# Plotting a plane using intercepts

I have a plane 2x + 5y + z = 10. I have determined that its intercepts are:

x = 5, y = 2, z = 10


I need to use Plot3D specifically to plot the plane formed by these three points. How might I go about doing that?

• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– user9660
Nov 11, 2014 at 21:42

ContourPlot3D[2 x + 5 y + z == 10, {x, 0, 6}, {y, 0, 3}, {z, 0, 11}, Mesh -> None] or

Graphics3D[Polygon[{{5, 0, 0}, {0, 2, 0}, {0, 0, 10}}],  BoxRatios -> 1, Axes -> True] or

Plot3D[ Piecewise[{{10 - 2 x - 5 y, 10 - 2 x - 5 y >= 0}},
Indeterminate], {x, 0, 6}, {y, 0, 3}, Mesh -> None, BoxRatios -> 1] You can use

InfinitePlane[{{5, 0, 0}, {0, 2, 0}, {0, 0, 10}}]


The plane will extend throughout the plot range:

Graphics3D[{
InfinitePlane[{{5, 0, 0}, {0, 2, 0}, {0, 0, 10}}]
},
PlotRange -> {{0, 10}, {0, 10}, {0, 10}},
Axes -> True]]


and

Graphics3D[{
InfinitePlane[{{5, 0, 0}, {0, 2, 0}, {0, 0, 10}}]
},
PlotRange -> 10,
Axes -> True]


preoduce, respectively, Plot3D[10 - 2 x - 5 y, {x, 0, 6}, {y, 0, 3},
RegionFunction -> Function[{x, y, z}, 0 < x && 0 < y && 0 < z],
Axes -> True, AxesOrigin -> {0, 0, 0}, Boxed -> False]


for view:

points = {{5, 0, 0}, {0, 2, 0}, {0, 0, 10}};
Show[Graphics3D[{Blue, Sphere[points, .2]}, Axes -> True,
Boxed -> False, AxesOrigin -> {0, 0, 0},
AxesStyle -> Directive[{Thick, Black}]],
Plot3D[10 - 2 x - 5 y, {x, 0, 6}, {y, 0, 3},
RegionFunction -> Function[{x, y, z}, 0 < x && 0 < y && 0 < z],
Mesh -> None]] 