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Unformatted text preview: b3 , then the cross product of a a 3 b2 , a 3 b1 a 1 b3 , a 1 b2 a 2 b1 This may seem like a strange way of deﬁning a product. The reason for the particular
form of Deﬁnition 1 is that the cross product deﬁned in this way has many useful properties, as we will soon see. In particular, we will show that the vector a b is perpendicular to both a and b.
In order to make Deﬁnition 1 easier to remember, we use the notation of determinants.
A determinant of order 2 is deﬁned by
a
c
2
6 For example, 1
4 b
d ad 24 bc 1 6 14 A determinant of order 3 can be deﬁned in terms of secondorder determinants as
follows:
2 a1
b1
c1 a2
b2
c2 a3
b3
c3 a1 b2
c2 b3
c3 a2 b1
c1 b3
c3 a3 b1
c1 b2
c2 5E13(pp 848857) 1/18/06 11:19 AM Page 851 SECTION 13.4 THE CROSS PRODUCT ❙❙❙❙ 851 Observe that each term on the right side of Equation 2 involves a number a i in the ﬁrst row
of the determinant, and a i is multiplied by the secondorder 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,
1
3
5 2
0
4 1
1
2 0
4 1 1
2
4 10 2 3
5 26 1
2 3
5 1 5 1 12 0
4 0 38 If we now rewrite Deﬁnition 1 using secondorder determinants and the standard basis
vectors i , j, and k, we see that the cross product of the vectors a a 1 i a 2 j a 3 k and
b b 1 i b 2 j b 3 k is a 3 a2
b2 b a3
i
b3 a1
b1 a3
j
b3 a1
b1 a2
k
b2 In view of the similarity between Equations 2 and 3, we often write a 4 ijk
a1 a2 a3
b1 b2 b3 b Although the ﬁrst row of the symbolic determinant in Equation 4 consists of vectors, if we
expand it as if it were an ordinary determinant using the rule in Equation 2, we obtain
Equation 3. The symbolic formula in Equation 4 is probably the easiest way of remembering and computing cross products.
EXAMPLE 1 If a a 1, 3, 4 and b
ij
13
27 b 3
7 5 , then k
4
5
4
i
5 15 1
2 28 i EXAMPLE 2 Show that a
SOLUTION If a 2, 7, a 4
j
5
5 8j 1
2 3
k
7
7 6k 43 i 13 j 0 for any vector a in V3. a 1, a 2 , a 3 , then a a i
a1
a1 j
a2
a2 a2a3
0i k
a3
a3
a3a2 i 0j 0k a1a3
0 a3a1 j a1a2 a2a1 k k 5E13(pp 848857) 852 ❙❙❙❙ 1/18/06 11:20 AM Page 852 CHAPTER 13 VECTORS A...
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 Fall '09
 hamrick

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