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# plotting planes in 3D

Posted 10 years ago
 Hi, I'm trying to plot the following planes in 3D with Alpha, but am having issues:2x + y - 2 = 0y + 2z - 4 = 0x + 3z - 6 = 0x = 4y = 2z = 5I do need the last 3 in 3D: x is the constant(4), but y and z vary; y is the constant(2), x and z vary, etc. Doesn't come out right, even with 3d plot. Trying this in matlab but I don't know the language very well and it's not working. Thanks.
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Posted 10 years ago
 I have not yet found a way to get WolframAlpha to plot two planes, or any other set of 3d functions, in the same graph, all attempts end up with it plotting each function in a different graph.I did find one possible way to get around this, first solve each equation for z and then plot3d the Max of the equations. If the intersection is obvious enough then this might provide the desired information.As an example, this works for your last two equations http://www.wolframalpha.com/input/?i=Plot3D%5BMax%5B-y%2F2+%2B+2%2C+-x%2F3%2B2%5D%5D and should work for any number of equations as long as each include z. If some do not include z then perhaps you can interpret those equations appropriately and still use this method.
Posted 10 years ago
 Take two - but it's Mma - not W|AShow[{ParametricPlot3D[\[Alpha] {0, 1, 3} + \[Beta] {1, 1/2, -1/3}, {\[Alpha], -2, 2}, {\[Beta], -3, 3},    DisplayFunction -> Identity],   ParametricPlot3D[\[Alpha] {1, -1/3, 1/5} + \[Beta] {-1/2, 1, -1}, {\[Alpha], -2, 2}, {\[Beta], -3, 3},    DisplayFunction -> Identity]}, DisplayFunction -> $DisplayFunction]in W|A you will find real entertainment, saying to itplot two planes through the originW|A displays a fighters ranking ... do not know about whattwo planes through the origingives valuable linguistic informationShow[{ParametricPlot3D[\ {0, 1, 3} + \ {1, 1/2, -1/3}, {\, -2, 2}, {\, -3, 3}, DisplayFunction -> Identity], ParametricPlot3D[\ {1, -1/3, 1/5} + \ {-1/2, 1, -1}, {\, -2, 2}, {\, -3, 3}, DisplayFunction -> Identity]}, DisplayFunction ->$DisplayFunction]says input too long ... so do yourself a favour and use Mma. Try thisplane intersects another onesays "... bites the dust" --- Freddy calls back ... great but not in the field of linear algebra or plane geometry.
Posted 10 years ago
 Still, if you doplot x+y+1, -x+y+3 in the same 3D plotin W|A you get two 3D plots
Posted 10 years ago
 Very informative -- thank you!
Posted 10 years ago
 You have an issue withRegionPlot3D[2 x + y - 2 == 0, {x, -3, 3}, {y, -4, 4}, {z, -5, 5}]because the graphics is empty because not one single damn triple of {x,y,z} which gets tested fulfills the linear equation exactly. You remove the issue by displaying the half space typingRegionPlot3D[2 x + y - 2 <= 0, {x, -3, 3}, {y, -4, 4}, {z, -5, 5}]If you really want to plot a plane, then you use ParametricPlot3D[], if one knows two base vectors (i.e. linear independent vectors) in the plane, then one typesParametricPlot3D[\[Alpha] {0, 1, 3} + \[Beta] {1, 1/2, -1/3}, {\[Alpha], -2, 2}, {\[Beta], -3, 3}]Please note, the vectors {0,1,3} and {1, 1/2, -1/3} are taken only for demonstration. You should compute the base vectors of the planes given to you on your own.
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