Triangle with two constraints, each corner on a given line

Question

Given:
- 3 3-dimensional straight lines: a, b, c. All lines are coninciding in a single point S.
- Point B on b.

I'm now looking for a point A on a and a point C on c, where AB and BC have the same length and the angle ABC is 90 degrees.

I think this should be solvable with the law of sines. I know the angle BSC and distance SB and should be able to set up 2 equations with 2 unknowns: Length BC and the angle SBC - if I can express the angle SBA with SBC and the 90 degrees of ABC. I'm not sure how to express this angle in this (3D) case, though.

Thanks a lot for any help,
Patrick

Answer 1

Instead of trigonometry I think a simpler approach is using parametric equations and dot products for orthogonality and distance constraint:

Let's place $S$ in the origin, without losing any generality.

The line $A$ can be defined as $A(s)=sa$ where $s$ is a real number and $a$ is a vector with $|a|=1$.

In analogy let's define $C$ to be $C(t)=tc$ with $t$ real and $|c|=1$.

The requirement of $A(s)$ and $C(t)$ being at the same distance from $B$ can be expressed as

\begin{equation} |A(s)-B|^2 = |C(t)-B|^2 \end{equation}

\begin{equation} (sa-B)\cdot(sa-B) = (tc-B)\cdot(tc-B) \end{equation}

\begin{equation} s^2|a|^2 + |B|^2 - 2s(a \cdot B) = t^2|c|^2 + |B|^2 - 2t(c \cdot B) \end{equation}

\begin{equation} s^2 - 2s(a \cdot B) = t^2 - 2t(c \cdot B) \end{equation}

The second requirement of perpendicularity can be expressed as

\begin{equation} (A(s) - B) \cdot (C(t) - B) = 0 \end{equation}

\begin{equation} (sa - B) \cdot (tc - B) = 0 \end{equation}

\begin{equation} st(a \cdot c) + |B|^2 -s(a\cdot B) -t(c\cdot B) = 0 \end{equation}

Solving this last equation for $s$ gives

\begin{equation} s = {t(c \cdot B) - |B|^2 \over t(a \cdot c) - (a \cdot B)} \end{equation}

Substituting this expression $s=s(t)$ into the equidistance condition finally gives the equation:

\begin{equation} \left({t(c \cdot B) - |B|^2 \over t(a \cdot c) - (a \cdot B)}\right)^2 - 2 \left({t(c \cdot B) - |B|^2 \over t(a \cdot c) - (a \cdot B)}\right)(a\cdot B) = t^2 - 2t(c \cdot B) \end{equation}

This is a 4th degree equation in $t$ than when solved provides the solutions. Experimentally I've been able to build cases in which there are 0, 1, 2, 3 or 4 real solutions and therefore I suppose that a closed formula is not going to be really simple.

For a case in which there are 4 distinct solutions consider for example:

a = (0.7832613199545745, 0.3457501646810926, -0.5166802960110005)
b = (0.22437668770758193, -0.7218976823125938, 0.6546135029810183)
c = (-0.4162213502972967, -0.02495946801502962, -0.9089206854908198)
B = (10.598100181328936, -34.09776673318595, 30.91969827848149)

The solutions are (approximately):

s = -324.127866359 , t = -337.34963584
    A = (-253.87682043855557, -112.06726317141957, 167.47048193587798)
    C = (140.41212095172617, 8.42006744563637, 306.62406225798975)
    |A - B| = 307.688710435
    |C - B| = 307.688710435
    (A - B)·(C - B) = -2.18278728426e-11

s = 16.2368805029 , t = -75.2329758241
    A = (12.717720454640398, 5.613904107782358, -8.389276224529821)
    C = (31.313570784400618, 1.8777750537577107, 68.38080795757674)
    |A - B| = 55.9169479457
    |C - B| = 55.9169479457
    (A - B)·(C - B) = 6.8212102633e-13

s = -31.0680366312 , t = -59.2027780354
    A = (-24.334391380153313, -10.741778781556734, 16.05224236309069)
    C = (24.641460215247356, 1.4776698447755863, 53.81062959490106)
    |A - B| = 44.5737858532
    |C - B| = 44.5737858532
    (A - B)·(C - B) = 0.0

s = -169.123690271 , t = 120.065787347
    A = (-132.46804487704648, -58.47454376258159, 87.38287835155742)
    C = (-49.973944134014694, -2.996778178983093, -109.13027773929545)
    |A - B| = 155.724885051
    |C - B| = 155.724885051
    (A - B)·(C - B) = 9.09494701773e-13

Answer 2

This is a complete rewrite, since my first solution involved an algebraic curve of degree three, and likely roots of a sixth degree polynomial. This here should be far easier.

You can describe the problem using three vectors $a,b,c\in\mathbb R^3$ such that each vector gives the direction of one of the lines, and furthermore $b=B-S$, i.e. it points exactly at point $B$. For the other two, the length of the vector does not matter. Then you can assume $A=S+\lambda\,a$ and $C=S+\mu\,c$ for some $\lambda,\mu\in\mathbb R$.

Now you have two conditions:

$$\langle A-B,C-B\rangle=0 \qquad \lVert A-B\rVert=\lVert C-B\rVert$$

where $\langle\cdot,\cdot\rangle$ denotes the dot product. You can write the first of these conditions, the right angle, as

\begin{align*} \langle A-B,C-B\rangle&=0 \\ \langle A,C\rangle - \langle A,B\rangle - \langle C,B\rangle + \langle B,B\rangle&=0 \\ \lambda\mu\,\langle a,c\rangle - \lambda\,\langle a,b\rangle - \mu\,\langle c,b\rangle + \langle b,b\rangle&=0 \\[2ex] (\lambda, \mu, 1)\cdot \begin{pmatrix} 0 & \langle a,c\rangle & -\langle a,b\rangle \\ \langle a,c\rangle & 0 & -\langle c,b\rangle \\ -\langle a,b\rangle & -\langle c,b\rangle & 2\,\langle b,b\rangle \end{pmatrix} \cdot\begin{pmatrix}\lambda\\\mu\\1\end{pmatrix} &=0 \end{align*}

The second condition, the equal length, can also be written as

\begin{align*} \lVert A-B\rVert^2 - \lVert C-B\rVert^2 &= 0 \\ \langle A-B,A-B\rangle - \langle C-B,C-B\rangle &= 0 \\ \langle A,A\rangle - 2\,\langle A,B\rangle - \langle C,C\rangle + 2\,\langle C,B\rangle &= 0 \\ \lambda^2\,\langle a,a\rangle - 2\lambda\,\langle a,b\rangle - \mu^2\,\langle c,c\rangle + 2\mu\,\langle c,b\rangle &= 0 \\[2ex] (\lambda, \mu, 1)\cdot \begin{pmatrix} \langle a,a\rangle & 0 & -\langle a,b\rangle \\ 0 & -\langle c,c\rangle & \langle c,b\rangle \\ -\langle a,b\rangle & \langle c,b\rangle & 0 \end{pmatrix} \cdot\begin{pmatrix}\lambda\\\mu\\1\end{pmatrix} &=0 \end{align*}

Now you can interpret each of the quadratic equations as a description of a conic section in the projective plane. You can then apply the machinery for intersecting conics, which will involve solving a cubic equation. In the end, you obtain up to four points of intersection, from which you can deduce possible solutions for $A=S+\lambda\,a$ and $C=S+\mu\,c$.