Example 1-6-3 - Maple Help

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Chapter 1: Vectors, Lines and Planes

Section 1.6: Lines

Example 1.6.3

The lines $R = A + t P$ and $R = B + s Q$ , are defined respectively by the parametric equations

$x = 3 - 2 t, y = 2 + 5 t, z = 6 + t$ and $x = 5 + 4 s, y = 7 + 2 s, z = 3 + 2 s$

a)	Show that these lines do not intersect and are not parallel (so they are skew lines).

b)	Find the common normal between them.

c)	Calculate the distance between them.

Solution

Mathematical Solution

Part (a)

•	Figure 1.6.3(a) shows the two skew lines, their common normal, and the two parallel planes in which the skew lines can be made to lie.

•	To show that the two lines do not intersect, solve the equations $x (t) = x (s)$ and $y (t) = y (s)$ , that is,

$3 - 2 t = 5 + 4 s, 2 + 5 t = 7 + 2 s$

for $s = - \frac{5}{6}, t = \frac{2}{3}$ , thereby determining $(x, y) = (5 / 3, 16 / 3)$ as the point in the $xy$ -plane above (or below) which the skew lines cross.

use plots in
module()
local p1,p2,p3,p4,p5,L;
p1 := plot3d([13/3+(1/3)*x+(1/3)*y,-1+(1/3)*x+(1/3)*y],x=-10..10,y=-5..15):
p2 := spacecurve([3-2*t, 2+5*t, 6+t], t = -2 .. 3, color = black, thickness = 3, numpoints = 2):
p3 := spacecurve([5+4*s, 7+2*s, 3+2*s], s = -4 .. 2, color = red, thickness = 3, numpoints = 2):
L := [21/11+(16/11)*m, 52/11+(16/11)*m, 72/11-(48/11)*m]:
p4:=spacecurve(L,m=0..1,color=green,thickness=3,numpoints=2):
p5:=display(p1,p2,p3,p4,axes=none,orientation=[50,55,-10],scaling=constrained);
print(p5);
end module:
end use:

Figure 1.6.3(a) Skew lines, common normal, and enclosing parallel planes

•	The corresponding $z$ -coordinates on the two lines are respectively $z (2 / 3) = 20 / 3$ and $z (- 5 / 6) = 4 / 3$ . Since these two values are different, the two lines do not intersect.

•	To show that the lines are not parallel, show that their direction vectors P and Q are not collinear, something that can be done by showing

$N = P \times Q = 8 [\begin{array}{c} 1 \\ 1 \\ - 3 \end{array}]$ $\neq 0$

	Part (b)
	The common normal between these lines is $N = 8 [\begin{array}{c} 1 \\ 1 \\ - 3 \end{array}]$ .

Part (c)

•	Figure 1.6.3(b) shows the skew lines, points A and B, the blue vector

$V = B - A$ = $[\begin{array}{c} 5 \\ 7 \\ 3 \end{array}] - [\begin{array}{c} 3 \\ 2 \\ 6 \end{array}]$ = $[\begin{array}{c} 2 \\ 5 \\ - 3 \end{array}]$

from A to B, and N, in green.

•	The distance between the lines is

$V \cdot N / ∥N∥ = 16 / \sqrt{11}$

the scalar projection of V on N.

Figure 1.6.3(b) Projecting V on N

Maple Solution - Interactive

Part (a)

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Define the two lines, naming them $L_{1}$ and $L_{2}$

•	Write the sequence A, P (or B, Q).

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

•	Context Panel: Assign to a Name≻L[1] (or L[2])

$[3, 2, 6], ⟨- 2, 5, 1⟩$ $\overset{make line}{\to}$ $<< Line 1 >>$ $\overset{assign to a name}{\to}$ $L_{1}$

$[5, 7, 3], ⟨4, 2, 2⟩$ $\overset{make line}{\to}$ $<< Line 2 >>$ $\overset{assign to a name}{\to}$ $L_{2}$

Query the lines

•	Write the sequence of names for the lines. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Skew (or Parallel, or Intersects)

$L_{1}, L_{2}$ = $<< Line 1 >>, << Line 2 >>$ $\overset{skew lines?}{\to}$ $true$

$L_{1}, L_{2}$ = $<< Line 1 >>, << Line 2 >>$ $\overset{parallel?}{\to}$ $false$

$L_{1}, L_{2}$ = $<< Line 1 >>, << Line 2 >>$ $\overset{intersect?}{\to}$ $false$

The first query indicates that the lines are skew; the second and third confirm this by declaring that the lines are neither parallel nor intersecting.

The traditional "textbook" approach to showing the lines do not intersect is to solve the equations $x (t) = x (s)$ and $y (t) = y (s)$ for the $(\overset{&Hat;}{s}, \overset{&Hat;}{t})$ -pair corresponding to the $(x, y)$ -point directly beneath the point in ${&reals;}^{3}$ where the two lines pass over each other. Then, showing that $z (\overset{&Hat;}{t}) \neq z (\overset{&Hat;}{s})$ is sufficient indication that the lines do not intersect.

That the lines don't intersect:

Obtain the "combined vector" form of each line

•	Write the name of the line. Context Panel: Evaluate and Display Inline

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

•	Context Panel: Assign to a Name≻R[k] (k = 1, 2) For the second line, set the parameter on the line to $s$ . (See Figure 1.6.1(c).)

$L_{1}$ = $<< Line 1 >>$ $\overset{representation}{\to}$ $\overset{assign to a name}{\to}$ $R_{1}$

$L_{2}$ = $<< Line 2 >>$ $\overset{representation}{\to}$ $\overset{assign to a name}{\to}$ $R_{2}$

Form the equations $x (t) = x (s)$ and $y (t) = y (s)$

•	Write the notation that equates the first and second components of the two position vectors.

