Example 1-7-5 - Maple Help

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Chapter 1: Vectors, Lines and Planes

Section 1.7: Planes

Example 1.7.5

Find an equation for $S_{3}$ , the plane that contains the point P: $(2, - 3, 1)$ and the line of intersection of the planes

$S_{1} &colon; 3 x - 7 y - 9 z = 8$ and $S_{2} &colon; 5 x + 4 y - 2 z = 6$

Solution

Mathematical Solution

•	Figure 1.7.5(a) shows the given planes $S_{1}$ and $S_{2}$ in red, and the solution plane $S_{3}$ in green.

•	The red arrow represents $N_{3}$ , the normal to plane $S_{3}$ .

•	The black line is the line of intersection of $S_{1}$ and $S_{2}$ , while the gold arrow represents V, the direction vector for the line of intersection.

•	The black arrow represents the vector $W = P - A$ , where A is an arbitrary point on the line of intersection.

•	The gold "dot" at the head of the black arrow marks the location of point P.

use plots, VectorCalculus in
module()
local p1,p2,p3,p4,p5,p6,M,H,N3:
M:=RootedVector(root=[74/47, -22/47, 0],<2,-3,1>-<74/47, -22/47, 0>):
H:=RootedVector(root=[74/47, -22/47, 0],<50/47, -39/47, 1>):
N3:=RootedVector(root=[74/47, -22/47, 0],<-80/47, 30/47, 110/47>):
p1:=implicitplot3d([3*x-7*y-9*z = 8,5*x+4*y-2*z = 6],x=-5..5,y=-5..5,z=-5..5,transparency=.9,style=surface,color=red):
p2:=spacecurve([74/47+(50/47)*t, -22/47-(39/47)*t, t],t=-3..3,color=black,thickness=5,numpoints=2):
p3:=implicitplot3d([14-8*x+3*y+11*z = 0],x=-5..5,y=-5..5,z=-5..5,style=surface,color=green):
p4:=PlotVector([M,H,N3],color=[black,gold,red],width=.3):
p5:=pointplot3d([2,-3,1],symbol=solidsphere,symbolsize=25,color=gold):
p6:=display(p1,p2,p3,p4,p5,scaling=constrained,axes=none,view=[0..5,-4..2,-2..5],lightmodel=none,orientation=[20,40,-20]):
print(p6);
end module:
end use:

Figure 1.7.5(a) Planes $S_{1}$ , $S_{2}$ (red), and $S_{3}$ (green), vectors V (gold), W (black) and $N_{3}$ (red)

•	Figure 1.7.5(a), which summarizes the path of the calculation, can be rotated with the mouse. The calculations themselves are as follows.

•	If $N_{1}$ and $N_{2}$ are respectively the normals to planes $S_{2}$ and $S_{2}$ , then

$V = N_{1} \times N_{2}$ = $| \begin{array}{c} i & j & k \\ 3 & - 7 & - 9 \\ 5 & 4 & - 2 \end{array} |$ = $[\begin{array}{c} 50 \\ - 39 \\ 47 \end{array}]$

•	An arbitrary point on the line of intersection can be found by setting $z = 0$ in the equations for planes $S_{1}$ and $S_{2}$ , and solving for A = $(74 / 47, - 22 / 47, 0)$ ; hence,

$W = P - A$ = $[\begin{array}{c} 2 \\ - 3 \\ 1 \end{array}] - [\begin{array}{c} 74 / 47 \\ - 22 / 47 \\ 0 \end{array}]$ = $[\begin{array}{c} 20 / 7 \\ - 119 / 47 \\ 1 \end{array}]$

•	The normal to plane $S_{3}$ is then

$N_{3} = V \times W$ = $| \begin{array}{c} i & j & k \\ 50 & - 39 & 47 \\ 20 / 47 & - 119 / 47 & 1 \end{array} |$ = $[\begin{array}{c} 80 \\ - 30 \\ - 110 \end{array}]$

•	Implementing $(R - A) \cdot N_{3} = 0$ as the equation of the plane $S_{3}$ leads to

$([\begin{array}{c} x \\ y \\ z \end{array}] - [\begin{array}{c} 74 / 47 \\ - 22 / 47 \\ 0 \end{array}]) \cdot [\begin{array}{c} 80 \\ - 30 \\ - 110 \end{array}]$ = $80 (x - 74 / 47) - 30 (y + 22 / 47) - 110 z = 0$

which simplify to $8 x - 3 y - 11 z = 14$ .

Maple Solution - Interactive

The tools for defining and manipulating lines and planes in the Student MultivariateCalculus package allow for a much shortened calculation. In effect, once the planes $S_{1}$ and $S_{2}$ are defined, $L$ , the line of their intersection becomes available. Then, plane $S_{3}$ is immediately defined by point P and line $L$ .

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Define planes $S_{1}$ and $S_{2}$

•	Control-drag the equation for plane $S_{1}$

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane

•	Context Panel: Assign to a Name≻S[k] ( $k = 1, 2$ )

$3 x - 7 y - 9 z = 8$ $\overset{make plane}{\to}$ $<< Plane 1 >>$ $\overset{assign to a name}{\to}$ $S_{1}$

$5 x + 4 y - 2 z = 6$ $\overset{make plane}{\to}$ $<< Plane 2 >>$ $\overset{assign to a name}{\to}$ $S_{2}$

Obtain $L$ , the line of intersection of planes $S_{1}$ and $S_{2}$

•	Write the sequence of names for planes $S_{1}$ and $S_{2}$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Line

•	Context Panel: Assign to a Name≻ $L$

$S_{1}, S_{2}$ = $<< Plane 1 >>, << Plane 2 >>$ $\overset{make line}{\to}$ $<< Line 1 >>$ $\overset{assign to a name}{\to}$ $L$

Obtain the equation of plane $S_{3}$ that contains line $L$ and point P: $(2, - 3, 1)$

•	Write the sequence consisting of point P (as a list) and the name $L$ .

