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I have two equations consisting of two variables and I'm plotting those two functions in 3D by defining the range of the two variables. I need to find the points of intersection of these two functions. I tried using Graphics'Mesh'FindIntersections but its not working for 3D plots. Can anyone help me with this? I need to get the data of the intersection points.

The code is as follows:

e1= -0.0492101+(0.00982664*l3*Sec[147.557*l3]^2*(Tan[147.557*l8]+3.34665*Tan[152.139*l8]))/(Tan[152.139*l3]*Tan[147.557*l8]-Tan[147.557*l3]*Tan[152.139*l8])+(0.00982664*l8*Sec[147.557*l8]^2*(Cot[147.557*l8]*Tan[147.557*l3]+3.34665*Cot[147.557*l8]*Tan[152.139*l3]))/(-Tan[152.139*l3]+Cot[147.557*l8]*Tan[147.557*l3]*Tan[152.139*l8])==0;
e2= 0.00629091 +(0.113333*l3*Sec[152.139*l3]^2*(Tan[147.557*l8]+3.34665*Tan[152.139*l8]))/(Tan[152.139*l3]*Tan[147.557*l8]-Tan[147.557*l3]*Tan[152.139*l8])+(0.113333*l8*Sec[152.139*l8]^2*(Cot[147.557*l8]*Tan[147.557*l3]+3.34665*Cot[147.557*l8]*Tan[152.139*l3]))/(-Tan[152.139 l3]+Cot[147.557*l8]*Tan[147.557*l3]*Tan[152.139*l8])==0 
plot=Plot3D[{e1,e2}, {l3, 0.008, 0.012}, {l8, 0.008, 0.012}]

I have got a 3D plot as shown here. plot

Any help is highly appreciated.

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  • $\begingroup$ The equation f[l3,l8]==0 can only define a contour in plane. $\endgroup$
    – cvgmt
    Mar 20, 2021 at 7:30
  • $\begingroup$ The intersection is symmetric about the diagonal line l3==l8. z=Numerator[Together[Rationalize[e1[[1]]-e2[[1]],0]]]; ContourPlot[z==0,{l3,0.008,0.012},{l8,0.008,0.012}] and FindRoot[Simplify[z==0/.l3->.0112],{l8,.0105}] finds an l8 given a good estimate of the location, but I do not trust that. Look at z==0/.l3->.0112 to see why I am skeptical. $\endgroup$
    – Bill
    Mar 20, 2021 at 7:34
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    – Michael E2
    Mar 20, 2021 at 18:15

1 Answer 1

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Perhaps ContourPlot helps to find the intersection points! Try

e1= -0.0492101+(0.00982664*l3*Sec[147.557*l3]^2*(Tan[147.557*l8]+3.34665*Tan[152.139*l8]))/(Tan[152.139*l3]*Tan[147.557*l8]-Tan[147.557*l3]*Tan[152.139*l8])+(0.00982664*l8*Sec[147.557*l8]^2*(Cot[147.557*l8]*Tan[147.557*l3]+3.34665*Cot[147.557*l8]*Tan[152.139*l3]))/(-Tan[152.139*l3]+Cot[147.557*l8]*Tan[147.557*l3]*Tan[152.139*l8]) ;

e2= 0.00629091 +(0.113333*l3*Sec[152.139*l3]^2*(Tan[147.557*l8]+3.34665*Tan[152.139*l8]))/(Tan[152.139*l3]*Tan[147.557*l8]-Tan[147.557*l3]*Tan[152.139*l8])+(0.113333*l8*Sec[152.139*l8]^2*(Cot[147.557*l8]*Tan[147.557*l3]+3.34665*Cot[147.557*l8]*Tan[152.139*l3]))/(-Tan[152.139 l3]+Cot[147.557*l8]*Tan[147.557*l3]*Tan[152.139*l8]);

plot = ContourPlot[ 
   e1 == e2  , {l3, 0.008, 0.012}, {l8, 0.008, 0.012}, 
   MaxRecursion -> 4, PlotPoints -> 50 ] // Quiet

enter image description here

intersectionpoints=plot[[1]] [[1]][[1]]
(*{{0.008, 0.0106924}, {0.00800158, 0.0106923}, {0.00800352,0.0106923}, {0.0080051, 0.0106923},...}*)
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  • $\begingroup$ Here are several other solutions: $$ [[\mathit{l3} = - 0.928876965234335, \mathit{l8} = 0.322423141027989] , [\mathit{l3} = 2.97284035794540, \mathit{l8} = 0.311944493392789] , [\mathit{l3} = - 21.6616722319277, \mathit{l8} = - 8.27489640495406] , [\mathit{l3} = - 55.0458668808167, \mathit{l8} = 30.2464029070779]] $$ $\endgroup$
    – user64494
    Mar 20, 2021 at 14:19
  • $\begingroup$ @user64494 Might be, I restricted the solution range to the range OP asked for! $\endgroup$
    – Ulrich Neumann
    Mar 20, 2021 at 15:40
  • $\begingroup$ Thank you so much. How can I substitute those (intersectionpoints) values back into the equations? And also I want to substitute all these points into another two expressions. I can use ReplaceAll by using /. {l3-> ,l8-> } but how to represent each point respectively in that case? $\endgroup$
    – sunanda
    Mar 22, 2021 at 4:59
  • $\begingroup$ Try Map["expression" /. {l3->#[[1]],l8->#[[2]]}&,intersectionpoints] $\endgroup$
    – Ulrich Neumann
    Mar 22, 2021 at 7:07
  • $\begingroup$ Perhaps it's clever to Interpolate every branch of the intersectionpoints... $\endgroup$
    – Ulrich Neumann
    Mar 22, 2021 at 7:10

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