solutions for homework 1

# solutions for homework 1 - 18.02 Problem Set 1 Part II...

• Homework Help
• crupani10
• 2

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

18.02 Problem Set 1, Part II Solutions 1. There are several ways to set up the tetrahedron for this problem. The simplest way is to inscribe it in the unit cube, so that it has vertices (0 , 0 , 0) , (1 , 1 , 0) , (0 , 1 , 1) , (1 , 0 , 1). If you didn’t think of this, you could also set it up with one face defined by (0 , 0 , 0) , (1 , 0 , 0) , (0 , p 3 / 2 , 0) and then figure out where the fourth vertex should be. In this solution we’ll use the sim- pler inscribed tetrahedron. In this case, we define the plane P 1 to contain (0 , 0 , 0) , (1 , 1 , 0) and (1 , 0 , 1); taking the cross product we see a normal to this plane is n 1 = h 1 , 0 , 1 i ⇥ h 1 , 1 , 0 i = h- 1 , 1 , 1 i . Similarly, we define the plane P 2 to contain (0 , 0 , 0), (1 , 1 , 0) and (0 , 1 , 1); a normal to this plane is n 2 = h 1 , 1 , 0 i ⇥ h 0 , 1 , 1 i = h 1 , - 1 , 1 i . The dihedral angle between these two planes is the smaller angle between n 1 and n 2 , with cos = n 1 · n 2 | n 1 || n 2 | 1 = . 3 (We use the absolute value to ensure that we get the cosine of the smaller angle.) We conclude that = cos - 1 (1 / 3) 1 . 23 or 70 . 5 deg . (Any other tetrahedron will be similar (in the geometric sense) to this one so it will have the same dihedral angle.) 2. (a) Calculating directly, we see | u + v | 2 = ( u + v ) · ( u + v ) = u · u + 2 u · v + v · v , since u v = v u . Also, · · | u - v | 2 = ( u - v ) · ( u - v ) = u · u - 2 u · v + v · v . Subtracting these two equations gives | u + v | 2 - | u - v | 2 = 4 u · v as desired. 1

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

(b) Since u and v have the same length (=1), u
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bisects the angle between u and v . We make this a unit vector: u + v . | u + v | See fgure. 3. a) See the sketch with part (b). It is clear that no matter what the angle ↵ the vector w 1 has a rightward component. b) In the sketch we’ve labeled the projection we want as w 2 . We see easily that | w 1 | = a cos ↵ and | w 2 | = cos β | w 1 | = a cos β cos ↵ . y Thus w 2 = a cos β cos ↵ h cos( ↵ + β ) , sin( ↵ + β ) i . l B l s The component oF w 2 in the i direction is a cos( ↵ ) cos( β ) cos( ↵ + β ). Since ↵ and β are between and ⇡ / 2 this component is negative exactly when w 2 ↵ + β > ⇡ / 2. (This is easily seen in the sketch as well.) w * 4 ↵ β > L w 1 S a 2...
View Full Document

• Spring '14
• FlorianEnescu
• Algebra, Dot Product, Vector Motors, dihedral angle, cos cos ↵hcos, 70.5 deg

{[ 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