Finding point of intersection between a line and a sphere

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André C. D.
André C. D. 2019-6-6
回答： Jason Der 2021-7-30
Hello all,
I have a MATLAB code that plots a 3D sphere using:
[xs,ys,zs] = sphere(10);
surface = surf(350*zs+1000,350*ys,350*xs);
I also have a line that represents the normal between three points (labeled P0, P1, P2) on a plane, which is plotted from the middle-point between all three points:
P0 = [tz1,ty1,tx1]; P1 = [tz2,ty2,tx2]; P2 = [tz3,ty3,tx3]; %represent a triangle
Pm = mean([P0;P1;P2]); %represents the midpoint between P0, P1 and P2
normal = cross(P0-P1,P0-P2);
cn = normal + Pm;
normal_vector = plot3([Pm(1) cn(1)],[Pm(2) cn(2)],[Pm(3) cn(3)],'k--'); %normal
What I am trying to do is find the coordinates of the point of intersection between the line "normal_vector" and the sphere "surface".
This is what the plot looks like:
untitled.jpg
The points P0, P1 and P2 are shown as coloured circles and are always inside the sphere, so their normal is always showing 'outwards' through the surface of the sphere.
Thank you in advance!

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采纳的回答

Torsten
Torsten 2019-6-6
编辑：Torsten 2019-6-6
Sphere:
(x-xs)^2 + (y-ys)^2 + (z-zs)^2 = R^2
Line:
[x y z] = Pm + l*normal
Thus
(Pm(1)+l*normal(1)-xs)^2 + (Pm(2)+l*normal(2)-ys)^2 + (Pm(3)+l*normal(3)-zs)^2 = R^2
Solve for (the positive) l.
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Torsten
Torsten 2019-6-7
Ah, I thought (xs,ys,zs) is the center of the sphere.
You have to solve
sol = solve((Pm1+(l*normal1) - xc)^2 + (Pm2+(l*normal2) - yc)^2 + (Pm3+(l*normal3) - zc)^2 == R^2,l)
where (xc,yc,zc) is the center of the sphere.

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更多回答（2 个）

EVELYN ROSSANA PARRA LOPEZ
The last line of code is summarized in replacing the terms x, y and z of the parametric equation of a line in space, in the equation that describes a sphere, and the variable to be found is the parameter, in this case l. I apply the same with a sphere and a known line, but the answer is as follows:
CODE LINES:
syms t
sol=solve((0.2118+t*0.8473-1).^2+(0.06883+t*0.2754-0.5).^2+(0.1135+t*0.4541-0.5).^2==((0.25).^2),t)
RESULT:
sol=
240523932/249992315 - (10^(1/2)*3160661400392057^(1/2))/999969260
(10^(1/2)*3160661400392057^(1/2))/999969260 + 240523932/249992315
Jason Der
Jason Der 2021-7-30
The accepted answer works, but I wasn't able to find a way to vectorize it for a large dataset.
However, I did read an elegant solution on eng-tips at the following url:
https://www.eng-tips.com/viewthread.cfm?qid=248900
Hope this helps someone one day.

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