Chapter 1: Vectors, Lines and Planes
Section 1.7: Planes
Example 1.7.1
Obtain an equation for the plane containing the points A:2,3,−1, B:5,−7,2, and C:3,6,1.
Solution
Mathematical Solution
Figure 1.7.1(a) details the traditional vector calculation for this example. The three blue arrows emanate from the origin, and represent the position vectors A, B, and C.
The green and red arrows respectively represent the vectors P=B−A and Q=C−A, while the gold arrow represents the normal vector N=P×Q.
If R is the position vector to the general point x,y,z, the vector R−A will lie in the plane if it is orthogonal to the normal vector N. Hence, the condition on x,y,z that constrains that point to lie in the plane is R−A·N=0.
use Student:-VectorCalculus, plots in
module()
local A,B,C,AB,AC,n,N,p1,p2,p3,R;
A:=RootedVector(root=[0,0,0],<2,3,-1>):
B:=RootedVector(root=[0,0,0],<5,-7,2>):
C:=RootedVector(root=[0,0,0],<3,6,1>):
R:=PositionVector([x,y,z]):
AB:=RootedVector(root=A,convert(B-A,Vector)):
AC:=RootedVector(root=A,convert(C-A,Vector)):
n:=CrossProduct(AC,AB/6):
N:=RootedVector(root=[2,3,-1],convert(n,Vector)):
p1:=VectorCalculus:-PlotVector([AB,AC,N,A,B,C],color=[red,green,gold,blue,blue,blue],width=.25):
p2:=implicitplot3d((R-A).convert(N,Vector),x=-1..8,y=-7..7,z=-5..5,style=surface,transparency=.8,color=red):
p3:=display(p1,p2,scaling=constrained,axes=none,orientation=[180,-30,150]);
print(p3)
end module:
end use:
Figure 1.7.1(a) Position vectors A, B, C, (blue), vectors P=B−A (green), Q=C−A (red), and N=P×Q (gold)
The requisite computations are then
P=5−72−23−1=3−103
Q=361−23−1=132
N=P×Q= |ijk3−103132| =−29−319
and
R−A·N = x−2y−3z+1·−29−319 = −29x−2−3y−3+19z+1=0
from which it follows that 29 x+3 y−19 z=58+9+19=86.
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Write a sequence of the three given points.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane (Click OK in the variable-choice dialog. See Figure 1.7.1(b).)
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation
Figure 1.7.1(b) Variable-choice dialog
2,3,−1,5,−7,2,3,6,1→make plane<< Plane 1 >>→representation−29x−3y+19z=−86
The traditional approach to obtaining this plane is vectorial, as shown below. If the three given points are A, B, and C, then define the position vectors A, B, and C, respectively, and define R, the generic position vector to the point x,y,z. The vectors P=B−A and Q=C−A lie in the plane for which N=P×Q is its normal. Since the vector R−A should be a general vector in the plane, the condition on R that restricts x,y,z to the plane is the orthogonality condition R−A·N=0.
Define the position vectors A, B, C, and R
Enter A as per Table 1.1.1.
Context Panel: Assign to a Name≻A
2,3,−1→assign to a nameA
Enter B as per Table 1.1.1.
Context Panel: Assign to a Name≻B
5,−7,2→assign to a nameB
Enter C as per Table 1.1.1.
Context Panel: Assign to a Name≻C
3,6,1→assign to a nameC
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
x,y,z→assign to a nameR
Define the vectors P and Q
Context Panel: Assign Name
P=B−A→assign
Q=C−A→assign
Obtain the normal N
N=P×Q→assign
Impose the orthogonality condition on R−A and N
Common Symbols palette: Dot product operator Press the Enter key.
R−A·N=0
−29x−3y+19z+86=0
For the reader's convenience, the vectors P, Q, and N are displayed here:
P,Q,N = 3−103,132,−29−319
A purely algebraic solution, which involves nothing more than solving three linear equations, is also available. The general form equation for the plane is a x+b y+c z+d=0. Since each of the points A, B, and C must satisfy this equation, a set of three linear equations in the four parameters a,b,c,d results. These three equations are solved for, say, a,b,c in terms of d, which is then "divided out" at the end. The full implementation of this approach is given below.
Define f, a scalar function of a vector argument
Write fv=a v1+b v2+c v3+d
Context Panel: Assign Function
fv=a v1+b v2+c v3+d→assign as functionf
Formulate and solve three equations in the unknowns a,b,c,d
Context Panel: Solve≻Solve for Variables≻a,b,c
Context Panel: Assign to a Name≻S
Context Panel: Substitute Into≻fR=0
Context Panel: Evaluate at a Point≻d=−86
fA=0,fB=0,fC=0
2a+3b−c+d=0,5a−7b+2c+d=0,3a+6b+c+d=0
→solve (specified)
a=−29d86,b=−3d86,c=19d86
→substitute into
−2986dx−386dy+1986dz+d=0
→evaluate at point
29x+3y−19z−86=0
Maple Solution - Coded
If the Student MultivariateCalculus package is loaded, then applying the GetRepresentation and Plane commands as shown below will yield the required plane.
GetRepresentationPlane2,3,−1,5,−7,2,3,6,1
−29x−3y+19z=−86
The traditional vector approach can be implemented as follows.
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
A,B,C,R≔2,3,−1,5,−7,2,3,6,1,x,y,z:
Apply the DotProduct and CrossProduct commands
DotProductR−A,CrossProductB−A,C−A=0
<< Chapter Overview Section 1.7 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
