Chapter 1: Vectors, Lines and Planes
Section 1.7: Planes
Obtain an equation for the plane that contains the points P:5,−3,7, and Q:1,2,−3, and that is parallel to the vector V=2 i−3 j+4 k.
Figure 1.7.3(a) suggests how the requisite plane can be found. Position vectors P and Q, emanating from the origin O, respectively have their heads at points P and Q, which lie in the plane.
Since the plane is parallel to the direction V, the vector V (in red) must lie in the plane. A second vector, W=Q−P, also lies in the plane, so the normal to the plane is N=V×W.
The equation of the plane now follows from the vector equation R−P·N=0.
Begin by obtaining the vectors
Figure 1.7.3(a) Position vectors P and Q (blue), direction vector V (red), W=Q−P (black), and N=V×W (green)
W=Q−P = 12−3−5−37=−45−10 and N=V×W = |ijk2−34−45−10| = 104−2
Conclude by implementing the vector equation of the plane, namely,
(xyz−5−37)·104−2 = 10x−5+4y+3−2z−7=0
from which follows 10 x+4 y−2 z=50−12−14=24. A simpler form for the equation is clearly
5 x+2 y−z=12
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Write the sequence of points P and Q, and vector V.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation
5,−3,7,1,2,−3,2,−3,4→make plane<< Plane 1 >>→representation5⁢x+2⁢y−z=12
The traditional vector approach to finding this plane would be as follows.
Define the position vectors P, Q, and R, and the direction vector V
Enter P as per Table 1.1.1.
Context Panel: Assign to a Name≻P
5,−3,7→assign to a nameP
Enter Q as per Table 1.1.1.
Context Panel: Assign to a Name≻Q
1,2,−3→assign to a nameQ
Enter R as per Table 1.1.1.
Context Panel: Assign to a Name≻R
x,y,z→assign to a nameR
Enter V as per Table 1.1.1.
Context Panel: Assign to a Name≻V
2,−3,4→assign to a nameV
Obtain the vector W=Q−P
Context Panel: Assign Name
Obtain the normal N=V×W
Implement the vector equation of the line
Common Symbols palette: Dot product operator
Press the Enter key.
Simplify the equation to the form 5 x+2 y−z=12 by the obvious manipulations.
In agreement with what is stated in the Mathematical Solution, the two calculated vectors are
W = −45−10
N = 104−2
The requisite plane can also be obtained by an algebraic calculation.
The general equation of the plane, namely, a x+b y+c z−d=0, must be satisfied by the coordinates of points P and Q; also, the normal (whose components are a,b,c) must be orthogonal to the vector V. These three conditions generate three equations in the four unknowns, namely, a,b,c,d. Hence a solution for a,b,c in terms of d is possible, and leads to the desired equation of the plane with a judicious choice for d. An interactive calculation along these lines is given below.
Write fv=n·v1,v2,v3− d
Context Panel: Assign Function
fv=n·v1,v2,v3−d→assign as functionf
Write n·V=0,fP=0,fQ=0 and press the Enter key.
Context Panel: Solve≻Solve for Variables≻a,b,c
Context Panel: Assign to a Name≻S
→assign to a name
Expression palette: Evaluation template
Press the Enter key
Context Panel: Simplify≻Assuming Real
Context Panel: Evaluate at a Point≻d=12
→evaluate at point
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the position vectors P, Q, R, and the direction vector V
Define the vectors P, Q, R, and V.
Implement the vector equation for the plane
Apply the DotProduct and CrossProduct commands.
<< Previous Example Section 1.7
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