I have an assignment that asks me to graphically solve the system of equations i.e. plot them in ?^3 :
x + 2y + 3z = 1
2x + 4y + 7z = 2
3x + 7y + 11z = 8
Currently, I’ve used ContourPlot3D for plotting my planes in a 3-dimensional space and then FindRoot in order to find the solution to the system.
ContourPlot3D[{x + 2 y + 3 z == 1, 2 x + 4 y + 7 z == 2,
3 x + 7 y + 11 z == 8}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]
FindRoot[{x + 2 y + 3 z == 1, 2 x + 4 y + 7 z == 2,
3 x + 7 y + 11 z == 8}, {x, 0}, {y, 0}, {z, 0}]
My question here is if I’ve correctly completed the assignement and if I did, why would I need to plot the three planes in ?^3 when it doesn’t really help me in solving the problem?
Is it mainly the creators of the assignement wanting us to familiarize ourselves with the different Mathematica functions?
No - simply by plotting you can see the solution, namely where all three planes intersect. You can set the focus on that spot:
sol = FindRoot[{x + 2 y + 3 z == 1, 2 x + 4 y + 7 z == 2, 3 x + 7 y + 11 z == 8}, {x, 0}, {y, 0}, {z, 0}];
reg = Sequence @@ ({#, # - 1 /. sol, # + 1 /. sol} & /@ {x, y, z});
ContourPlot3D[{x + 2 y + 3 z == 1, 2 x + 4 y + 7 z == 2, 3 x + 7 y + 11 z == 8}, Evaluate@reg]
I see.
I’ll remember that you can actually focus the view on the intersection point like you illustrated. Makes it way faster than rotating the graph and looking for an intersection.