•	Press the Enter key.

•	Context Panel: Solve≻Solve

•	Context Panel: Assign to a Name≻ST

$R_{1} [1] = R_{2} [1], R_{1} [2] = R_{2} [2]$

$3 - 2 t = 5 + 4 s, 2 + 5 t = 7 + 2 s$

$\overset{solve}{\to}$

$\{s = - \frac{5}{6}, t = \frac{2}{3}\}$

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

$ST$

Show that $z (\overset{&Hat;}{t}) \neq z (\overset{&Hat;}{s})$

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

$\binom{R_{1} [3]}{} | \binom{}{ST}$ = $\frac{20}{3}$

$\binom{R_{2} [3]}{} | \binom{}{ST}$ = $\frac{4}{3}$

The traditional "textbook" approach to showing the lines are not parallel is vector-based.

That the lines are not parallel:

•	Write the direction vector for the first line.

•	Context Panel: Assign to a Name≻P

$⟨- 2, 5, 1⟩$ $\overset{assign to a name}{\to}$ $P$

•	Write the direction vector for second line.

•	Context Panel: Assign to a Name≻Q

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

Obtain the cross product of the two direction vectors

•	Write $N = P \times Q$ .

•	Context Panel: Assign Name

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

Exhibit N, the cross product of P and Q

•	Write N. Context Panel: Evaluate and Display Inline

•	Since $N \neq 0$ , P and Q are not parallel.

$N$ =

	Part (b)
	The common normal is the vector $N$ = obtained at the end of Part (a).

Part (c)

•	Write the sequence of names for the line structures. Context Panel: Evaluate and Display Inline

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

$L_{1}, L_{2}$ = $<< Line 1 >>, << Line 2 >>$ $\overset{distance}{\to}$ $\frac{16}{11} \sqrt{11}$

The traditional "textbook" approach to finding the distance between skew lines is vector-based. An arbitrary vector between the two lines is projected onto the common normal. The length of this projection is the distance between the lines. Since points A and B are respectively on each line, the vector $V = B - A$ is a convenient vector to project onto the common normal.

•	Write the position vector A.

•	Context Panel: Assign to a Name≻A

$⟨3, 2, 6⟩$ $\overset{assign to a name}{\to}$ $A$

•	Write the position vector B.

•	Context Panel: Assign to a Name≻B

$⟨5, 7, 3⟩$ $\overset{assign to a name}{\to}$ $B$

•	Write $V = B - A$

•	Context Panel: Assign Name

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

Obtain the length of the projection of V on N

•	Write the sequence of two vectors.

•	Context Panel: Evaluate and Display Inline

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

•	Context Panel: Norm≻Euclidean

$V, N$ = $\overset{projection}{\to}$ $\overset{Euclidean-norm}{\to}$ $\frac{16}{11} \sqrt{11}$

Alternate calculation of the scalar projection

•	Write $V \cdot N / {∥N∥}_{2}$ Context Panel: Evaluate and Display Inline

$V \cdot N / {∥N∥}_{2}$ = $\frac{16}{11} \sqrt{11}$

Maple Solution - Coded

Part (a)

If the Student MultivariateCalculus package has been loaded, and the lines $L_{1}$ and $L_{2}$ defined, then the following commands will determine that these lines are skew.

•	Apply the AreSkew command.

$AreSkew (L_{1}, L_{2})$ = $true$

•	Apply the AreParallel command.

$AreParallel (L_{1}, L_{2})$ = $false$

•	Apply the Intersects command.

$Intersects (L_{1}, L_{2})$ = $false$

A solution from first principles:

•	Apply the solve command to obtain the solution of the equations $x (t) = x (s)$ and $y (t) = y (s)$ .

$q ≔ solve (\{3 - 2 t = 5 + 4 s, 2 + 5 t = 7 + 2 s\})$

$\{s = - \frac{5}{6}, t = \frac{2}{3}\}$

•	Apply the eval command to obtain $z (\overset{&Hat;}{t})$ .

$eval (6 + t, q)$ = $\frac{20}{3}$

•	Apply the eval command to obtain $z (\overset{&Hat;}{s})$ .

$eval (3 + 2 s, q)$ = $\frac{4}{3}$

Since $z (\overset{&Hat;}{t}) = 20 / 3 \neq 4 / 3 = z (\overset{&Hat;}{s})$ , the lines do not intersect. That they are not parallel is established by showing the cross product of the direction vectors is not the zero vector.

That the lines are not parallel:

Show $N = P \times Q \neq 0$
$N ≔ CrossProduct (⟨- 2, 5, 1⟩, ⟨4, 2, 2⟩)$ =

	Part (b)
	The common normal is the vector $N$ = obtained at the end of Part (a).

Part (c)

If V is an arbitrary vector from line $L_{1}$ to line $L_{2}$ , then the distance between the lines is the magnitude of the projection of V onto N, the common normal. The following calculations use the Norm, DotProduct, and Projection commands from the Student MultivariateCalculus package.

•	Define the position vectors A and B.

$A, B ≔ ⟨3, 2, 6⟩, ⟨5, 7, 3⟩ &colon;$

•	Define V, the vector from A to B.

$V ≔ B - A &colon;$

Use the Norm and Projection commands to calculate the length of the projection of V on N

$Norm (Projection (V, N))$ = $\frac{16}{11} \sqrt{11}$

Obtain the scalar projection of V on N from first principles

•	Apply the DotProduct and Norm commands.

$\frac{DotProduct (V, N)}{Norm (N)}$ = $\frac{16}{11} \sqrt{11}$

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