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2 vectors with cylindrical polar coordinates

  1. Jan 12, 2016 #1
    Hi this isn't my homework, but it is taken from a worksheet for a Maths course(trying to refresh my rusty math), so I hope it fits in here.

    1. The problem statement, all variables and given/known data

    two cylindrical polar vectors with same origin:
    P(2,55°,3); Q(4,25°,6) units in m

    2. Relevant equations
    a) Express in cartesian coordinates
    b) Express in unit vectors
    c) Find the distance from origin described by each P and Q
    d) Find the distance between P and Q and the vector displacement between P and Q
    e) Express the distance between P and Q in degrees


    3. The attempt at a solution
    a)
    P(2*cos(55), 2*sin(55), 3)
    Q(4*cos(25), 4*sin(25), 6)

    b)
    P:
    i= (cos(55), sin(55), 0)
    j= (-sin(55), cos(55), 0)
    k=(0,0,1)

    Q:
    i= (cos(25), sin(25), 0)
    j= (-sin(25), cos(25), 0)
    k=(0,0,1)

    c) That would just be a matter of calculating the magnitude, correct?

    d)
    is this correct?
    sqrt( Px - Qx, Py - Qy, Pz - Qz)
    using the cartesian coordinates.
    Would you please point me in the right direction as to what the difference is between calculating the vector displacement and the distance between the two points?


    I'd appreciate your input and hope I got some of it right at least!
     
  2. jcsd
  3. Jan 12, 2016 #2

    SteamKing

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    You seem t have omitted something from the expression above.
    Remember the Pythagorean relation: c2 = a2 + b2
     
  4. Jan 12, 2016 #3
    For c) Arent polar coordinates (angle, radius)? .. radius as in distance to origin ? ...
     
  5. Jan 12, 2016 #4
    Oh my, right, that was silly.
    It should be like so, right?
    sqrt( (Px - Qx)^2 + (Py - Qy)^2 + (Pz - Qz)^2 )

    would you please comment on part a) and b) as well?

    you are so right, WrongMan! Thanks for that.
    so if I'm not mistaken:

    c)
    P: 2m
    Q: 4m

    e)
    With all lenghts of an imaginary triangle known (distance to P, distance to Q and distance between both of them), I can just use the Cosine law, correct?
     
  6. Jan 12, 2016 #5

    haruspex

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    I don't think you have understood what is being asked for in part b. Perhaps I'm wrong, but I think it is just asking you to take the Cartesian form and write it out as a sum of unit vectors, where ##\vec i## (or maybe ##\hat i##) stands for (1,0,0) etc.

    In part c, don't forget the z contribution.

    For part e, an alternative would be to use the dot product of the vectors.
     
    Last edited: Jan 12, 2016
  7. Jan 12, 2016 #6
    oh right, cylindrical... missed that bit
     
  8. Jan 13, 2016 #7
    ok, dot product would be
    =Px*Qx + Py*Qy + Pz*Qz using the cartesian coordinates, correct?

    for c) that would be
    for P: Sqrt(2^2+3^2)
    and
    for Q: SQRT( 4^2+6^2)
    ?
    I'll have another look at e) later.

    Thanks so much this really helped already.
    I think I have to read up more on unit vectors again, as I'm a little confused regarding your suggestion haruspex.

    Also, I am still uncertain what the vector displacement is, and how it is different from the distance between the two.
     
  9. Jan 13, 2016 #8

    haruspex

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    Right, but how does that relate to the magnitudes of the vectors and the angle between them?
    Yes.
    Displacement is the vector difference. Distance is its magnitude.
    In the usual i, j, k notation, ##\hat i, \hat j, \hat k## are the unit vectors (1,0,0), (0,1,0), (0,0,1). A vector (a,b,c) you can write as ##a\hat i+b\hat j+ c\hat k##.
     
  10. Jan 14, 2016 #9
    So, I'd could divide by the product of their magnitudes to get the cosine of the angle?

    Ok, that is straight-forward, thanks!

    -------
    EDIT:
    Also thanks for clarifying vector displacement vs distance!
     
  11. Jan 14, 2016 #10

    haruspex

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    Yes.
     
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