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Vector projections

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data

    a. Is it possible to have u ↓ v undefined?
    b. Is it possible to have u ↓ v = v ↓ u ?
    c. Explain why u ↓( v ↓ w ) = u ↓ w .

    2. Relevant equations

    3. The attempt at a solution

    I know a is possible if the length of vector v is 0.
    I think be is false, but not sure.
    And c, Im not sure.
    I'm really just unsure how to demonstrate a, b, and c mathematically, and the question implies I use diagrams to demonstrate my answers. Any advice or assistance is appreciated..
     
  2. jcsd
  3. Feb 19, 2013 #2

    LCKurtz

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    You might define your notation. Does u ↓ v mean the vector projection of u on v? Or what?? In any case, the way to show your results mathematically is to use the formula for the vector projection.
     
  4. Feb 21, 2013 #3
    You are right u ↓ v mean the vector projection of u on v. In terms of part b though, it is possible for u ↓ v = v ↓ u if u=v right? In all other cases, they would not equal. Is this right?

    And also for part c, I am unsure how to start the question, could you please give me a hint.
     
  5. Feb 21, 2013 #4
    I just want to confirm with you: is u↓v equivalent to the formula:

    ##proj_{\vec v} \vec u = \displaystyle \frac{\vec u \cdot \vec v}{||\vec v||^2} \vec v## ?
     
  6. Feb 21, 2013 #5

    LCKurtz

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    Why don't you show us your argument for these statements using the formulas? Just stating them and asking if they are right doesn't cut it.
     
  7. Feb 21, 2013 #6
    Yeah karnage1993 they are of equivalent forms.

    For, a. I figured it out, and know how to demonstrate it using diagrams and mathematically.
    (if the vector your projecting onto has length of 0, then the vector projection is undefined-basic idea).

    For b. If you set u=v, then u ↓ v = v ↓ u is always true, as the vector projection of a vector onto itself, is just itself.
    u ↓ v= v, if u=v (I understand how to do this segment)
    So this is the only case in which the above statement is true, otherwise its always false right (I just really want to confirm what I believe to be right). And I understand how to demonstrate this visually using diagrams, so thats not a problem.

    For c., this is really the question I am unsure about. Is there a particular condition in which this statement is true, and only true if that condition is in place. ex. like if vectors v=w, then the statement would be true.
     
  8. Feb 21, 2013 #7

    LCKurtz

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    I have been trying to get you to use the formulas. First, write down the formula for ##u\downarrow w##. Then write down the formula for ##v\downarrow w## and then ##u\downarrow (v\downarrow w)##. Simplify it carefully and see if you get the same thing you got for ##u\downarrow w##.

    [Edit added later] If that seems too complicated, try proving the simpler fact that ##u\downarrow kv = u\downarrow v## for any nonzero constant ##k## and use that.
     
    Last edited: Feb 21, 2013
  9. Feb 22, 2013 #8

    HallsofIvy

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    Your "basic idea" is that if a vector has length 0 it is "undefined"? What about the 0 vector?

    What is true when two vectors are perpendicular?

    What is the direction of v projected onto w? What information about w is important in deciding what v, or u, projected onto w is?
     
  10. Feb 22, 2013 #9
    If you project v onto w, and then project u onto the projection of v on w, it will give you a vector that is parallel (or more appropriately 'a scalar multiple') to the projection of u onto w. And in terms of actually expressing u ↓( v ↓ w ), I understand how to show v ↓ w, but how do I demonstrate u ↓( v ↓ w )??
     
  11. Feb 22, 2013 #10

    HallsofIvy

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    Okay, you know that v↓w is a vector in the direction of w. So that u↓(v↓w) is a vector in the direction of (u↓w)- i.e in the direction of w. If by "demonstrate" it, draw there 'vectors' representing u, v, and w. Draw a perpendicular from v to w, then draw a vector from u to that line- which is just the same as drawing a line form u to w to begin with. As for length, do you understand that the length of u↓v is independent of the length of v? It depends only on the length of u and the angle between u and v.
     
  12. Feb 22, 2013 #11
    Thanks alot. The diagram makes complete sense.

    And for the second part of your statement, is it geared towards a, b or c?
    Otherwise, I do understand the notion that length of u ↓ v is independent of the length of v, since u cosθ =(compvU) Note: compvU meaning the scalar projection of u on v
    And thus it only depends on θ and u.
     
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