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Chapter 1: Vectors, Lines and Planes

Section 1.5: Applications of Vector Products

Example 1.5.14

If $A \neq 0$ , under what conditions on B and C can $A \times B = A \times C$ not imply that $B = C$ ?

Solution

Mathematical Solution

•	Figure 1.5.14(a) illustrates an example where $A \times B = A \times C$ but $B \neq C$ . The vectors A, B, and C must necessarily lie in the same plane and A must be to the right (or left) of both B and C.

•	The following analysis is the basis for constructing any such example.

•	Start with vectors A, B, C, respectively given by

$[\begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array}], [\begin{array}{c} b_{1} \\ b_{2} \\ b_{3} \end{array}], [\begin{array}{c} c_{1} \\ c_{2} \\ c_{3} \end{array}]$

Figure 1.5.14(a) $A \times B = A \times C$ but $B \neq C$

•	For $A \times B = A \times C$ to hold, the following three equations must likewise hold.

$a_{2} b_{3} - a_{3} b_{2}$	$= a_{2} c_{3} - a_{3} c_{2}$
$- a_{1} b_{3} + a_{3} b_{1}$	$= - a_{1} c_{3} + a_{3} c_{1}$
$a_{1} b_{2} - a_{2} b_{1}$	$= a_{1} c_{2} - a_{2} c_{1}$

•	Depending on which component of A is nonzero, there are three solutions for B, namely,

$B_{1} = [\begin{array}{c} b_{1} \\ \frac{a_{2}}{a_{1}} (b_{1} - c_{1}) + c_{2} \\ \frac{a_{3}}{a_{1}} (b_{1} - c_{1}) + c_{3} \end{array}]$ , $B_{2} = [\begin{array}{c} \frac{a_{1}}{a_{2}} (b_{2} - c_{2}) + c_{1} \\ b_{2} \\ \frac{a_{3}}{a_{2}} (b_{2} - c_{2}) + c_{3} \end{array}]$ , $B_{3} = [\begin{array}{c} \frac{a_{1}}{a_{3}} (b_{3} - c_{3}) + c_{1} \\ \frac{a_{2}}{a_{3}} (b_{3} - c_{3}) + c_{2} \\ b_{3} \end{array}]$

•

If $a_{1} \neq 0$ , the vector $B_{1}$ satisfies $A \times B_{1} = A \times C$ . In order that $B_{1} \neq C$ , it is necessary that $b_{1} \neq c_{1}$ and at least one of $a_{2}$ and $a_{3}$ be nonzero. Any vector $B_{1}$ satisfying those conditions will satisfy $A \times B_{1} = A \times C$ and will not equal C.

•	A similar analysis can be made for each of the other two cases.

•	The vectors shown in Figure 1.5.14(a), namely, A, B, C, and $A \times B$ (or $A \times C$ ) are respectively

$[\begin{array}{r} 1 \\ 2 \\ 3 \end{array}], [\begin{array}{c} \frac{1}{3} \\ \frac{2}{3} \\ 5 \end{array}], [\begin{array}{r} 1 \\ 2 \\ 7 \end{array}], [\begin{array}{r} 8 \\ - 4 \\ 0 \end{array}]$

The solution for B was obtained from $B_{3}$ after choosing A and C arbitrarily, and setting $b_{3} = 5$ .

•

An alternate, but less productive, approach to this example equates the lengths of the cross products, so that $∥A ∥∥ B∥$ $\sin (θ_{1})$ = $∥A∥$ $∥C∥$ $\sin (θ_{2})$ , where $θ_{1}$ is the angle between A and B, and $θ_{2}$ is the angle between A and C. All that can be determined from this approach is

$\frac{∥B∥}{&par; C &par;}$ = $\frac{\sin (θ_{2})}{\sin (θ_{1})}$

Maple Solution - Interactive

Initialize

•	Enter A as per Table 1.1.1.

•	Context Panel: Assign to a Name≻A

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

•	Enter B as per Table 1.1.1.

•	Context Panel: Assign to a Name≻B

$⟨b_{1}, b_{2}, b_{3}⟩$ $\overset{assign to a name}{\to}$ $B$

•	Enter C as per Table 1.1.1.

•	Context Panel: Assign to a Name≻C

$⟨c_{1}, c_{2}, c_{3}⟩$ $\overset{assign to a name}{\to}$ $C$

Equate corresponding components of the vectors $A \times B$ and $A \times C$

•	Write the sequence of cross products and press the Enter key.

