# Finding Intersections Between Arbitrary Surface and A Line

I have a self-intersecting surface H defined as follows:

M = 0; phi = Pi/2;

b1 = {{-Sqrt/2}, {3/2}};
b2 = {{-Sqrt/2}, {-3/2}};
b3 = {{Sqrt}, {0}};

hx[kx_, ky_] := 1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2]
hy[kx_, ky_] := Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2]
hz[kx_, ky_] :=
M - 2 Sin[
phi] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] + Sin[{kx, ky}.b3])
H[kx_, ky_] =
Flatten[{hx[kx, ky], hy[kx, ky], hz[kx, ky]}, 1] // Simplify;


and I wish to find the number of times a line from the origin to some point {10, 10, 10} intersects with this surface.

What I have tried:

1. Going off this reference - Graphics3D: Finding intersection of 3d objects and lines - I attempted to define a Region and line as follows:

R = ParametricRegion[{H[u, v], 1.8 <= u <= 5.5 && -2.2 <= v <= 2.2}, {u,v}];
TheLine = Line[{{0,0,0}, {10,10,10}}];


and continue with the method given. However, Mathematica does not correctly give me the number of intersections, as it does not identify H as a properly defined Region. Additionally, my arbitrary surface is apparently "not a proper Graphics3D primitive or directive".

2. To tackle this numerically, I tried populating a table of coordinates for the line and comparing it with values of my surface to see where the difference between coordinates for rows would be least (after ordering rows in ascending order). The line now goes as:

p1 = {0, 0, 0};
p2 = {10, 0, 10};
v = p2 - p1;
TheLine[t_] := p1 + t v


However, this method is problematic because I miss several potential points. I read online that this could have worked if the third component of H is a function of the first two: H = {x,y,z(x,y)}. However, this is not the case.

3. I tried solving the following system of 'equations':

Solve[{x, y, z} \[Element] R && {x, y, z} \[Element] theLine, {x, y, z}, Reals]


but this takes forever to run. I doubt this would work.

Additionally, I don't think that I can solve for a system of equations like https://www.wolfram.com/mathematica/new-in-10/basic-and-formula-regions/surface-intersection.html in my case because a line cannot have a stand-alone equation of the form l(x,y,z) because a line is the intersection of two planes.

Perhaps I am missing something trivial, but none of the other potential solutions on this website seem to work. Therefore I would appreciate insight on any approach that would work! Thanks!

(I managed to borrow someone else's computer for a few minutes...)

Before anything else: using actual vectors instead of $$2\times 1$$ matrices will make constructing your parametric equations much easier. Thus:

M = 0; ϕ = π/2;
b1 = {-Sqrt/2, 3/2}; b2 = {-Sqrt/2, -3/2}; b3 = {Sqrt, 0};


Set up the parametric equations:

eqs = Thread[{x, y, z} == {1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2],
Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2],
M - 2 Sin[ϕ] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] +
Sin[{kx, ky}.b3])}];


Now, we can use GroebnerBasis[] to derive the implicit Cartesian equation:

impl = GroebnerBasis[Join[TrigExpand[eqs],
{Cos[(Sqrt kx)/2]^2 + Sin[(Sqrt kx)/2]^2 == 1,
Cos[ky/2]^2 + Sin[ky/2]^2 == 1}], {x, y, z},
{Cos[(Sqrt kx)/2], Sin[(Sqrt kx)/2],
Cos[ky/2], Sin[ky/2]}][] // FullSimplify
(3 + (-4 + x) x + y^2)^2 (-3 + (-2 + x) x + y^2) + ((-1 + x)^2 + y^2) z^2


Deriving the intersection points is then as easy as

Solve[impl == 0, {x, y, z} ∈ InfiniteLine[{{0, 0, 0}, {10, 10, 10}}]] // FullSimplify
{{x -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1],
y -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1],
z -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1]},
{x -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2],
y -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2],
z -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2]}}

N[%]
{{x -> -0.814231, y -> -0.814231, z -> -0.814231},
{x -> 1.30996, y -> 1.30996, z -> 1.30996}}


Show the geometry:

Show[ParametricPlot3D[{1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2],
Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2],
M - 2 Sin[ϕ] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] +
Sin[{kx, ky}.b3])},
{kx, 1.8, 5.5}, {ky, -2.2, 2.2},
Mesh -> False, PlotStyle -> Opacity[1/2]],
Graphics3D[{Tube[Line[{{0, 0, 0}, {10, 10, 10}}]],
{Red, Sphere[{x, y, z}, 1/12] /. %}}]] • THIS IS SUCH A BEAUTIFUL SOLUTION!!! Thank you. (Minor note: I changed InfiniteLine to Line in my own code, to count the intersections of the ray with H in only one direction). – TribalChief Oct 16 '18 at 4:33
• Yes, that's fine. You might also be interested in HalfLine[] for representing rays as opposed to line segments with Line[]. – J. M. is away Oct 16 '18 at 5:01
• GroebnerBasis is new to me. Can I ask you how you wrote the expression for impl, so that I can play with this for other surfaces? The trig equalities, I mean? I see the Sqrt term, and so I know that the values of b are somehow in there. Thanks! – TribalChief Oct 16 '18 at 5:07
• GroebnerBasis[] is a pretty general function for eliminating variables in sets of algebraic equations. In this case, since you have algebraic combinations of trigonometric functions, it fits perfectly here for eliminating those parameters and leaving only x, y, z. As a simpler example of how it works, look at the result of GroebnerBasis[{x == Cos[t], y == Sin[t], Cos[t]^2 + Sin[t]^2 == 1}, {x, y}, {Cos[t], Sin[t]}], which should yield the usual implicit equation for the unit circle. – J. M. is away Oct 16 '18 at 5:12
• J. M., I know that I am going too far with asking for advice, but if you have the time, I posted a question on GroebnerBasis here: mathematica.stackexchange.com/questions/184069/… – TribalChief Oct 17 '18 at 3:52