The reason for the particular form of denition 1 is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online