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

Section 12.4 The Cross Product 815
Hamilton
The cross product was invented by
the Irish mathematician Sir William
Rowan Hamilton (1805–1865), who
had created a precursor of vectors,
called quaternions. When he was five
years old Hamilton could read Latin,
Greek, and Hebrew. At age eight he
added French and Italian and when
ten he could read Arabic and Sanskrit.
At the age of 21, while still an undergraduate
at Trinity College in Dublin,
Hamilton was appointed Professor
of Astronomy at the university and
Royal Astronomer of Ireland!
4 Definition If a − ka 1 , a 2 , a 3 l and b − kb 1 , b 2 , b 3 l, then the cross product
of a and b is the vector
a 3 b − ka 2 b 3 2 a 3 b 2 , a 3 b 1 2 a 1 b 3 , a 1 b 2 2 a 2 b 1 l
Notice that the cross product a 3 b of two vectors a and b, unlike the dot product, is
a vector. For this reason it is also called the vector product. Note that a 3 b is defined
only when a and b are three-dimensional vectors.
In order to make Definition 4 easier to remember, we use the notation of determinants.
A determinant of order 2 is defined by
Z
a
c
b
d Z − ad 2 bc
(Multiply across the diagonals and subtract.) For example,
Z
2
26
1
4
Z − 2s4d 2 1s26d − 14
A determinant of order 3 can be defined in terms of second-order determinants as
follows:
5
Z
a
1
b 1
a 2 a 3
b 2 b 3
c 3
Z
b
− a 1 Z
2 b 3
c 2 c 3
Z 2 a 2 Z b 1 b 3
c 1 c 3
Z 1 a 3 Z b 1 b 2
c 1 c 2
Z
Observe that each term on the right side of Equation 5 involves a number a i in the first
row of the determinant, and a i is multiplied by the second-order determinant obtained
from the left side by deleting the row and column in which a i appears. Notice also the
minus sign in the second term. For example,
Z
1
3
25
2
0
4
21
1
2
Z
0
− 1 Z
4
1 3
2
Z 2 2 Z
25
1
3
2
Z 1 s21d Z
25
0
4
Z
− 1s0 2 4d 2 2s6 1 5d 1 s21ds12 2 0d − 238
If we now rewrite Definition 4 using second-order determinants and the standard basis
vectors i, j, and k, we see that the cross product of the vectors a − a 1 i 1 a 2 j 1 a 3 k and
b − b 1 i 1 b 2 j 1 b 3 k is
6 a 3 b − Z
a 2
b 2
a 3
b 3
Z
i 2 Z a 1 a 3
b 1 b 3
Z j 1 Z a 1 a 2
b 1 b 2
Z k
In view of the similarity between Equations 5 and 6, we often write
Z
i j k
Z
7 a 3 b − a 1 a 2 a 3
b 3
b 1
Although the first row of the symbolic determinant in Equation 7 consists of vectors, if
we expand it as if it were an ordinary determinant using the rule in Equation 5, we obtain
b 2
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