18
pset1sol[1]

# pset1sol[1] - 18.02 Spring 2009 Problem Set 1 Solutions 1...

• Notes
• 4

This preview shows pages 1–2. Sign up to view the full content.

18.02 Spring 2009 Problem Set 1 Solutions 1. As in the hint, let A = --→ OP and B = --→ OR . Let M be the midpoint of the segment PQ . Then we have: --→ OM = A + 1 2 B. (1) The diagonal vector -→ PR from P to R is -→ PR = B - A. (2) Let N be the point 1 / 3 of the way from P to R . Then: --→ ON = A + 1 3 -→ PR = 2 3 A + 1 3 B = 2 3 --→ OM. (3) Because --→ ON is a scalar multiple of --→ OM , the vectors --→ ON and --→ OM are parallel. This shows that the point N lies on the line segment from O to M , which is what we wanted to show. 2. (a) The “geometric” definition of the dot product is A · B = | A || B | cos θ . Because | cos θ | ≤ 1 for all θ , we have: | A · B | = | A || B || cos θ | ≤ | A || B | . (4) Equality holds when | cos θ | = 1. This means that either θ = 0 or θ = π . The case θ = 0 means that A and B point in the same direction, while the case θ = π means A and B point in opposite directions. (b) The inequality in components becomes simpler if we square both sides: | A · B | 2 ≤ | A | 2 | B | 2 (5) In 2D, where A = a 1 ˆ ı + a 2 ˆ , B = b 1 ˆ ı + b 2 ˆ : ( a 1 b 1 + a 2 b 2 ) 2 ( a 2 1 + a 2 2 )( b 2 1 + b 2 2 ) (6) In 3D, where A = a 1 ˆ ı + a 2 ˆ + a 3 ˆ k , B = b 1 ˆ ı + b 2 ˆ + b 3 ˆ k : ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) 2 ( a 2 1 + a 2 2 + a 2 3 )( b 2 1 + b 2 2 + b 2 3 ) (7) (c) Following the hint, define: f ( t ) = ( A + tB ) · ( A + tB ) . (8)

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

This is the end of the preview. Sign up to access the rest of the document.
• Fall '08
• Auroux

{[ 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