that satisfies (2). Then (3) ax 0 + by 0 + cz 0 = d.

Transcription

1 Planes.nb 1 Plotting Planes in Mathematica Copright 199, 1997, 1 b James F. Hurle, Universit of Connecticut, Department of Mathematics, Unit 39, Storrs CT All rights reserved. This notebook discusses planes in 3-space, and the Mathematica code to generate plots of them. 1. Normal Form. The best algebraic representation of planes is the normal-form equation, whose geometric basis is simple: a plane is uniquel determined b one point P on it and a normal vector n. Definition. The normal-form equation of the plane through P(,, ) with the normal vector n = ai + bj + ck = (a, b, c) is (1) n ( ) =, where = (,, ) = i + j + k = OQ for an point Q(,, ) on the plane and = OP = i + j + k =(,, ) is the vector from the origin to the point P. Equation (1) sas that a point Q(,, ) lies on the plane if and onl if the plane's normal vector n is perpendicular to the vector from the given point P to Q. The vector equation leads easil to a scalar equation. The definition of dot product and (1) give (a i + b j + c k) [( ) i + ( L j + ( L k] =, a( ) + b( ) + c( ) = a + b + c = a + b + c () a + b + c = d, where d = n. Conversel, an linear equation of the form () is the equation of a plane in 3-space, provided that not all three of a, b, and c are. To see wh, let P(,, ) be an point

2 Planes.nb that satisfies (). Then (3) a + b + c = d. Let n = ai + bj + ck = (a, b, c) and = OP = i + j + k = (,, ). Then () and (3) give n = n = d n ( ) =, which is the normal-form equation of a plane. To plot the graph of a plane in Mathematica, the simplest approach is to plot its equation as that of a (simple flat) surface with equation = f(, ). That is possible unless the variable does not appear in the equation (1). To plot a plane with equation in the form (1), where the coefficient c of is not, solve (1) for the variable : () = (a + b + c a b)/c The following routine illustrates how to plot the graph of the plane b plotting the equation = f(, ), where f(, ) is the right side of (). The values of a, b, c, and d come from the plane with normal vector n = 3i + j + k = (3, 1, ) through the point P(, 1, 3). As usual, to generate the plot eecute the routine b placing the cursor at the end of the last blue line and pressing the Enter ke. In[11]:= (* Mathematica Routine to plot graph of a plane a + b + c = d, where c is not *) a := 3; b := 1; c := ; := -; := 1; := 3; F[_, _] := (a + b + c - a - b )/c; Plot3D[ F[, ], {, -, }, {, -, 1}, AesLabel -> {,, } ]

3 Planes.nb Out[19]= Ü SurfaceGraphics Ü As before, it is possible to add coordinate aes, at the epense of complicating the code. The following routine outputs the original rendering first, and then repeats that with coordinate aes. Use whichever routine that gives our ee the better image. In[1]:= (* Mathematica Routine to plot graph of a plane a + b + c = d, where c is not with coordinate aes drawn in *) a := 3; b := 1; c := ; := -; := 1; := 3; F[_, _] := (a + b + c - a - b )/c; planeplot = Graphics3D[ Plot3D[ F[, ], {, -, }, {, -, 1}, AesLabel -> {,, } ] ]; coordaes = Graphics3D[{ {RGBColor[1,, ], Line[{{-,, }, {6,, }}], Tet["", {7,, }]}, {RGBColor[1,, ], Line[{{, -3, }, {, 1, }}], Tet["", {, 13, }]}, {RGBColor[1,, ], Line[{{,, -3}, {,, 9}}], Tet["", {,, 9}]} }]; Show[planeplot, coordaes]

4 Planes.nb Out[31]= What if c =, that is, the plane is vertical? Then the equation of the plane involves onl the two variables and, so it is not possible to plot it as the graph of an equation = F(, ). The best approach is to treat the plane as a simple eample of a parametric surface. If the normal vector n = ai + bj + k = (a, b, ), then the normal-form equation (1) becomes (ai + bj) [( )i + ( )j + ( )k] =, a( ) + b( ) = a + b = a + b () a + b = d, where d = n. In (), not both a and b can be ero, because the normal vector n is nonero. Suppose for eample that b is nonero. Then = d/b a/b = (d a)/b. Since the equation of the plane does not involve, it can var over all real numbers. An point on the plane then satisfies the triple of parametric equations = u, = (d au)/b, = v, where u (, + ) and v (, + ). Note that there are two parameters, reflecting the fact that a plane is two dimensional. The following Mathematica routine plots the plane through P(1, 3, ) with normal vector n = 3 i j = (3, 1, ). Since d = n, for this plane d = (3, 1, ) (1, 3, ) = 3

5 Planes.nb 3 =. The plane is vertical, so there is no need to add the coordaes routine to displa the coordinate aes. The default plot's bo displas the nature of the plane quite well. In[33]:= (* Mathematica Routine to plot graph of a plane a + b = d, where b is not *) a := 3; b := -1; c := ; := 1; := 3; := ; d := Dot[{a, b, c}, {,, }] ParametricPlot3D[ {u, (d - a u)/b, v}, {u, -, }, {v, -, 8}, AesLabel -> {,, } ] Out[1]=. Vector cross product and planes. Euclidean geometr arose man centuries before vectors. The classical ideal of a plane in 3-space was the surface determined b three points P, Q, and R that are non-collinear (that is, do not lie in a common line) or b two non-skew lines (lines that are either parallel or intersect). One tool makes it eas to find the normal-form equation of a plane with either description. That tool is the vector cross product of 3-dimensional vectors. It produces a vector perpendicular to two given vectors, sa v and w. The definition of v w is somewhat involved, so it is worth noting first that finding a vector perpendicular to each of two given vectors is indeed enough to determine the equation of a plane through three non-collinear points or two non-skew lines. Consider

