Also i whereas i i j 0 j 0 so the associative law for

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Unformatted text preview: i Thus, the cross product is not commutative. Also i i j i k 0 j j whereas i i j 0 So the associative law for multiplication does not usually hold; that is, in general, a b c a b c However, some of the usual laws of algebra do hold for cross products. The following theorem summarizes the properties of vector products. 8 Theorem If a, b, and c are vectors and c is a scalar, then 1. a 2. 3. 4. 5. 6. b b b (c a) c (a a ( b c) a (a b) c a abc a a bc a a b) b c b cb a a b c (c b) c c a bc These properties can be proved by writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises. Proof of Property 5 If a b c b1, b2 , b3 , and c a 1 b2 c3 b3 c2 a 2 b3 c1 b1 c3 a 1 b2 c3 a 1 b3 c2 a 2 b3 c1 a 2 b1 c3 a 2 b3 9 a a 1, a 2 , a 3 , b a 3 b2 c1 a 3 b1 a b a 1 b3 c2 c1, c2 , c3 , then a 3 b1 c2 a 3 b1 c2 a 1 b2 b2 c1 a 3 b2 c1 a 2 b1 c3 c The product a b c that occurs in Property 5 is called the scalar triple product of the vectors a, b, and c. Notice from Equation 9 that we can write the scalar triple product as a determinant: 10 a b c a1 b1 c1 a2 b2 c2 a3 b3 c3 5E-13(pp 848-857) 1/18/06 11:21 AM Page 855 SECTION 13.4 THE CROSS PRODUCT ❙❙❙❙ 855 The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c (Figure 3). The area of the base parallelogram is A b c . If is the angle between a and b c, then the height h of the parallelepiped is h a cos . (We must use cos instead of cos in case 2.) Therefore, the volume of the parallelepiped is bxc h¨a c b V FIGURE 3 Ah b c a cos a b c Thus, we have proved the following formula. 11 The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product: V a b c If we use the formula in (11) and discover that the volume of the parallelepiped determined by a, b, and c is 0, then the vectors must lie in the same plane...
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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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