13.ReviewEx22-38

13.ReviewEx22-38 - 436 C H A P T E R 13 V E C T O R G E O M...

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436 CHAPTER 13 VECTOR GEOMETRY (ET CHAPTER 12) = −→ AB + 1 2 ( BC + CA ) = + 1 2 BA = 1 2 = 1 2 Therefore, EF is a constant multiple of , which implies that and are parallel vectors. In Exercises 22–27, let v =h 1 , 3 , 2 i and w 2 , 1 , 4 i . 22. Compute v · w . SOLUTION Using the deFnition of the dot product we have v · w 1 , 3 , 2 i·h 2 , 1 , 4 i= 1 · 2 + 3 · ( 1 ) + ( 2 ) · 4 = 2 3 8 =− 9 23. Compute the angle between v and w . The cosine of the angle θ between v and w is cos = v · w ± v ±± w ± (1) We compute the lengths of the vectors: ± v ±=±h 1 , 3 , 2 i± = q 1 2 + 3 2 + ( 2 ) 2 = 14 ± w 2 , 1 , 4 i± = q 2 2 + ( 1 ) 2 + 4 2 = 21 In the previous exercise we found that v · w 9. Substituting these values in (1) gives cos = 9 14 · 21 = 9 7 6 ≈− 0 . 5249 The solution for 0 π is = 2 . 123 rad . 24. Compute v × w . We use the deFnition of the cross product as a “determinant”: v × w = ¯ ¯ ¯ ¯ ¯ ¯ ij k 13 2 2 14 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 3 2 ¯ ¯ ¯ ¯ i ¯ ¯ ¯ ¯ 1 2 24 ¯ ¯ ¯ ¯ j + ¯ ¯ ¯ ¯ 2 1 ¯ ¯ ¯ ¯ k = ( 12 2 ) i ( 4 + 4 ) j + ( 1 6 ) k = 10 i 8 j 7 k 10 , 8 , 7 i 25. ±ind the area of the parallelogram spanned by v and w . The parallelogram spanned by v and w has area ± v × w ± . In the previous exercise, we found that v × w = h 10 , 8 , 7 i . Therefore the area A of the parallelogram is A v × w 10 , 8 , 7 i± = q 10 2 + ( 8 ) 2 + ( 7 ) 2 = 213 14 . 59 26. ±ind the volume of the parallelepiped spanned by v , w ,and u = h 1 , 2 , 6 i . The volume V of the parallelepiped spanned by v , w and u is the following determinant: V = ¯ ¯ ¯ ¯ ¯ ¯ det v w u ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ 2 2 126 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 1 · ¯ ¯ ¯ ¯ 26 ¯ ¯ ¯ ¯ 3 ¯ ¯ ¯ ¯ 16 ¯ ¯ ¯ ¯ 2 ¯ ¯ ¯ ¯ 2 1 12 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =| 1 · ( 6 8 ) 3 ( 12 4 ) 2 ( 4 + 1 ) |= 48 27. ±ind all the vectors orthogonal to both v and w . A vector u a , b , c i is orthogonal to v and to w if the dot products u · v and u · w are zero. That is, u · v = 0a n d u · w = 0 . We compute the dot products: u · v a , b , c 1 , 3 , 2 a + 3 b 2 c
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Chapter Review Exercises 437 u · w =h a , b , c i·h 2 , 1 , 4 i= 2 a b + 4 c We obtain the following equations: a + 3 b 2 c = 0 2 a b + 4 c = 0 The Frst equation implies a = 2 c 3 b . Substituting in the second equation and solving for b in terms of c gives 2 ( 2 c 3 b ) b + 4 c = 0 4 c 6 b b + 4 c = 0 8 c 7 b = 0 b = 8 7 c We Fnd a in terms of c , using the relation a = 2 c 3 b : a = 2 c 3 · 8 7 c = 2 c 24 7 c =− 10 7 c .
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13.ReviewEx22-38 - 436 C H A P T E R 13 V E C T O R G E O M...

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