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Chapter 4: Partial Differentiation

Section 4.8: Unconstrained Optimization

Example 4.8.7

Line $L_{1}$ passes through the point P: $(1, 2, 3)$ and has direction $V_{1} = 2 i - 5 j + 4 k$ . Line $L_{2}$ passes through the point Q: $(3, - 1, 2)$ and has direction $V_{2} = 3 i + 7 j - 6 k$ . Show that the lines are skew, and find the minimum distance between them. Hint: Parametrize each line with a different parameter and minimize the square of the distance between an arbitrary point on each line.

Solution

Mathematical Solution

•	Figure 4.8.7(a) shows line $L_{1}$ in black, line $L_{2}$ in red, their common normal in green, and the points between which the distance from one line to the other is measured.

•	Since $V_{1} \times V_{2}$ = $[\begin{array}{c} 2 \\ - 5 \\ 4 \end{array}] \times [\begin{array}{c} 3 \\ 7 \\ - 6 \end{array}] = [\begin{array}{c} 2 \\ 24 \\ 29 \end{array}] \neq 0$ , the lines are not parallel.

•	Parametric equations for lines $L_{1}$ and $L_{2}$ are, respectively

use plots in
module()
local p1,p2,p3,p4,p5,A,B;
A:=[3287/1421,-1823/1421,7995/1421];
B:=[3093/1421,-593/203, 5182/1421];
p1:=spacecurve([1+2*t,2-5*t,3+4*t],t=-.4..1.6,color=black,numpoints=2,thickness=3):
p2:=spacecurve([3+3*s,-1+7*s,2-6*s],s=-.8..0.2,color=red,numpoints=2,thickness=3):
p3:=spacecurve([A,B],color=green,numpoints=2):
p4:=pointplot3d([A,B],symbol=solidsphere,symbolsize=20,color=gold);
p5:=display(p1,p2,p3,p4,scaling=constrained,axes=none,labels=[x,y,z],orientation=[-45,55,-15]);
print(p5);
end module:
end use:

Figure 4.8.7(a) Skew lines and common normal

$x_{1} (t) = 1 + 2 t, y_{1} (t) = 2 - 5 t, z_{1} (t) = 3 + 4 t$ and $x_{2} (s) = 3 + 3 s, y_{2} (s) = - 1 + 7 s, z_{2} (s) = 2 - 6 s$ .

•	The equations $x_{1} (t) = x_{2} (s), y_{1} (t) = y_{2} (s), z_{1} (t) = z_{2} (s)$ are inconsistent, so the lines do not intersect.

•	Hence, the lines are skew.

•	The distance between the lines is given by

$d$	$= \sqrt{{(x_{1} (t) - x_{2} (s))}^{2} + {(y_{1} (t) - y_{2} (s))}^{2} + {(z_{1} (t) - z_{2} (s))}^{2}}$
	$= \sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}$

•	To minimize $d$ , solve the equations

$\frac{\partial d}{\partial t} = \frac{1}{2} \frac{188 s + 106 t - 18}{\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}} = 0$

$\frac{\partial d}{\partial s} = \frac{1}{2} \frac{106 s + 90 t - 30}{\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}} = 0$

for $\overset{&Hat;}{t} = \frac{933}{1421}, \overset{&Hat;}{s} = - \frac{390}{1421}$ .

•	The minimum of $d$ is then $97 \sqrt{29} / 203 ≐ 2.57$ .

