Chapter 13

# 7 b sin in fact that is exactly how corollary two

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Unformatted text preview: right-hand rule, and whose length is a physicists deﬁne a b. 7 b sin . In fact, that is exactly how b 0 Proof Two nonzero vectors a and b are parallel if and only if sin b sin ¨ ¨ FIGURE 2 853 Corollary Two nonzero vectors a and b are parallel if and only if a b ❙❙❙❙ 0, so a b 0 and therefore a b 0 or . In either case 0. The geometric interpretation of Theorem 6 can be seen by looking at Figure 2. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base a , altitude b sin , and area A a a ( b sin ) a b Thus, we have the following way of interpreting the magnitude of a cross product. The length of the cross product a determined by a and b. b is equal to the area of the parallelogram EXAMPLE 3 Find a vector perpendicular to the plane that passes through the points P 1, 4, 6 , Q 2, 5, 1 , and R 1, 1, 1 . l l l l SOLUTION The vector PQ PR is perpendicular to both PQ and PR and is therefore perpendicular to the plane through P, Q, and R. We know from (13.2.1) that l PQ 2 1i 5 4j 1 6k 3i j 7k l PR 1 1i 1 4j 1 6k 5 j 5k We compute the cross product of these vectors: l PR i 3 0 j 1 5 5 l PQ 35 i k 7 5 15 0j 15 0k 40 i 15 j 15 k So the vector 40, 15, 15 is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as 8, 3, 3 , is also perpendicular to the plane. EXAMPLE 4 Find the area of the triangle with vertices P 1, 4, 6 , Q and R 1, 2, 5, 1, 1, 1 . l l PR 40, 15, 15 . The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product: SOLUTION In Example 3 we computed that PQ l PQ l PR s 40 2 15 2 15 2 5s82 The area A of the triangle PQR is half the area of this parallelogram, that is, 5 s82. 2 5E-13(pp 848-857) 854 ❙❙❙❙ 1/18/06 11:21 AM Page 854 CHAPTER 13 VECTORS AND THE GEOMETRY OF SPACE If we apply Theorems 5 and 6 to the standard basis vectors i , j, and k using we obtain i j j k i j k k k i j k i i i k 2, j j Observe that i j j...
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