Below example is equation of a plane example I would know how to determine the range of values for x, y and z respectively.

f = 5 x - 2 y + 7 z == 15;
 {x, -7, 7}, {y, -7, 7}, {z, -7, 10},
 Mesh -> None,
 ContourStyle -> Opacity[1],
 AspectRatio -> 1]


  • 1
    $\begingroup$ I would know how to determine the range of values for x, y and z respectively It is not clear what you are asking. The range of values are the ones you gave in the command itself {x, -7, 7}, {y, -7, 7}, {z, -7, 10}. These are the values you decided to use. $\endgroup$
    – Nasser
    Apr 11 '20 at 6:45

The question gave the shifted point-normal-form of a plane in 3D.


n={5,-2,7}. d=np={5,-2,7}.{p1,p2,p2}=15.

A vector p is in general constructed from the normal

px={3,0,0}, py={0,-2/15,0} and pz={0,0,7/15}.

For graphical representation purposes, Mathematica has a built-in Hyperplane. That is in this question to be used in 3D. Just enter the normal vector to get a parallel plane through the origin.

Graphics3D[Hyperplane[{5, -2, 7}]]

The input for the shifted plane is

Graphics3D[Hyperplane[{5, -2, 7}, 15]]

A plot with the normal on the plane is

ill = Graphics3D[{Arrowheads[Medium], Thick, 
    Arrow[{{0, 0, 0}, {5, -2, 7}}]}, PlotRange -> 20, Axes -> True];
Show[ill, Graphics3D[Hyperplane[{5, -2, 7}, 15]]]

plane with normal

The range is 20 selected ad hoc. This corresponds to the setting Mathematica selects for the first input presented.

Mathematica can do more on planes: RegionMember:

RegionMember[Hyperplane[{5, -2, 7}, 15], {x, y, z}]

(x | y | z) \[Element] Reals && 5 x - 2 y + 7 z == 15

RegionDistance[Hyperplane[{5, -2, 7}, 15], {3, 0, 0}]

The point is on the plane.

There is no centroid of a plane. The plane is extended infinitely in two dimensions, directions. So range always arbitrary. Mathematica always shows a cube in three dimensions the standard view and viewport.

Have a look at the documentation of Hyperplane in Mathematica.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.