ch13 - 13.1 SOLUTIONS 907 CHAPTER THIRTEEN Solutions for...

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Unformatted text preview: 13.1 SOLUTIONS 907 CHAPTER THIRTEEN Solutions for Section 13.1 Exercises 1. 4 ~ i + 2 ~ j- 3 ~ i + ~ j = ~ i + 3 ~ j 2. ~ i + 2 ~ j- 6 ~ i- 3 ~ j =- 5 ~ i- ~ j 3.- 4 ~ i + 8 ~ j- . 5 ~ i + 0 . 5 ~ k =- 4 . 5 ~ i + 8 ~ j + 0 . 5 ~ k 4. (0 . 9 ~ i- 1 . 8 ~ j- . 02 ~ k )- (0 . 6 ~ i- . 05 ~ k ) = 0 . 3 ~ i- 1 . 8 ~ j + 0 . 03 ~ k 5. k ~v k = p 1 2 + (- 1) 2 + 3 2 = 11 . 6. k ~v k = p 1 2 + (- 1) 2 + 2 2 = 6 . 7. k ~v k = p 1 . 2 2 + (- 3 . 6) 2 + 4 . 1 2 = 31 . 21 5 . 6 . 8. k ~v k = p 7 . 2 2 + (- 1 . 5) 2 + 2 . 1 2 = 58 . 5 7 . 6 . 9. 5 ~ b = 5(- 3 ~ i + 5 ~ j + 4 ~ k ) =- 15 ~ i + 25 ~ j + 20 ~ k . 10. ~a + ~ z = (2 ~ j + ~ k ) + ( ~ i- 3 ~ j- ~ k ) = (0 + 1) ~ i + (2- 3) ~ j + (1- 1) ~ k = ~ i- ~ j 11. 2 ~ c + ~x = 2( ~ i + 6 ~ j ) + (- 2 ~ i + 9 ~ j ) = (2 ~ i + 12 ~ j ) + (- 2 ~ i + 9 ~ j ) = (2- 2) ~ i + (12 + 9) ~ j = 21 ~ j . 12. k ~ z k = p (1) 2 + (- 3) 2 + (- 1) 2 = 1 + 9 + 1 = 11 . 13. k ~ y k = p (4) 2 + (- 7) 2 = 16 + 49 = 65 . 14. 2 ~a + 7 ~ b- 5 ~ z = 2(2 ~ j + ~ k ) + 7(- 3 ~ i + 5 ~ j + 4 ~ k )- 5( ~ i- 3 ~ j- ~ k ) = (4 ~ j + 2 ~ k ) + (- 21 ~ i + 35 ~ j + 28 ~ k )- (5 ~ i- 15 ~ j- 5 ~ k ) = (- 21- 5) ~ i + (4 + 35 + 15) ~ j + (2 + 28 + 5) ~ k =- 26 ~ i + 54 ~ j + 35 ~ k . 15. (a) See Figure 13.1. (b) k ~v k = 5 2 + 7 2 = 74 = 8 . 602 . (c) We see in Figure 13.2 that tan = 7 5 and so = 54 . 46 . ~v 5- 7 x y Figure 13.1 7 5 x y Figure 13.2 16. Resolving ~v into components gives ~v = 8 cos(40 ) ~ i- 8 sin(40 ) ~ j = 6 . 13 ~ i- 5 . 14 ~ j . Notice that the component in the ~ j direction must be negative. 17. ~a =- 2 ~ j , ~ b = 3 ~ i , ~ c = ~ i + ~ j , ~ d = 2 ~ j , ~e = ~ i- 2 ~ j , ~ f =- 3 ~ i- ~ j . 908 Chapter Thirteen /SOLUTIONS 18. The vector we want is the displacement from Q to P , which is given by ~ QP = (1- 4) ~ i + (2- 6) ~ j =- 3 ~ i- 4 ~ j 19. ~a = ~ b = ~ c = 3 ~ k , ~ d = 2 ~ i + 3 ~ k , ~e = ~ j , ~ f =- 2 ~ i 20. ~u = ~ i + ~ j + 2 ~ k and ~v =- ~ i + 2 ~ k . Problems 21. To determine if two vectors are parallel, we need to see if one vector is a scalar multiple of the other one. Since ~u =- 2 ~w , and ~v = 1 4 ~ q and no other pairs have this property, only ~u and ~w , and ~v and ~ q are parallel. 22. The length of the vector ~ i- ~ j +2 ~ k is p 1 2 + (- 1) 2 + 2 2 = 6 . We can scale the vector down to length 2 by multiplying it by 2 6 . So the answer is 2 6 ~ i- 2 6 ~ j + 4 6 ~ k . 23. (a) The displacement from P to Q is given by-- PQ = (4 ~ i + 6 ~ j )- ( ~ i + 2 ~ j ) = 3 ~ i + 4 ~ j . Since k-- PQ k = p 3 2 + 4 2 = 5 , a unit vector ~u in the direction of-- PQ is given by ~u = 1 5-- PQ = 1 5 (3 ~ i + 4 ~ j ) = 3 5 ~ i + 4 5 ~ j ....
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This note was uploaded on 08/26/2009 for the course MATH 1ABCD taught by Professor All during the Winter '08 term at Foothill College.

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ch13 - 13.1 SOLUTIONS 907 CHAPTER THIRTEEN Solutions for...

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