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Please can you advise me how to speed up the rendering of the following two planes? When I click and drag to rotate the plot, its movement is quite jerky. Presumably it is working harder than I need it to. The matrix equation A.X = B represents two flat planes embedded in 3D space. The solution of the system is the line of intersection of the two planes. I believe ContourPlot3D interprets the matrix equation as a list of the two planes. In fact, A.X == B evaluates as {{x + y + z}, {x - y + 2 z}} == {{4}, {0}}.

A = {{1, 1, 1}, {1, -1, 2}};
B = {{4}, {0}};
X = {{x}, {y}, {z}};
ContourPlot3D[A.X == B, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, 
ContourStyle -> Directive[Opacity[.6]], MeshStyle -> Gray]

I have tried using Evaluate on the expression A.X == B, and on the expression A.X, but to no avail.

You might think it better to separate the two planes, and use Plot3D command for each, but I would like my code to work automatically for general linear systems, i.e., for different choices of A, B and X, and I would like it to accomodate vertical planes, to which I guess Plot3D would not apply.

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You need to get rid of some extraneous List wrappers

A = {{1, 1, 1}, {1, -1, 2}};
B = {4, 0};
X = {x, y, z};

and then thread the two sides of your matrix equation over Equals.

ContourPlot3D[Evaluate @ Thread[A.X == B], {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, 
  ContourStyle -> Opacity[.6], MeshStyle -> Gray]

plot

The graphics shown above rotate in a nice, smooth fashion.

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  • $\begingroup$ Thank you very much, @m_goldberg ! That is great. I know Mathematica's Dot command cleverly interprets vectors as row or column vectors depending on the context. I had hoped to insist that my column vectors be represented as nx1 matrices but I guess it is not important. I tried forcing this using Evaluate[Thread[Map[Flatten, Map[Transpose, A.X == B]]]] for the function in the first argument of ContourPlot3D but ContourPlot3D still didn't want to play ball. Do you have any idea why ? $\endgroup$ – Simon Sep 24 '15 at 12:34

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