Chapter 1: Vectors, Lines and Planes
Section 1.4: Cross Product
Example 1.4.2
For the vectors A=3 i−2 j+4 k, B=5 i+7 j−6 k, and C=4 i−3 j−5 k,
Verify the distributive property A×B+C=A×B+A×C.
Verify the distributive property A+B×C=A×C+B×C.
Show that A×B×C≠A×B×C.
Verify the identity A×B×C+B×C×A+C×A×B=0.
Verify the identity A×B×C=A·CB−A·BC.
Verify the identity A×B×C=A·CB−B·CA.
Solution
Mathematical Solution
Part (a)
Since B+C=9 i+4 j−11 k, A×B+C is given by
|ijk3−2494−11| = 22−16−(−33−36)12−(−18) = 66930
Since A×B = ijk3−2457−6 = −163831 and A×C = ijk3−244−3−5 = 2231−1, then
A×B+A×C = −163831+2231−1 = 66930.
Part (b)
Since A+B = 8 i+5 j−2 k, A+B×C is given by
|ijk85−24−3−5| = −25−6−(−40−(−8))−24−20 = −3132−44
Since B×C = |ijk57−64−3−5| = −531−43 and A×C = 2231−1, then
A×C+B×C = 2231−1+−531−43 = −3132−44.
Part (c)
A×B×C= |ijk3−24−531−43| = 86−4−(−129−(−212))3−106 = 82−83−103
A×B×C = |ijk−1638314−3−5| = −190−(−93)−(80−124)48−152 = −9744−104
Since 82−83−103≠−9744−104, it is clear that A×B×C≠A×B×C.
Part (d)
From Part (c), A×B×C=82−83−103 and A×B×C=−9744−104, so
C×A×B= −A×B×C= −−9744−104.
Again, from Part (a), A×C=2231−1, so C×A=−A×C=−2231−1. Hence
B×C×A= |ijk57−6−22−311| = 7−186−(5−132)−155+154 = −179127−1
and A×B×C+B×C×A+C×A×B becomes
82−83−103+−179127−1+97−44104=000=0
Part (e)
From Part (c), A×B×C= 82−83−103.
Since A·C=12+6−20= −2, and A·B=15−14−24= −23, it follows that
A·CB−A·BC = −257−6−−234−3−5=82−83−103
Part (f)
From Part (d), A×B×C=−9744−104.
Since A·C=12+6−20= −2, and B·C=20−21+30=29, it follows that
A·CB−B·CA = −257−6−293−24=−9744−104
Maple Solution - Interactive
Initialization
Enter A as per Table 1.1.1.
Context Panel: Assign to a Name≻A
3,−2,4→assign to a nameA
Enter B as per Table 1.1.1.
Context Panel: Assign to a Name≻B
5,7,−6→assign to a nameB
Enter C as per Table 1.1.1.
Context Panel: Assign to a Name≻C
4,−3,−5→assign to a nameC
Common Symbols palette: Cross-product operator
Context Panel: Evaluate and Display Inline
Left-hand Side
Right-hand Side
A×B+C =
A×B+A×C =
A+B×C =
A×C+B×C =
A×B×C =
A×B×C+B×C×A+C×A×B =
Common Symbols palette: Cross-product and dot-product operators
A·CB−A·BC =
A·CB−B·CA =
Maple Solution - Coded
Tools≻Load Package: Student Multivariate Calculus
withStudent:-MultivariateCalculus:
Define the vectors A, B, and C.
A,B,C≔3,−2,4,5,7,−6,4,−3,−5:
Apply the CrossProduct command.
CrossProductA,B+C
CrossProductA,B+CrossProductA,C
CrossProductA+B,C
CrossProductA,C+CrossProductB,C
CrossProductA,CrossProductB,C =
CrossProductCrossProductA,B,C =
CrossProductA,CrossProductB,C+CrossProductB,CrossProductC,A+CrossProductC,CrossProductA,B
Apply the DotProduct and CrossProduct commands.
DotProductA,C⋅B−DotProductA,B⋅C =
DotProductA,C⋅B−DotProductB,C⋅A =
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