•	This minimum distance is measured between the points

$(x_{1} (\overset{&Hat;}{t}), y_{1} (\overset{&Hat;}{t}), z_{1} (\overset{&Hat;}{t})) = (\frac{3287}{1421}, - \frac{1823}{1421}, \frac{7995}{1421}) ≐ (2.3132, - 1.2829, 5.6263)$

and

$(x_{2} (\overset{&Hat;}{s}), y_{2} (\overset{&Hat;}{s}), z_{2} (\overset{&Hat;}{s})) = (\frac{3093}{1421}, - \frac{593}{203}, \frac{5182}{1421}) ≐ (2.1766, - 2.9212, 3.6467)$

Maple Solution - Interactive

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Define vectors $V_{1}$ and $V_{2}$

•	Context Panel: Assign Name

$V_{1} = ⟨2, - 5, 4⟩$ $\overset{assign}{\to}$

$V_{2} = ⟨3, 7, - 6⟩$ $\overset{assign}{\to}$

Define the points P and Q

•	Context Panel: Assign Name

$P = [1, 2, 3]$ $\overset{assign}{\to}$

$Q = [3, - 1, 2]$ $\overset{assign}{\to}$

Form lines $L_{1}$ and $L_{2}$

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

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

•	Context Panel: Assign to a Name≻L[1] or L[2], as appropriate

$P, V_{1}$ = $[1, 2, 3], [\begin{array}{c} 2 \\ −5 \\ 4 \end{array}]$ $\overset{make line}{\to}$ $<< Line 1 >>$ $\overset{assign to a name}{\to}$ $L_{1}$

$Q, V_{2}$ = $[3, −1, 2], [\begin{array}{c} 3 \\ 7 \\ −6 \end{array}]$ $\overset{make line}{\to}$ $<< Line 4 >>$ $\overset{assign to a name}{\to}$ $L_{2}$

Show lines $L_{1}$ and $L_{2}$ are skew lines

•	Write the sequence of names for lines $L_{1}$ and $L_{2}$ . Context Panel: Evaluate and Display Inline

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

$L_{1}, L_{2}$ = $<< Line 10 >>, << Line 11 >>$ $\overset{skew lines?}{\to}$ $true$

Obtain the distance between lines $L_{1}$ and $L_{2}$

•	Write the sequence of names for lines $L_{1}$ and $L_{2}$ . Context Panel: Evaluate and Display Inline

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

•	Context Panel: Approximate≻5 (digits)

$L_{1}, L_{2}$ = $<< Line 1 >>, << Line 2 >>$ $\overset{distance}{\to}$ $\frac{97}{203} \sqrt{29}$ $\overset{at 5 digits}{\to}$ $2.5732$

The distance between the lines can also be found from first principles by minimizing the distance between points on each line.

Obtain the general point on each line

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

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Representation≻combined_vector (Use $t$ as the parameter on one line, and $s$ as the parameter on the other. See Figure 4.8.7(b).)

•	Context Panel: Conversions≻To List

•	Context Panel: Assign to a Name≻ $p$ (first line) and $q$ (second line)

Figure 4.8.7(b) Line representation dialog

$L_{1}$ = $<< Line 10 >>$ $\overset{representation}{\to}$ $[\begin{array}{c} 1 + 2 t \\ 2 - 5 t \\ 3 + 4 t \end{array}]$ $\overset{to list}{\to}$ $[1 + 2 t, 2 - 5 t, 3 + 4 t]$ $\overset{assign to a name}{\to}$ $p$

$L_{2}$ = $<< Line 11 >>$ $\overset{representation}{\to}$ $[\begin{array}{c} 3 + 3 s \\ - 1 + 7 s \\ 2 - 6 s \end{array}]$ $\overset{to list}{\to}$ $[3 + 3 s, - 1 + 7 s, 2 - 6 s]$ $\overset{assign to a name}{\to}$ $q$

Obtain the distance between points $p$ and $q$

•	Write the sequence of names for the generic points and press the Enter key.

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

•	Context Panel: Simplify≻Simplify

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

$p, q$

$[1 + 2 t, 2 - 5 t, 3 + 4 t], [3 + 3 s, - 1 + 7 s, 2 - 6 s]$

$\overset{distance}{\to}$

$\sqrt{{(- 2 + 2 t - 3 s)}^{2} + {(3 - 5 t - 7 s)}^{2} + {(1 + 4 t + 6 s)}^{2}}$

$\overset{simplify}{=}$

$\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}$

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

$d$

Determine the values of $s$ and $t$ that minimize the distance $d$

•	Write $d$ and press the Enter key.

