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Unformatted text preview: 18.02 HOMEWORK #1, DUE SEPTEMBER 16, 2010 BJORN POONEN Page numbers refer to Edwards & Penney. Problem numbers like 1A1 refer to the Sup plementary Notes. “(After Sept. 10)” means that the math needed for the problem should have been covered in lecture by that date. Make sure that you know how to do all exercises listed, but hand in only those labelled with point values . 1. Part A (After Sept. 9) p. 777: 21, 31 (After Sept. 9) p. 786: 13* (for the vectors in 1 only) (5 pts.), 39, 58* (10 pts.), 70* (5 pts.) Solution. (to p. 786, 13*) By definition, comp a b = b · a  a  = h 1 , 2 , 3 i · h 2 , 5 , 4 i p 2 2 + 5 2 + ( 4) 2 = (1)(2) + ( 2)(5) + ( 3)( 4) √ 45 = 4 3 √ 5 = 4 15 √ 5 comp b a = a · b  b  = 4  b  = 4 √ 14 = 4 √ 14 14 = 2 7 √ 14 . Solution. (to p. 786, 39) They are parallel since b = 3 2 a . Solution. (to p. 786, 58*) If a or b is , then both sides are 0 and the inequality holds. Otherwise, let θ be the angle between a and b . Then a · b =  a  b  cos θ , and taking the 1 absolute value of both sides yields  a · b  =  a  b  cos θ  ≤  a  b  1 =  a  b  . (Note: sometimes   means absolute value of a real number, and sometimes it means length of a vector. The meaning has to be deduced from context.) Solution. (to p. 786, 70*) Let A = h , , i , B = h 1 , 1 , i , and C = h 1 / 2 , 1 / 2 , 1 / 2 i . We want the angle α between the vectors from the center to two of the hydrogen atoms: these vectors are A C = h 1 / 2 , 1 / 2 , 1 / 2 i and B C = h 1 / 2 , 1 / 2 , 1 / 2 i . We have cos α = ( A C ) · ( B C )  A C  B C  = 1 / 4 1 / 4 + 1 / 4 p 3 / 4 p 3 / 4 = 1 / 3 , so α = cos 1 ( 1 / 3) ≈ 1 . 91 (in radians)....
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This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.
 Fall '08
 Auroux
 Math, Multivariable Calculus

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