6 Planes.nb 6 first that two lines: l 1 with vector equation = + tv and l with equation = 1 + sw. Then v and w are direction vectors in the directions of the respective lines. So a normal vector n to the plane determined b v and w is indeed perpendicular to both v and w. Similarl, given three points P, Q, and R not on a common line, the vectors v = PQ and w = PR are in the plane determined b P, Q, and R. So once again, a normal vector to that plane is perpendicular to both v and w. The following definition thus provides the promised tool for finding the equation of a plane with either description. Definition. The cross product of two vectors v = (v 1, v, v 3 ) = v 1 i + v j + v 3 k and w = (w 1, w, w 3 ) = w 1 i + w j + w 3 k in Euclidean 3-space is (6) v w = (v w 3 v 3 w )i + (v 3 w 1 v 1 w 3 )j + (v 1 w v w 1 )k. The compleit of the formulas for the coordinates results from solving the sstem of equations v n = and w n = for n = (,, ). (See the tet for details.) Mathematica has a command for calculating cross products. The following short routine illustrates not onl that but also Mathematica's capacit to carr out smbolic algebra. In[3]:= (* Mathematica Routine to illustrate smbolic computation of the cross product of two vectors *) v = {v1, v, v3}; w = {w1, w, w3}; Cross[v, w] Out[6]= 8-v3 w + v w3, v3 w1 - v1 w3, -v w1 + v1 w< For a numerical eample, simpl input a couple of specific vectors v and w. In[7]:= (* Mathematica Routine to illustrate numerical computation of the cross product of two vectors *) v = {, -1, 3}; w = {1,, -3}; Cross[v, w] Out[]= 8-1, 9, 11< From this it follows that the plane through the two lines = (,, 3) + s(, 1, 3) and = (, 3, 6) + t(1,, 3) has normal vector n = ( 1, 9, 11). Its normal-form equation is therefore

7 Planes.nb 7 ( 1, 9, 11) (,, 3) = 1( ) ( 3) =, which simplifies to = 9. (Note: both points (,, 3) and (, 3, 6) from the two given equations satisf this equation, which confirms that there has been no error in the calculation. It is alwas advisable to substitute a known point from each line into the final equation of the plane, to verif that the lines are not skew. If the are, one of the points will satisf the final equation, but the other will not.) Mathematica can plot two vectors and their cross product. The following modification of a routine from VectPlot illustrates that. Eecute it to view the plot. In[1]:= (* Mathematica Routine to illustrate graphicall the cross product of two vectors *) v = {, -1, 3}; w = {1,, -3}; k = Cross[v, w]; q = Sqrt[v.v]; r = Sqrt[w.w]; s = Sqrt[(v + w).(v + w)]; coordaes = Graphics3D[{ {RGBColor[, 1, ], Line[{{-,, }, {6,, }}], Tet["", {8,, }]}, {RGBColor[, 1, ], Line[{{, -3, }, {, 1, }}], Tet["", {, 1, }]}, {RGBColor[, 1, ], Line[{{,, -3}, {,, 9}}], Tet["", {,, 1}]} }]; vplot = Graphics3D[{ {Line[{{,, }, v}], Polgon[{v,{.8 v[[1]],.8 v[[]],.8 v[[3]] - /q}, {.8 v[[1]],.8 v[[]],.8 v[[3]] + /q} }], Tet[FontForm["v", {"Times-Bold", 1}], {1. v[[1]], 1. v[[]], 1. v[[3]]}] }, {Line[{{,, }, w}], Polgon[{w, {.8 w[[1]],.8 w[[]],.8 w[[3]] - /r}, {.8 w[[1]],.8 w[[]],.8 w[[3]] + /r} }], Tet[FontForm["w", {"Times-Bold", 1}], {1. w[[1]], 1. w[[]], 1. w[[3]]}] }, {RGBColor[1,, 1], Line[{{,, }, Cross[v, w]}], Polgon[{k,{.9 (k)[[1]] - 1,.9 (k)[[]] + 1,.9 (k)[[3]] - 1}, {.9 (k)[[1]] + 1,.9 (k)[[]] + 1,.9 (k)[[3]] - 1} }], Tet[FontForm["v w", {"Times-Bold", 1}], {.8 (k)[[1]],.8 (k)[[]],

8 Planes.nb 8.7 (k)[[3]]}] } }] Show[vplot, coordaes, Aes -> True, Lighting -> False] Out[9]= 1 1 vw v w -1 - Out[6]= No new routine is needed to plot the plane: just modif the earlier routine for plotting the normal-form equation. There is no tet figure with which to compare this plot, but the net edition will have one! In[61]:= (* Mathematica Routine to plot graph of a plane a + b + c = d, where c is not *) a = -1; b = 9; c = 11; = ; = ; = 3; F[_, _] := (a + b + c - a - b )/c; planeplot = Graphics3D[ Plot3D[ F[, ], {, -, }, {, -, 1}, AesLabel -> {"", "", ""} ] ]; coordaes = Graphics3D[{{RGBColor[1,, ], Line[{{-,, }, {,, }}], Tet["", {6,, }]}, {RGBColor[1,, ], Line[{{, -, }, {, 1, }}], Tet["", {, 11, }]},

9 Planes.nb 9 {RGBColor[1,, ], Line[{{,, -}, {,, 9}}], Tet["", {,, 11}]} }]; Show[planeplot, coordaes] Out[71]=