So I have an equation given by \begin{equation} 2rvz+v^2+x^2+y^2+z^2=1 \end{equation} which is a hypersurface and I want to intersect it with another surface \begin{equation} 1+v+rz=0 \end{equation} which is just a 3D surface. I think that if I do that I'll get a simple curve. How can I do this in Mathematica? I don't need to plot both curves but I need a plot of their intersection. Any help would be much appreciated.

  • $\begingroup$ See if this Simplify[Reduce[{2 r v z+v^2+x^2+y^2+z^2==1, 1+v+r z==0}, {x,y,z},Reals]] gives you any insight into what the intersection might be. IF you have any domain information about any of your variables then that might let someone give a better suggestion. $\endgroup$ – Bill Apr 6 '16 at 3:50
  • $\begingroup$ It says this.(v == -1 && r == 0 && (Sqrt[-x^2 - y^2] + z == 0 || Sqrt[-x^2 - y^2] == z)) || (r != 0 && (y == -(Sqrt[-(1 + v)^2 + r^2 (1 + 2 v + v^2 - x^2)]/r) || y == Sqrt[-(1 + v)^2 + r^2 (1 + 2 v + v^2 - x^2)]/r) && z == -((1 + v)/r))... No, idea what it means. Newbie to Mathematica. $\endgroup$ – Janus Apr 6 '16 at 4:01
  • $\begingroup$ In that && stands for "and" where expressions on both sides must be true while || stands for "or" where at least one expression on one side of that must be true. So that says one point of intersection is where (v= -1, r=0, x=0, y=0, z=0) all at the same time. You can plug those numbers in and check that both equations are true. The next part in ( ) is another point of intersection with x=0 and y=0. The next part looks like it might be more interesting with a range of values for x, y, z. If you forget MMA for a moment, remember && and || and (), separate into pieces and it is the intersection. $\endgroup$ – Bill Apr 6 '16 at 5:34
  • $\begingroup$ @Janus . Read wolfram.com/language/fast-introduction-for-programmers $\endgroup$ – QuantumDot Apr 6 '16 at 9:45

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