Chapter 1: Vectors, Lines and Planes
Section 1.4: Cross Product
Example 1.4.1
For the vectors A=3 i−2 j+4 k and B=5 i+7 j−6 k, and the scalar c,
Obtain A×B.
Obtain θ, the angle between A and B.
Verify computationally that A×A=0.
Verify computationally that A×B=A B sin(θ).
Verify computationally that B×A= −A×B.
Verify that both A and B are perpendicular to A×B.
Verify that A·B2+A×B2 = A B2.
Verify that A·B2−A×B2 = A B2cos(2 θ).
Show that A×A×B≠A×A×B=0.
Verify that c A×B=cA×B=A×c B.
Verify that A−B×A+B=2A×B.
Solution
Mathematical Solution
Part (a)
A×B = |ijk3−2457−6|
= i −247−6−j |345−6|+k |3−257|
=i 12−28−j −18−20+k 21+10
=−16 i+38 j+31 k
Part (b)
The angle between two vectors A and B can be obtained from the following expression.
cosθ=A·BA B=A·BA·A B·B
The necessary numbers are then
A·B=3−24·57−6
= 15−14−24= −23
A·A=3−24·3−24
= 9+4+16=29
B·B=57−6·57−6
= 25+49+36=110
Finally, the angle θ is given by
θ=cos−1−2329 110 = π−cos−12329 110 ≐ 1.99 radians
Part (c)
A×A=ijk3−243−24 = −2⋅4−4⋅(−2)−(3⋅4−4⋅3)3⋅(−2)−(−2)⋅3 = 000=0
Part (d)
The left-hand side:
A×B=−162+382+312=2661
The right-hand side:
A B cosθ
=29 110 sincos−1−2329 110
=29 110 266129 110
=2661
With appropriate care for quadrants, the evaluation of x=sincos−1a/b is effected by setting y=cos−1a/b so that cosy=a/b and x=siny.
Then, in the right triangle drawn in Figure 1.4.1(a), x=siny=b2−a2/b.
Figure 1.4.1(a) sincos−1a/b
Part (e)
Interchanging two adjacent rows in a determinant negates the value of the determinant. However, the relevant computation is shown below.
B×A = |ijk57−63−24| = 7⋅4−(−6)⋅(−2)−(5⋅4−(−6)⋅3)5⋅(−2)−7⋅3 = 16−38−31 = −−163831 = −A×B
Part (f)
If the dot product of two vectors is zero, the vectors are perpendicular.
A·A×B = 3−24·−163831 = −48−76+124=0
B·A×B = 57−6·−163831 = −80+266−186=0
Part (g)
The relevant calculations are summarized below.
Left-hand Side
A·B2+A×B2
= −232+26612=529+2661=3190
Right-hand Side
A B2
= 29 1102=29⋅110=3190
Part (h)
A·B2− A×B2
= −232− 26612=529− 2661=−2132
A B2cos(2 θ)
= 29 1102−10661595=−2132
Since cosθ=−2329 110, the double-angle formula cos2 θ=2 cos2θ−1 is used to obtain cos2 θ=−10661595.
Part (i)
The relevant calculations are shown below.
A×A×B= |ijk3−24−163831| = −2⋅31−4⋅38−(3⋅31−4⋅(−16))3⋅38−(−2)⋅(−16) = −214−15782
Using the result of Part (c),
A×A×B = |ijk00057−6| = 0⋅(−6)−0⋅7−(0⋅(−6)−0⋅5)0⋅7−0⋅5 = 000 = 0
Part (j)
c A×B
= |ijk3 c−2 c4 c57−6| = (−2 c)⋅(−6)−4 c⋅7−(3 c⋅(−6)−4 c⋅5)3 c⋅7−(−2 c)⋅5 = −16 c38 c31 c= c−163831 = cA×B