•	Context Panel: Equate

•	Context Panel: Solve≻Solve for Variables≻b[2],b[3]

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

$A \times B, A \times C$

$\overset{equate}{\to}$

$[a_{2} b_{3} - a_{3} b_{2} = a_{2} c_{3} - a_{3} c_{2}, - a_{1} b_{3} + a_{3} b_{1} = - a_{1} c_{3} + a_{3} c_{1}, a_{1} b_{2} - a_{2} b_{1} = a_{1} c_{2} - a_{2} c_{1}]$

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

$\{b_{2} = \frac{a_{1} c_{2} + a_{2} b_{1} - a_{2} c_{1}}{a_{1}}, b_{3} = \frac{a_{1} c_{3} + a_{3} b_{1} - a_{3} c_{1}}{a_{1}}\}$

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

$S_{1}$

Obtain the vector B that corresponds to this solution

•	Expression palette: Evaluation template

•	Context Panel: Evaluate and Display Inline

•	Type in as $B_{1}$ , a factored form of the vector B

$\binom{B}{} | \binom{}{S_{1}}$ =

$B_{1} = [\begin{array}{c} b_{1} \\ \frac{a_{2}}{a_{1}} (b_{1} - c_{1}) + c_{2} \\ \frac{a_{3}}{a_{1}} (b_{1} - c_{1}) + c_{3} \end{array}]$

Cyclically permute the indices in $B_{1}$ to obtain alternate forms of the vector B

$B_{2} = [\begin{array}{c} \frac{a_{1}}{a_{2}} (b_{2} - c_{2}) + c_{1} \\ b_{2} \\ \frac{a_{3}}{a_{2}} (b_{2} - c_{2}) + c_{3} \end{array}]$

$B_{3} = [\begin{array}{c} \frac{a_{1}}{a_{3}} (b_{3} - c_{3}) + c_{1} \\ \frac{a_{2}}{a_{3}} (b_{3} - c_{3}) + c_{2} \\ b_{3} \end{array}]$

A similar analysis can be made for each of the other two cases.

Maple Solution - Coded

Initialize

•	Install the Student MultivariateCalculus package.

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

•	Define the vectors A, B, and C.

$A, B, C ≔ ⟨a_{1}, a_{2}, a_{3}⟩, ⟨b_{1}, b_{2}, b_{3}⟩, ⟨c_{1}, c_{2}, c_{3}⟩ &colon;$

Equate corresponding components of the vectors $A \times B$ and $A \times C$

•	Apply the CrossProduct and Equate commands.

$q ≔ Equate (CrossProduct (A, B), CrossProduct (A, C))$

$[a_{2} b_{3} - a_{3} b_{2} = a_{2} c_{3} - a_{3} c_{2}, - a_{1} b_{3} + a_{3} b_{1} = - a_{1} c_{3} + a_{3} c_{1}, a_{1} b_{2} - a_{2} b_{1} = a_{1} c_{2} - a_{2} c_{1}]$

Solve for B under the assumptions that $a_{1} \neq 0$ , $a_{2} \neq 0$ , or $a_{3} \neq 0$

•	Apply the solve command.

$S_{1} ≔ solve (q, \{b_{2}, b_{3}\}) &semi; S_{2} ≔ solve (q, \{b_{1}, b_{3}\}) &semi; S_{3} ≔ solve (q, \{b_{1}, b_{2}\})$

$\{b_{2} = \frac{a_{1} c_{2} + a_{2} b_{1} - a_{2} c_{1}}{a_{1}}, b_{3} = \frac{a_{1} c_{3} + a_{3} b_{1} - a_{3} c_{1}}{a_{1}}\}$

Obtain the vector B that corresponds to each solution

•	Apply the eval command.

•	Type in a factored form of each vector.

$eval (B, S_{1})$ =

$B_{1} = [\begin{array}{c} b_{1} \\ \frac{a_{2}}{a_{1}} (b_{1} - c_{1}) + c_{2} \\ \frac{a_{3}}{a_{1}} (b_{1} - c_{1}) + c_{3} \end{array}]$

$eval (B, S_{2})$ =

$B_{2} = [\begin{array}{c} \frac{a_{1}}{a_{2}} (b_{2} - c_{2}) + c_{1} \\ b_{2} \\ \frac{a_{3}}{a_{2}} (b_{2} - c_{2}) + c_{3} \end{array}]$

$eval (B, S_{3})$ =

$B_{3} = [\begin{array}{c} \frac{a_{1}}{a_{3}} (b_{3} - c_{3}) + c_{1} \\ \frac{a_{2}}{a_{3}} (b_{3} - c_{3}) + c_{2} \\ b_{3} \end{array}]$

A similar analysis can be made for each of the other two cases.

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