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I have the 3d cone:

Plot3D[-Sqrt[x^2 + y^2], {x, -20, 20}, {y, -20, 20}, Mesh -> None, 
 BoxRatios -> {1, 1, 1}]

I need a slanted plane intersecting it. Can somebody help me?

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    $\begingroup$ Add a plane to your plot? e.g. Plot3D[{-Sqrt[x^2 + y^2], -1 - x - y}, {x, -1, 1}, {y, -1, 1}] $\endgroup$
    – 2012rcampion
    Commented Feb 26, 2015 at 0:10
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    $\begingroup$ I don't quite get what you're after. You know the equation of a cone but not of a plane? Any linear function of x and y yields a plane. $\endgroup$
    – Michael E2
    Commented Feb 26, 2015 at 0:19

2 Answers 2

Reset to default
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Graphics3D[
 {
  {Opacity[0.5], Cone[{{0,0,0}, {0,0,3}}, 1]},
  {Yellow, Opacity[0.5], Polygon[{{-1,-1,1}, {-1,1,1}, {1,1,2}, {1,-1,2}, {-1,-1,1}}]}
  }
 ]

enter image description here

Or play around with this:

Manipulate[
 Graphics3D[
  {
   {LightBlue, Opacity[0.5], Cone[{{0, 0, 0}, {0, 0, 3}}, rcone]},
   {Yellow, Opacity[0.5], 
    Polygon[{{-1,-1,1}, {-1,1,1}, {1,1,m}, {1,-1,m}, {-1,-1,1}}]}
   }
  ],
 {rcone, .5, 2}, {m, 1, 3}
 ]
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let's say you have a plane:

myPlane=-8-x-2y

you can use Plot3D like so;

Plot3D[{-Sqrt[x^2 + y^2], myPlane}, {x, -15, 15}, {y, -15, 15}, 
Mesh -> None, BoxRatios -> {1, 1, 1}, PlotLegends -> "Expressions"]

enter image description here

you can pimp your plot with RegionFunction:

Plot3D[{-Sqrt[x^2 + y^2], myPlane}, {x, -15, 15}, {y, -15, 15}, 
 RegionFunction -> Function[{x, y, z}, -10 < z < 10], BoxRatios -> 1, 
 PlotLegends -> "Expressions"]

enter image description here

and or and specify ViewPoint for visualization:

enter image description here

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