If you gave me a particular vector A as an arrow of some length A and

If you gave me a particular vector a as an arrow of

This preview shows page 35 - 38 out of 461 pages.

If you gave me a particular vector A as an arrow of some length A and orientation θ relative to the x -axis, what do I use for A x and A y ? You can see from trigonometry that A x = A cos θ (2.6) A y = A sin θ . (2.7) Conversely, given the components, the length and angle are A = A 2 x + A 2 y (2.8) θ = tan 1 A y A x . (2.9) Eqns. 2.6 to Eqn. 2.9 will be invoked often. So please commit them to memory.
Image of page 35
Free ebooks ==> Motion in Higher Dimensions 21 If you give me a pair of numbers, ( A x , A y ), that’s as good as giving me this arrow, because I can find the length of the arrow by Pythago- ras’ theorem and I can find the orientation from tan θ = A y A x . You have the option of either working with the two components of A or with the arrow. In practice, most of the time we work with these two numbers, ( A x , A y ). In particular, if we are describing a particle whose location is the position vector r , then we write it in terms of its components as r = i x + j y . (2.10) The changes in r are the displacement vectors and examples are A and B in Figure 2.2 that described the two hikes. I have not given you any other example of vectors besides the dis- placement vector, but at the moment, we’ll define a vector to be any object that looks like some multiple of i plus some multiple of j . If I tell you to add two vectors A and B , you have got two options. You can draw the arrow corresponding to A and attach to its end an arrow corresponding to B , and then add them, as in Figure 2.2. But you can also do the bookkeeping without drawing any pictures as follows: A + B = i A x + j A y + i B x + j B y (2.11) = i ( A x + B x ) + j ( A y + B y ) (2.12) so that the sum C is the vector with components ( A x + B x , A y + B y ). In the above, I have used the fact that vectors can be added in any order. So I grouped the things involving just i and likewise j . Then I argued that since i A x and i B x are vectors along i , their sum is a vector of length A x + B x also along i . I did the same for j . In summary if A + B = C (2.13) then C x = A x + B x (2.14) C y = A y + B y (2.15) which can be summarized as follows:
Image of page 36
Free ebooks ==> 22 Motion in Higher Dimensions To add two vectors, add their respective components. An important result is that A = B is possible only if A x = B x and A y = B y . You cannot have two vectors equal without having exactly the same x component and exactly the same y component. If two arrows are equal, one cannot be longer in the x direction and correspondingly shorter in the y direction. Everything has to match completely. The vector equation A = B is actually a shorthand for two equations: A x = B x and A y = B y . 2.4 Choice of axes and basis vectors I have in mind a vector whose components are 3 and 5. Can you draw the vector for me? If you immediately said, “It is 3 i + 5 j ,” you’re making the assumption that I am writing the vector in terms of i and j . I agree i and j point along two natural directions. For most of us, given that the blackboard or notebook is oriented this way, it is very natural to line up our axes with it. But there is no reason why somebody else couldn’t come along and say, “I want to use a different set of axes. The
Image of page 37
Image of page 38

You've reached the end of your free preview.

Want to read all 461 pages?

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes