Chapter 13

# The reason for the particular form of denition 1 is

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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 second-order 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 5E-13(pp 848-857) 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 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, 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 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 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 5E-13(pp 848-857) 852 ❙❙❙❙ 1/18/06 11:20 AM Page 852 CHAPTER 13 VECTORS A...
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## This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.

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