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ISM-Chp_6

Linear Algebra and Its Applications (3rd Edition)

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335 6.1 SOLUTIONS Notes : The first half of this section is computational and is easily learned. The second half concerns the concepts of orthogonality and orthogonal complements, which are essential for later work. Theorem 3 is an important general fact, but is needed only for Supplementary Exercise 13 at the end of the chapter and in Section 7.4. The optional material on angles is not used later. Exercises 27–31 concern facts used later. 1 . Since 1 2 ± ª º « » ¬ ¼ u and 4 , 6 ª º « » ¬ ¼ v 2 2 ( 1) 2 5 ± ² u u , v u = 4(–1) + 6(2) = 8, and 8 . 5 v u u u 2 . Since 3 1 5 ª º « » ± « » « » ± ¬ ¼ w and 6 2 , 3 ª º « » ± « » « » ¬ ¼ x 2 2 2 3 ( 1) ( 5) 35 ² ± ² ± w w , x w = 6(3) + (–2)(–1) + 3(–5) = 5, and 5 1 . 35 7 x w w w 3 . Since 3 1 , 5 ª º « » ± « » « » ± ¬ ¼ w 2 2 2 3 ( 1) ( 5) 35 ² ± ² ± w w , and 3/35 1 1/35 . 1/7 ª º « » ± « » « » ± ¬ ¼ w w w 4 . Since 1 , 2 ± ª º « » ¬ ¼ u 2 2 ( 1) 2 5 ± ² u u and 1/5 1 . 2/5 ± ª º « » ¬ ¼ u u u 5 . Since 1 2 ± ª º « » ¬ ¼ u and 4 , 6 ª º « » ¬ ¼ v u v = (–1)(4) + 2(6) = 8, 2 2 4 6 52, ² v v and 4 8/13 2 . 6 12/13 13 ª º ª º § · ¨ ¸ « » « » © ¹ ¬ ¼ ¬ ¼ u v v v v 6 . Since 6 2 3 ª º « » ± « » « » ¬ ¼ x and 3 1 , 5 ª º « » ± « » « » ± ¬ ¼ w x w = 6(3) + (–2)(–1) + 3(–5) = 5, 2 2 2 6 ( 2) 3 49, ² ± ² x x and 6 30/ 49 5 2 10/ 49 . 49 3 15/ 49 ª º ª º § · « » « » ± ± ¨ ¸ « » « » © ¹ « » « » ¬ ¼ ¬ ¼ x w x x x
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336 CHAPTER 6 Orthogonality and Least Squares 7 . Since 3 1 , 5 ª º « » ± « » « » ± ¬ ¼ w 2 2 2 || || 3 ( 1) ( 5) 35. ² ± ² ± w w w 8 . Since 6 2 , 3 ª º « » ± « » « » ¬ ¼ x 2 2 2 || || 6 ( 2) 3 49 7. ² ± ² x x x 9 . A unit vector in the direction of the given vector is 2 2 30 30 3/5 1 1 40 40 4/5 50 ( 30) 40 ± ± ± ª º ª º ª º « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ± ² 10 . A unit vector in the direction of the given vector is 2 2 2 6/ 61 6 6 1 1 4 4 4/ 61 61 ( 6) 4 ( 3) 3 3 3 61 ª º ± ± ± ª º ª º « » « » « » « » « » « » « » ± ² ² ± « » « » ± ± ± ¬ ¼ ¬ ¼ « » ¬ ¼ 11 . A unit vector in the direction of the given vector is 2 2 2 7/ 69 7/ 4 7/ 4 1 1 1/ 2 1/ 2 2/ 69 69/16 (7/ 4) (1/ 2) 1 1 1 4/ 69 ª º ª º ª º « » « » « » « » « » « » « » ² ² « » « » ¬ ¼ ¬ ¼ « » ¬ ¼ 12 . A unit vector in the direction of the given vector is 2 2 8/3 8/3 4/5 1 1 2 2 3/5 100/9 (8/3) 2 ª º ª º ª º « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ² 13 . Since 10 3 ª º « » ± ¬ ¼ x and 1 , 5 ± ª º « » ± ¬ ¼ y 2 2 2 || || [10 ( 1)] [ 3 ( 5)] 125 ± ± ± ² ± ± ± x y and dist ( , ) 125 5 5. x y 14 . Since 0 5 2 ª º « » ± « » « » ¬ ¼ u and 4 1 , 8 ± ª º « » ± « » « » ¬ ¼ z 2 2 2 2 || || [0 ( 4)] [ 5 ( 1)] [2 8] 68 ± ± ± ² ± ± ± ² ± u z and dist ( , ) 68 2 17. u z 15 . Since a b = 8(–2) + (–5)( –3) = –1 0, a and b are not orthogonal.
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