•	Context Panel: Student Multivariate Calculus≻Differentiate≻Gradient

•	Context Panel: Conversions≻To List

•	Context Panel: Solve≻Solve

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

$d$

$\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}$

$\overset{gradient}{\to}$

$\overset{to list}{\to}$

$[\frac{1}{2} \frac{188 s + 106 t - 18}{\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}}, \frac{1}{2} \frac{106 s + 90 t - 30}{\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}}]$

$\overset{solve}{\to}$

$\{s = - \frac{390}{1421}, t = \frac{933}{1421}\}$

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

$S$

Evaluate $d$ at the values in $S$

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

$\binom{d}{} | \binom{}{S}$ = $\frac{1}{1421} \sqrt{9409} \sqrt{1421}$ $\overset{simplify}{=}$ $\frac{97}{203} \sqrt{29}$

Finally, the two points that are closest, are found by substituting into the generic points $p$ and $q$ , the values in $S$ .

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

$\binom{p}{} | \binom{}{S}$ = $[\frac{3287}{1421}, - \frac{1823}{1421}, \frac{7995}{1421}]$

$\binom{q}{} | \binom{}{S}$ = $[\frac{3093}{1421}, - \frac{593}{203}, \frac{5182}{1421}]$

Maple Solution - Coded

Initialize

•	Install the Student MultivariateCalculus package.

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

•	Define the directions of the two lines.

$V_{1}, V_{2} ≔ ⟨2, - 5, 4⟩, ⟨3, 7, - 6⟩ &colon;$

•	Define the points through which each line passes.

$P, Q ≔ [1, 2, 3], [3, - 1, 2] &colon;$

•	Use the Line command to define the two given lines.

$L_{1}, L_{2} ≔ Line (P, V_{1}), Line (Q, V_{2}) &colon;$

Show the lines are skew

•	Use the AreSkew command to test for skewness.

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

Calculate the distance between the lines

•	Apply the simplify command to the result of the Distance command.

$simplify (Distance (L_{1}, L_{2}))$ = $\frac{97}{203} \sqrt{29}$

Work from first principles

•	Apply the GetRepresentation command to each line, returning the equations as vectors assigned to the names A and B, respectively. These names then represent generic points on the lines.

$A ≔ GetRepresentation (L_{1}, form = combined_vector, parameter = t)$ =

$B ≔ GetRepresentation (L_{2}, form = combined_vector, parameter = s)$ =

Obtain an expression for the distance between points A and B

Apply the convert/list command to A and B, and to this, simplify the Distance between them.

$d ≔ simplify (Distance (convert (A, list), convert (B, list)))$

$\sqrt{94 s^{2} + 106 s t + 45 t^{2} - 18 s - 30 t + 14}$

Minimize $d$ by forming and solving the equations $\nabla d = 0$

•	Obtain $\nabla d$ with the Gradient command, and the Equate command to form the componentwise equations in $\nabla d = 0$ .

•	Apply the solve command to obtain the minimizing values of $s$ and $t$ .

•	Assign the solution to the name $S$ .

$S ≔ solve (Equate (Gradient (d, [t, s]), ⟨0, 0⟩))$ = $\{s = - \frac{390}{1421}, t = \frac{933}{1421}\}$

By substitution, obtain the value of the minimum distance

Use the eval and simplify commands to obtain the exact value of the minimum distance, and the evalf command to obtain this distance as a floating-point number.

$MinDist ≔ simplify (eval (d, S))$ = $\frac{97}{203} \sqrt{29}$

$evalf (MinDist)$ = $2.573206829$

By substitution, obtain the coordinates of the points between which the minimum distance occurs

•	Change the vectors A and B to lists with the convert/list command, and make the substitution with the eval command.

$convert (eval (A, S), list)$

$[\frac{3287}{1421}, - \frac{1823}{1421}, \frac{7995}{1421}]$

$convert (eval (B, S), list)$

$[\frac{3093}{1421}, - \frac{593}{203}, \frac{5182}{1421}]$

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