Chapter 1: Vectors, Lines and Planes
Section 1.5: Applications of Vector Products
If A≠0, under what conditions on B and C can A×B=A×C not imply that B=C?
Figure 1.5.14(a) illustrates an example where A×B=A×C but B≠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
Figure 1.5.14(a) A×B=A×C but B≠C
For A×B=A×C to hold, the following three equations must likewise hold.
Depending on which component of A is nonzero, there are three solutions for B, namely,
B1=b1a2a1(b1−c1)+c2a3a1(b1−c1)+c3, B2=a1a2(b2−c2)+c1b2a3a2(b2−c2)+c3, B3=a1a3(b3−c3)+c1a2a3(b3−c3)+c2b3
If a1≠0, the vector B1 satisfies A×B1=A×C. In order that B1≠C, it is necessary that b1≠c1 and at least one of a2 and a3 be nonzero. Any vector B1 satisfying those conditions will satisfy A×B1=A×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×B (or A×C) are respectively
The solution for B was obtained from B3 after choosing A and C arbitrarily, and setting b3=5.
An alternate, but less productive, approach to this example equates the lengths of the cross products, so that A B sinθ1 = A C sinθ2, where θ1 is the angle between A and B, and θ2 is the angle between A and C. All that can be determined from this approach is
B∥C∥ = sinθ2sinθ1
Maple Solution - Interactive
Enter A as per Table 1.1.1.
Context Panel: Assign to a Name≻A
a1,a2,a3→assign to a nameA
Enter B as per Table 1.1.1.
Context Panel: Assign to a Name≻B
b1,b2,b3→assign to a nameB
Enter C as per Table 1.1.1.
Context Panel: Assign to a Name≻C
c1,c2,c3→assign to a nameC
Equate corresponding components of the vectors A×B and A×C
Write the sequence of cross products and press the Enter key.
Context Panel: Equate
Context Panel: Solve≻Solve for Variables≻b,b
Context Panel: Assign to a Name≻S
→assign to a name
Obtain the vector B that corresponds to this solution
Expression palette: Evaluation template
Context Panel: Evaluate and Display Inline
Type in as B1, a factored form of the vector B
Bx=a|f(x)S1 = b1a1⁢c2+a2⁢b1−a2⁢c1a1a1⁢c3+a3⁢b1−a3⁢c1a1
Cyclically permute the indices in B1 to obtain alternate forms of the vector B
Maple Solution - Coded
Install the Student MultivariateCalculus package.
Define the vectors A, B, and C.
Apply the CrossProduct and Equate commands.
Solve for B under the assumptions that a1≠0, a2≠0, or a3≠0
Apply the solve command.
Obtain the vector B that corresponds to each solution
Apply the eval command.
Type in a factored form of each vector.
evalB,S1 = b1a1⁢c2+a2⁢b1−a2⁢c1a1a1⁢c3+a3⁢b1−a3⁢c1a1
evalB,S2 = a1⁢b2−a1⁢c2+a2⁢c1a2b2a2⁢c3+a3⁢b2−a3⁢c2a2
evalB,S3 = a1⁢b3−a1⁢c3+a3⁢c1a3a2⁢b3−a2⁢c3+a3⁢c2a3b3
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