma$ and that plane $\pi$.

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I cast the problem using these symbols: the pyramid edges are the vectors $\vec{a}=OA$, $\vec{b}=OB$, $\vec{c}=OC$; and their unknown lengths are $a = |\vec{a}|, b = |\vec{b}|, c = |\vec{c}|$.

I use the law of the cosines to express the fact that the faces of the pyramid are generic triangles:

$ a^2 - 2 \cos \hat{\alpha\beta}\, ab + b^2 = \overline{AB}^2$
$ b^2 - 2 \cos \hat{\beta\gamma}\, bc + c^2 = \overline{BC}^2$
$ c^2 - 2 \cos \hat{\gamma\alpha}\, ca + a^2 = \overline{CA}^2$

and the Pythagorean theorem to state that the base of the pyramid is a right triangle:

$\overline{AB}^2 + \overline{BC}^2 = \overline{AC}^2 $

I have three nonlinear equations in the three variables $a, b, c$ and one condition. I lack a strategy to solve this system and obtain closed-form formulas for the solution.

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My questions are

• As a sanity check, does the formulation express the problem well in the first place?

• How can I then solve for $a,b,c$ in the system above?

Answer 1

The nonlinear system of equations can be solved successfully and easily with the Newton-Raphson multivariate method. The vector function is

$ f(a,b,c) = \begin{bmatrix} a^2 + b^2 - 2 a b \cos t_1 - X^2 \\ a^2 + c^2 - 2 a c \cos t_2 - Y^2 \\ b^2 + c^2 - 2 b c \cos t_3 - (X^2 + Y^2) \end{bmatrix}$

Where $t_1$ is the angle between $v_1$ and $v_2$ and $t_2$ is the angle between $v_1$ and $v_3$ and $t_3$ is the angle between $v_2$ and $v_3$. $X$ is the length of the first leg of the right triangle and $Y$ is the length of the second leg.

Newton-Raphson method uses the Jacobian matrix of the vector function $f$. The ${ij}$-th entry of this matrix is the partial derivative of $i$-th function with respect to the $j$-th variable. Thus the Jacobian is given by

$J(a,b,c) = \begin{bmatrix} 2 a - 2 b \cos t_1 && 2 b - 2 a \cos t_1 && 0 \\ 2a - 2 c \cos t_2 && 0 && 2 c - 2 a \cos t_2 \\ 0 && 2 b - 2 c \cos t_3 && 2 c - 2 b \cos t_3 \end{bmatrix}$

Start with a good guess of the solution $x_0 = (a,b,c)$, then update your guess with the Newton-Raphson recursive update formula

$ x_{k+1} = x_k - J^{-1} f(x_k) $ , $k = 0, 1, 2, ... $

This method is very robust and will lead to a solution if it exists within $4-5$ iterations for this small number of variables (3 variables in this problem).

The only catch is you need to have a routine to find the inverse of a matrix (because you need to calculate $J^{-1}$).

Answer 2

Let $\vec{OA}=\vec a=\lambda u$, $\vec{OB}=\vec b=\vec v$ and $\vec{OC}=\vec c=\mu\vec w$ representing the three rays. Any other solution may be achieved by scaling.

Please note that there's not always a solution, e.g., if $\vec u$, $\vec v$, and $\vec w$ are pairwise orthogonal. And you can't choose the lengths of $AB$ and $BC$ independently.

For a right angle at $B$ we must have $$\langle\vec{BA},\vec{BC}\rangle= \langle\lambda\vec u-v,\mu\vec w-u\rangle=0.$$ Solving for $\mu$ gives $$ \mu=\frac{\lambda\langle u,v\rangle-\langle\vec v,\vec v\rangle}{\lambda\langle u,v\rangle-\langle\vec v,\vec w\rangle}. $$ Now if for example $$ \vec u=\begin{pmatrix}1\\1\\0\end{pmatrix}, \vec v=\begin{pmatrix}1\\0\\1\end{pmatrix}, \vec w=\begin{pmatrix}0\\1\\1\end{pmatrix} $$ and $\lambda=3$ you'll get $\mu=1/2$ and $$ \vec{BA}=\begin{pmatrix}3\\2\\-1\end{pmatrix}, \vec{BC}=\begin{pmatrix}-1\\1/2\\-1/2\end{pmatrix}. $$ If the length of $BA$ is $1$, let $\vec u$, $\vec v$, and $\vec w$ be unit vectors. As $\vec{BA}=\lambda\vec u -\vec v$ we get $$ 1=\|\vec{BA}\|^2=\|\lambda\vec u-\vec v\| =\lambda^2-2\lambda\langle\vec u,\vec v\rangle+1$$ and from here $$\lambda=\frac{\langle\vec u,\vec v\rangle}{2}.$$ Now compute $\mu$ and finally $\vec{OC}$. If $AB$ has a length different from $1$, scale accordingly.