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Plane

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation

$[2, - 3, 1], L$ $\overset{make plane}{\to}$ $<< Plane 3 >>$ $\overset{representation}{\to}$ $8 x - 3 y - 11 z = 14$

The traditional, but more tedious vector-based solution, is implemented as follows.

Define $N_{1}$ and $N_{2}$ , the normals to planes $S_{1}$ and $S_{2}$ , respectively

•	Context Panel: Assign to a Name≻N[1]

$⟨3, - 7, - 9⟩$ $\overset{assign to a name}{\to}$ $N_{1}$

•	Context Panel: Assign to a Name≻N[2]

$⟨5, 4, - 2⟩$ $\overset{assign to a name}{\to}$ $N_{2}$

Obtain V, the direction of the line of intersection

•	Common Symbols palette: Cross-product operator

•	Context Panel: Assign Name

$V = N_{1} \times N_{2}$ $\overset{assign}{\to}$

Obtain (as the position vector A), a point on the line of intersection

•	Form a sequence of equations for planes $S_{1}$ , $S_{2}$ Press the Enter key.

•	Context Panel: Evaluate at a Point≻ $z = 0$

•	Context Panel: Solve≻Solve

•	Context Panel: Assign to a Name≻ $s$

•	Expression palette: Evaluation template Evaluate $⟨x, y, 0⟩$ at the solution $s$

•	Context Panel: Assign to a Name≻A

$3 x - 7 y - 9 z = 8, 5 x + 4 y - 2 z = 6$

$\overset{evaluate at point}{\to}$

$3 x - 7 y = 8, 5 x + 4 y = 6$

$\overset{solve}{\to}$

$\{x = \frac{74}{47}, y = - \frac{22}{47}\}$

$\overset{assign to a name}{\to}$

$s$

$\binom{⟨x, y, 0⟩}{} | \binom{}{s}$ = $\overset{assign to a name}{\to}$ $A$

Define the position vectors P and R

•	Context Panel: Assign to a Name≻P

$⟨2, - 3, 1⟩$ $\overset{assign to a name}{\to}$ $P$

•	Context Panel: Assign to a Name≻R

$⟨x, y, z⟩$ $\overset{assign to a name}{\to}$ $R$

Obtain $W = P - A$ , a second vector in plane $S_{3}$

•	Context Panel: Assign Name

$W = P - A$ $\overset{assign}{\to}$

Obtain $N_{3} = V \times W$ , the normal to plane $S_{3}$

•	Common Symbols palette: Cross product operator

•	Context Panel: Assign Name

$N_{3} = V \times W$ $\overset{assign}{\to}$

Implement the vector form of the equation of a plane

•	Common Symbols palette: Dot product operator Press the Enter key.

$(R - A) \cdot N_{3} = 0$

$80 x - 30 y - 110 z - 140 = 0$

Elementary algebraic manipulation leads finally to $8 x - 3 y - 11 z = 14$ .

Maple Solution - Coded

Using the "lines and planes" tools in the Student MultivariateCalculus package, the equation of plane $S_{3}$ would be obtained as per the calculations in Table 1.7.5(a).

•	Use the Plane command to define plane $S_{1}$ .

$S_{1} ≔ Plane (3 x - 7 y - 9 z = 8) &colon;$

•	Use the Plane command to define plane $S_{2}$ .

$S_{2} ≔ Plane (5 x + 4 y - 2 z = 6) &colon;$

•	Use the Line command to define $L$ , the line of intersection of planes $S_{1}$ and $S_{2}$ .

$L ≔ Line (S_{1}, S_{2}) &colon;$

•	Apply the Plane command to point P and line $L$ to obtain plane $S_{3}$ .

$S_{3} ≔ Plane ([2, - 3, 1], L) &colon;$

•	Apply the GetRepresentation command to $S_{3}$ to obtain the equation of this plane.

$GetRepresentation (S_{3})$

$8 x - 3 y - 11 z = 14$

Table 1.7.5(a) The equation of plane $S_{3}$ via the "lines and planes" tools

The traditional vector-based calculation can build on the results in Table 1.7.5(a).

•	Install the Student MultivariateCalculus package.

$with (Student :- MultivariateCalculus) &colon;$

Obtain $V = N_{1} \times N_{2}$

•	Obtain normals to planes $S_{1}$ and $S_{2}$ via the GetNormal command; then apply the CrossProduct command

$V ≔ CrossProduct (GetNormal (S_{1}), GetNormal (S_{2}))$ =

Obtain $W = P - \overset{&Hat;}{A}$ , where $\overset{&Hat;}{A}$ is a random point on line $L$

•	Apply the GetPoint command to line $L$ to obtain an arbitrary point on this line. Use the Vector command to convert the list to a vector.

$W ≔ P - Vector (GetPoint (L))$

Obtain $N_{3} = V \times W$ as the normal to plane $S_{3}$

•	Apply the CrossProduct command.

$N_{3} ≔ CrossProduct (V, W)$

Implement the vector form of the plane $S_{3}$ as $N_{3} \cdot (R - P) = 0$

•	Apply the DotProduct command.

$DotProduct (N_{3}, R - P) = 0$

$80 x - 30 y - 110 z - 140 = 0$

<< Previous Example Section 1.7 Next Example >>

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