Copyright © 1997, 2001 by James F. Hurley, University of Connecticut, Department of Mathematics, Unit 3009, Storrs CT 06269-3009. All rights reserved.

**1. Planes whose equations are solvable for ***z*** as a function of ***x*** and ***y**. *As the text mentions, if *c ~= *0 it is easy to solve the equation *ax + by + cz = d *of a plane for *z* as a function of *x* and *y*: *z = (d -- ax -- by)/d. Mathematica* can easily plot such a plane, as the following routine illustrates for the plane with equation --12*x* + 9*y + *11*z* = 9. Note that it passes through the point (2, 0, 3) and has normal vector **n** = (--12, 9, 11) = --12 **i** + 9 **j** + 11 **k**.To see the plot, hit the Enter key at the extreme lower right of the keyboard, or press the Shift and Return keys simultaneously.

**(* ***Mathematica*** Routine to plot graph of a plane **

b = 9;

c = 11;

x0 = 2;

y0 = 0;

z0 = 3;

F[x_, y_] := (a x0 + b y0 + c z0 - a x - b y)/c;

Plot3D[ F[x, y], {x, -2, 5}, {y, -2, 10},

AxesLabel -> {"x", "y", "z"} ]

The following more elaborate routine adds coordinate axes (in red).

**(* ***Mathematica*** Routine to plot graph of a plane **

coordinate axes *)

b = 9;

c = 11;

x0 = 2;

y0 = 0;

z0 = 3;

F[x_, y_] := (a x0 + b y0 + c z0 - a x - b y)/c;

planeplot = Graphics3D[

Plot3D[ F[x, y], {x, -2, 5}, {y, -2, 10},

AxesLabel -> {"x", "y", "z"} ]

];

coordaxes = Graphics3D[{

{RGBColor[1, 0, 0],

Line[{{-2, 0, 0}, {5, 0, 0}}],

Text["x", {6, 0, 0}]},

{RGBColor[1, 0, 0],

Line[{{0, -2, 0}, {0, 10, 0}}],

Text["y", {0, 11, 0}]},

{RGBColor[1, 0, 0],

Line[{{0, 0, -2}, {0, 0, 9}}],

Text["z", {0, 0, 11}]}

}];

Show[planeplot, coordaxes]

**2. Plotting a vertical plane.** The equation of a plane perpendicular to the *xy*-plane has no *z*-term. Thus, it is not possible to plot such a plane as the graph of an equation * z = F*(*x*, *y*). However, the following simple *parametric* approach produces a *Mathematica* plot, and provides a preview of the important idea of parametric representation of surfaces. The plane has normal vector **n** = *a ***i** + *b ***j** + 0 **k** = (*a*, *b*, 0), so the normal-form equation of the plane is

(*a ***i** + *b ***j) ****· **[(*x -- x** ***i **+ (*y -- y** ***j** + (*z -- z** ***k**] = 0 , * *

(1)

In (1), not both *a* and *b* are zero, because the normal vector **n** is nonzero. Thus, one of them is expressible in terms of the other. Suppose that *b* is nonzero. Then *y = d*/*b* -- *ax/b *= (*d -- ax*)/*b*. Since the equation of the plane does not involve *z*, it can vary over all real numbers. The following *triple* of parametric equations thus describes the plane:

*x = u, y = *(*d -- au*)/*b*, *z = v,* where *u *∈* *(-- ∞, +∞) and *v* ∈ (-- ∞, +∞).

Note that there are *two* parameters, reflecting the fact that a plane has two dimensions, rather than the single dimension of lines and curves. The following *Mathematica* routine plots the plane through *P*(1, 3, 2) with normal vector **n** = 3 **i** -- **j = **(3, -- 1, 0). Since *d = ***n** **· ** for this plane *d = *(3, -- 1, 0) · (1, 3, 2) = 3 -- 3 = 0. Execute the routine to see the graph of the plane!

**(* ***Mathematica* Routine to plot graph of a plane

*ax *+* by *=* d, *where *b* is not 0 *)

a := 3;

b := -1;

c := 0;

x0 := 1;

y0 := 3;

z0 := 2;

d := Dot[{a, b, c}, {x0, y0, z0}]

ParametricPlot3D[{u, (d - a u)/b, v}, {u, -4, 4},

{v, -2, 8},AxesLabel -> {x, y, z} ]

Converted by