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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 $N = P \times Q$ .

•	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 $(R - A) \cdot N = 0$ .

>	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 $N = P \times Q$ (gold)

The requisite computations are then

$P = [\begin{array}{c} 5 \\ - 7 \\ 2 \end{array}] - [\begin{array}{c} 2 \\ 3 \\ - 1 \end{array}] = [\begin{array}{c} 3 \\ - 10 \\ 3 \end{array}]$

$Q = [\begin{array}{c} 3 \\ 6 \\ 1 \end{array}] - [\begin{array}{c} 2 \\ 3 \\ - 1 \end{array}] = [\begin{array}{c} 1 \\ 3 \\ 2 \end{array}]$

$N = P \times Q = | \begin{array}{c} i & j & k \\ 3 & - 10 & 3 \\ 1 & 3 & 2 \end{array} | = [\begin{array}{c} - 29 \\ - 3 \\ 19 \end{array}]$

and

$(R - A) \cdot N$ = $[\begin{array}{c} x - 2 \\ y - 3 \\ z + 1 \end{array}] \cdot [\begin{array}{c} - 29 \\ - 3 \\ 19 \end{array}]$ = $- 29 (x - 2) - 3 (y - 3) + 19 (z + 1) = 0$

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

$[2, 3, - 1], [5, - 7, 2], [3, 6, 1]$ $\overset{make plane}{\to}$ $<< Plane 1 >>$ $\overset{representation}{\to}$ $- 29 x - 3 y + 19 z = −86$

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 $N = P \times Q$ 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 $(R - A) \cdot N = 0$ .

Define the position vectors A, B, C, and R

•	Enter A as per Table 1.1.1.

•	Context Panel: Assign to a Name≻A

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

•	Enter B as per Table 1.1.1.

•	Context Panel: Assign to a Name≻B

$⟨5, - 7, 2⟩$ $\overset{assign to a name}{\to}$ $B$

•	Enter C as per Table 1.1.1.

•	Context Panel: Assign to a Name≻C

$⟨3, 6, 1⟩$ $\overset{assign to a name}{\to}$ $C$

•	Enter R as per Table 1.1.1.

•	Context Panel: Assign to a Name≻R

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

Define the vectors P and Q

•	Context Panel: Assign Name

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

•	Context Panel: Assign Name

$Q = C - A$ $\overset{assign}{\to}$

Obtain the normal N

•	Context Panel: Assign Name

$N = P \times Q$ $\overset{assign}{\to}$

Impose the orthogonality condition on $R - A$ and N

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

$(R - A) \cdot N = 0$

$- 29 x - 3 y + 19 z + 86 = 0$

For the reader's convenience, the vectors P, Q, and N are displayed here:

$P, Q, N$ = $[\begin{array}{c} 3 \\ −10 \\ 3 \end{array}], [\begin{array}{c} 1 \\ 3 \\ 2 \end{array}], [\begin{array}{c} −29 \\ −3 \\ 19 \end{array}]$

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 $f (v) = a v_{1} + b v_{2} + c v_{3} + d$

•	Context Panel: Assign Function

$f (v) = a v_{1} + b v_{2} + c v_{3} + d$ $\overset{assign as function}{\to}$ $f$

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≻ $f (R) = 0$

•	Context Panel: Evaluate at a Point≻ $d = - 86$

$f (A) = 0, f (B) = 0, f (C) = 0$

$2 a + 3 b - c + d = 0, 5 a - 7 b + 2 c + d = 0, 3 a + 6 b + c + d = 0$

$\overset{solve (specified)}{\to}$

$\{a = - \frac{29 d}{86}, b = - \frac{3 d}{86}, c = \frac{19 d}{86}\}$

$\overset{substitute into}{\to}$

$- \frac{29}{86} d x - \frac{3}{86} d y + \frac{19}{86} d z + d = 0$

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

$29 x + 3 y - 19 z - 86 = 0$

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.

$GetRepresentation (Plane ([2, 3, - 1], [5, - 7, 2], [3, 6, 1]))$

$- 29 x - 3 y + 19 z = −86$

The traditional vector approach can be implemented as follows.

•	Install the Student MultivariateCalculus package.

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

Define the position vectors A, B, C, and R

$A, B, C, R ≔ ⟨2, 3, - 1⟩, ⟨5, - 7, 2⟩, ⟨3, 6, 1⟩, ⟨x, y, z⟩ &colon;$

Apply the DotProduct and CrossProduct commands

$DotProduct (R - A, CrossProduct (B - A, C - A)) = 0$

$- 29 x - 3 y + 19 z + 86 = 0$

<< Chapter Overview Section 1.7 Next Example >>

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