# h1s - 18.02 HOMEWORK#1 DUE BJORN POONEN Page numbers refer...

This preview shows pages 1–3. Sign up to view the full content.

18.02 HOMEWORK #1, DUE SEPTEMBER 16, 2010 BJORN POONEN Page numbers refer to Edwards & Penney. Problem numbers like 1A-1 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 | = 1 , - 2 , - 3 · 2 , 5 , - 4 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 0 , 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = 0 , 0 , 0 , B = 1 , 1 , 0 , and C = 1 / 2 , 1 / 2 , 1 / 2 . We want the angle α between the vectors from the center to two of the hydrogen atoms: these vectors are A - C = - 1 / 2 , - 1 / 2 , - 1 / 2 and B - C = 1 / 2 , 1 / 2 , - 1 / 2 . We have cos α = ( A - C ) · ( B - C ) | A - C || B - C | = - 1 / 4 - 1 / 4 + 1 / 4 3 / 4 3 / 4 = - 1 / 3 , so α = cos - 1 ( - 1 / 3) 1 . 91 (in radians). (After Sept. 9) 1A-1b, 1A-5, 1A-8, 1A-12* (10 pts.) Solution. (to 1A-12*) Set up a coordinate system with O at the origin. Let P , Q , and R be the position vectors of P , Q , and R , respectively. Then Q = P + R . The midpoint M of PQ has position vector M = 1 2 ( P + Q ) = P + 1 2 R . The point 1 / 3 of the way from P to R has position vector obtained by starting with P and adding 1 / 3 of R - P , so it is P + 1 3 ( R - P ) = 2 3 P + 1 3 R . This equals 2 3 M
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern