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homework_10_solutions - MA 265 HOMEWORK ASSIGNMENT#10...

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MA 265 HOMEWORK ASSIGNMENT #10 SOLUTIONS #1. Page 297; Exercise 2. In Exercises 1 and 2, find the length of each vector. (a) 0 - 2 0 (b) - 1 - 3 - 4 (c) 1 - 2 4 Solution: (a) We define the 3-vector v = 0 - 2 0 = || v || = p 0 2 + ( - 2) 2 + 0 2 = 4 = 2 (b) We define the 3-vector v = - 1 - 3 - 4 = || v || = p ( - 1) 2 + ( - 3) 2 + ( - 4) 2 = 26 (c) We define the 3-vector v = 1 - 2 4 = || v || = p 1 2 + ( - 2) 2 + 4 2 = 21 #2. Page 297; Exercise 6. In Exercises 5 and 6, find the distance between u and v . (a) u = - 1 - 2 - 3 , v = 4 5 6 (b) u = 0 1 - 1 , v = 1 2 0 . Solution: (a) We have the two 3-vectors u = - 1 - 2 - 3 and v = 4 5 6 = || v - u || = q (4 - ( - 1)) 2 + (5 - ( - 2)) 2 + (6 - ( - 3)) 2 = p 5 2 + 7 2 + 9 2 = 155 1
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2 MA 265 HOMEWORK ASSIGNMENT #10 SOLUTIONS (b) We have the two 3-vectors u = 0 1 - 1 and v = 1 2 0 = || v - u || = q (1 - 0) 2 + (2 - 1) 2 + (0 - ( - 1)) 2 = p 1 2 + 1 2 + 1 2 = 3 #3. Page 297; Exercise 8. In Exercises 7 and 8, determine all values of c so that each given condition is satisfied. || u || = 1 for u = 1 c 2 c - 2 c Solution: The length of the vector is || u || = q ( 1 c ) 2 + ( 2 c ) 2 + ( - 2 c ) 2 = q 1 c 2 + 4 c 2 + 4 c 2 = q 9 c 2 = 3 | c | . Since we want || u || = 1, we must have | c | = 3, i.e., c = ± 3. #4. Page 297; Exercise 10. For each pair of vectors in Exercise 6, find the cosine of the angle θ between u and v . Solution: (a) We have the two 3-vectors u = - 1 - 2 - 3 and v = 4 5 6 = cos θ = u · v || u || || v || = ( - 1) (4) + ( - 2) (5) + ( - 3) (6) p ( - 1) 2 + ( - 2) 2 + ( - 3) 2 4 2 + 5 2 + 6 2 = - 32 14 77 (b) We have the two 3-vectors u = 0 1 - 1 and v = 1 2 0 = cos θ = u · v || u || || v || = (0) (1) + (1) (2) + ( - 1) (0) p 0 2 + 1 2 + ( - 1) 2 1 2 + 2 2 + 0 2 = 2 2 5 #5. Page 298; Exercise 18. Which of the vectors v 1 = 1 - 1 - 2 , v 2 = 3 - 1 2 , v 3 = 2 4 - 1 , v 4 = 1 2 0 1 4 , v 5 = 1 2 - 1 2 - 1 , v 6 = - 2 3 - 4 3 1 3 . (a) orthogonal? (b) in the same direction? (c) in opposite directions?
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MA 265 HOMEWORK ASSIGNMENT #10 SOLUTIONS 3 Solution: First we compute the length of each of these vectors: || v 1 || = p 1 2 + ( - 1) 2 + ( - 2) 2 = 6 || v 4 || = q ( 1 2 ) 2 + 0 2 + ( 1 4 ) 2 = 5 4 || v 2 || = p 3 2 + ( - 1) 2 + 2 2 = 14 || v 5 || = q ( 1 2 ) 2 + ( - 1 2 ) 2 + ( - 1) 2 = 6
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