Chapter 1: Vectors, Lines and Planes
Section 1.4: Cross Product
Example 1.4.8
Starting with the length, orthogonality, and right-handedness properties in Table 1.4.1, derive the expression for the cross product stated in Definition 1.4.1.
Solution
Mathematical Solution
Let the cross product of A=a1 i+a2 j+a3 k and B=b1 i+b2 j+b3 k be given by the vector C=c1 i+c2 j+c3 k. Solve the three equations
A·C=a1⁢c1+a2⁢c2+a3⁢c3=0
B·C=b1⁢c1+b2⁢c2+b3⁢c3=0
c12+c22+c32=a12+a22+a32⁢b12+b22+b32⁢1−a1⁢b1+a2⁢b2+a3⁢b32a12+a22+a32⁢b12+b22+b32
for
c1c2c3=a2⁢b3−a3⁢b2a3⁢b1−a1⁢b3a1⁢b2−a2⁢b1 and c1c2c3=−a2⁢b3+a3⁢b2a1⁢b3−a3⁢b1−a1⁢b2+a2⁢b1.
The first two equations express the orthogonality of A and B with C, and the third equation expresses the length condition C = A B sinθ, where θ=cos−1A·BA B is the angle between A and B.
Only one of the two solutions, namely the first, satisfies the right-hand rule. Hence, that is the expression in Definition 1.4.1 for the cross product.
The simplest way to decide which of the two solutions obeys the right-hand rule is to note that the unit basis vectors i,j,k form a right-handed system. Hence, i×j=k, and it is just the first of the two solutions shown above that satisfies this relation. (The second solution would yield i×j= −k.)
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
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
Obtain and solve: three equations expressing two orthogonality conditions and the length condition
Common Symbols palette: Dot-product operator
Typeset math notation for the norms.
Context Panel: Solve≻Solve for Variables≻c1,c2,c3
A·C=0,B·C=0,∥C∥ = A⋅∥B∥⋅sin(arccos(A·B∥A∥⋅B))
a1⁢c1+a2⁢c2+a3⁢c3=0,b1⁢c1+b2⁢c2+b3⁢c3=0,c12+c22+c32=a12+a22+a32⁢b12+b22+b32⁢1−a1⁢b1+a2⁢b2+a3⁢b32a12+a22+a32⁢b12+b22+b32
→solve (specified)
c1=a2⁢b3−a3⁢b2,c2=−a1⁢b3+a3⁢b1,c3=a1⁢b2−a2⁢b1,c1=−a2⁢b3+a3⁢b2,c2=a1⁢b3−a3⁢b1,c3=−a1⁢b2+a2⁢b1
Evaluate C for each of the two solutions found
Expression palette: Evaluation template Reference solutions via equation label and the selection brackets.
Cx=a|f(x)1 =
Cx=a|f(x)2 =
Only one of these two solutions obeys the right-hand rule; the one that does is then the cross product.
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the vectors A, B, and C.
A,B,C≔a1,a2,a3,b1,b2,b3,c1,c2,c3:
Obtain the equations expressing the orthogonality of A and B with C
Apply the DotProduct command.
q1≔DotProductA,C:
q2≔DotProductB,C:
Obtain θ, the angle between vectors A and B
Apply the Angle command.
θ≔AngleA,B:
Obtain an equation expressing the length condition
Apply the Norm command.
q3≔NormC=NormA⋅NormB⋅sinθ:
Solve the three relevant equations for the components of C
Apply the solve command.
S≔solveq1,q2,q3,c1,c2,c3:
Obtain the two possible solutions for C
Use the eval command to evaluate C for each of the two solutions obtained by solve.
evalC,S1 =
evalC,S2 =
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