James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

816 Chapter 12 Vectors and the Geometry of Space
Equation 6. The symbolic formula in Equation 7 is probably the easiest way of remembering
and computing cross products.
ExamplE 1 If a − k1, 3, 4l and b − k2, 7, 25l, then
Z Z
i j k
a 3 b − 1 3 4
2 7 25
− Z 3 7
4
25
Z i 2 Z 1 2
4
25
Z j 1 Z 1 2
3
7
Z k
− s215 2 28d i 2 s25 2 8d j 1 s7 2 6d k − 243i 1 13j 1 k
■
ExamplE 2 Show that a 3 a − 0 for any vector a in V 3 .
SOLUTION If a − k a 1 , a 2 , a 3 l, then
Z Z
i j k
a 3 a − a 1 a 2 a 3
a 1
a 2
a 3
− sa 2 a 3 2 a 3 a 2 d i 2 sa 1 a 3 2 a 3 a 1 d j 1 sa 1 a 2 2 a 2 a 1 d k
− 0 i 2 0 j 1 0 k − 0 ■
We constructed the cross product a 3 b so that it would be perpendicular to both a
and b. This is one of the most important properties of a cross product, so let’s emphasize
and verify it in the following theorem and give a formal proof.
8 Theorem The vector a 3 b is orthogonal to both a and b.
Proof In order to show that a 3 b is orthogonal to a, we compute their dot product as
follows:
sa 3 bd ? a − Z a 2
b 2
a 3
b 3
Z
a 1 2 Z a 1 a 3
b 1 b 3
Z a 2 1 Z a 1 a 2
b 1 b 2
Z a 3
− a 1 sa 2 b 3 2 a 3 b 2 d 2 a 2 sa 1 b 3 2 a 3 b 1 d 1 a 3 sa 1 b 2 2 a 2 b 1 d
axb
n
− a 1 a 2 b 3 2 a 1 b 2 a 3 2 a 1 a 2 b 3 1 b 1 a 2 a 3 1 a 1 b 2 a 3 2 b 1 a 2 a 3
− 0
a
¨
b
A similar computation shows that sa 3 bd ? b − 0. Therefore a 3 b is orthogonal to
both a and b.
■
FIGURE 1
The right-hand rule gives the direction
of a 3 b.
If a and b are represented by directed line segments with the same initial point (as
in Figure 1), then Theorem 8 says that the cross product a 3 b points in a direction
perpendicu lar to the plane through a and b. It turns out that the direction of a 3 b is
given by the right-hand rule: if the fingers of your right hand curl in the direction of
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