Example 1-7-7 - Maple Help

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Chapter 1: Vectors, Lines and Planes

Section 1.7: Planes

Example 1.7.7

Find the distance between the parallel planes $2 x + 3 y + 4 z = 5$ and $2 x + 3 y + 4 z = 7$ .

Solution

Mathematical Solution

•	Figure 1.7.7(a) shows the parallel planes $S_{1}$ and $S_{2}$ .

•	The red arrow represents the common normal N; the green arrow, the vector $W = B - A$ , where A and B are position vectors to the arbitrary points A and B, respectively; and the black arrow represents $W_{N}$ , the component of W along N.

•	The arbitrary points on the two planes are taken as

A: $(0, 0, 5 / 4)$ and B: $(0, 0, 7 / 4)$

•	The vectors N and W are then

Figure 1.7.7(a) Distance between parallel planes: $d = ∥W_{N}∥$

$N = [\begin{array}{c} 2 \\ 3 \\ 4 \end{array}]$ and $W = B - A = [\begin{array}{c} 0 \\ 0 \\ 7 / 4 \end{array}] - [\begin{array}{c} 0 \\ 0 \\ 5 / 4 \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 1 / 2 \end{array}]$

$d = ∥W_{N}∥$ = $\frac{W \cdot N}{\sqrt{N \cdot N}}$ = $\frac{2}{\sqrt{4 + 9 + 16}} = \frac{2}{\sqrt{29}}$

Maple Solution - Interactive

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Define the two planes and name them $S_{1}$ and $S_{2}$ , respectively

•	Control-drag the equation of each plane.

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

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

$2 x + 3 y + 4 z = 5$ $\overset{make plane}{\to}$ $<< Plane 1 >>$ $\overset{assign to a name}{\to}$ $S_{1}$

$2 x + 3 y + 4 z = 7$ $\overset{make plane}{\to}$ $<< Plane 2 >>$ $\overset{assign to a name}{\to}$ $S_{2}$

Obtain the distance between the two planes

•	Form the sequence of the two names of the planes. Context Panel: Evaluate and Display Inline

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

•	Context Panel: Approximate≻5 (digits)

$S_{1}, S_{2}$ = $<< Plane 1 >>, << Plane 2 >>$ $\overset{distance}{\to}$ $\frac{2}{29} \sqrt{29}$ $\overset{at 5 digits}{\to}$ $0.37140$

The traditional vector-based calculation relies on projection, and is implemented in Table 1.7.7(a), below.

Define position vectors A and B, and the common normal N

•	Enter A as per Table 1.1.1.

•	Context Panel: Assign to a Name≻A

$⟨0, 0, 5 / 4⟩$ $\overset{assign to a name}{\to}$ $A$

•	Enter B as per Table 1.1.1.

•	Context Panel: Assign to a Name≻B

$⟨0, 0, 7 / 4⟩$ $\overset{assign to a name}{\to}$ $B$

•	Enter N as per Table 1.1.1.

•	Context Panel: Assign to a Name≻N

$⟨2, 3, 4⟩$ $\overset{assign to a name}{\to}$ $N$

Obtain $W = B - A$ , a vector from point A to point B , and obtain its scalar projection on N

•	Context Panel: Assign Name

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

•	Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Approximate≻5 (digits)

$\frac{W \cdot N}{\sqrt{N \cdot N}}$ = $\frac{2}{29} \sqrt{29}$ $\overset{at 5 digits}{\to}$ $0.37140$

Table 1.7.7(a) Traditional vector-based calculation of distance between parallel planes

Maple Solution - Coded

Having defined the planes $S_{k}, k = 1, 2$ , within the framework of the "Lines & Planes" tools of the Student MultivariateCalculus package, the distance between the planes is obtained with the Distance command.

$Distance (S_{1}, S_{2})$ = $\frac{2}{29} \sqrt{29}$

With these same tools, the traditional vector-based calculation is implemented in Table 1.7.7(b). The GetPoint command is used to return an arbitrary point on each plane; the convert command changes the resulting list to a Vector, which is then projected onto the normal, N, via the Projection command. The GetNormal command returns N, the common normal. The Norm command returns the length of the projection, which is the requisite distance.

$WN ≔ Projection (convert (GetPoint (S_{2}) - GetPoint (S_{1}), Vector), GetNormal (S_{1}))$

$Norm (WN)$ = $\frac{2}{29} \sqrt{29}$

Table 1.7.7(b) Traditional vector-based calculation via the "Lines & Planes" tools

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