Subtracting vectors from different origins

Discussion in 'Math' started by tquiva, Sep 4, 2011.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
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    Problem:

    [​IMG]

    Attempt:

    (attached)

    I've been googling for examples on the internet how to do so, and my book does not show any examples either. What I've done was set both vectors at the same origin in two scenarios with the two given origins. I then accounted the change based on the origin (0,0,0) and counted points from there. For the origin (1, pi/2, 0) for instance, based on (0,0,0), I found the second component of A-B by subtracting one point from the y-axis. So instead of -7-(-4), I got -6-(-3) which gave -3 (correct from answer in back).

    This method worked for some components, but not for some, so now I am lost. Is there a proper way to doing this?
     
  2. someonesdad

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    Jul 7, 2009
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    Vector addition is defined to be the addition of the corresponding vector components in some common basis. Note you can't add the spherical components because the spherical basis vectors change direction with position. Hint: these are probably free vectors; what does this imply that you can do with their Cartesian coordinates?
     
  3. tquiva

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    Oct 19, 2010
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    So I should convert the vectors to Cartesian form and proceed from there?
     
  4. someonesdad

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    Couldn't hurt... :p Also, don't they teach students to make models any more (I'm assuming you're a student)? You could make a simple model from some wire and see the vectors and their relationship in space. Then the result of their subtraction should be pretty obvious. Or, just make a sketch like you were taught in your basic calculus class (NB: they probably don't teach that stuff anymore either :p).
     
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  5. tquiva

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    Oct 19, 2010
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    now I'm a bit confused from converting spherical to cartesian.

    I know the formulas are:

    x=rsin(θ)cos(\phi)
    y=rsin(θ)sin(\phi)
    z=rcos(θ)

    But how would I find out what r is? Would that be the magnitude of the given vector? And how would I find theta and phi? Is that the given origin of (1, π/2, 0°) ?
     
  6. someonesdad

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    Read the problem statement, Luke -- your answers are staring you in the face.
     
  7. tquiva

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    Oct 19, 2010
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    I'm sorry, I just want to clarify. According to the problem statement for the origin of A,

    r = 1
    θ = pi/2
    phi = 0

    Is that right?
     
  8. someonesdad

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    That's what I would assume. I interpret your spherical coordinates as (r, phi, θ) where phi is the same as the cylindrical coordinate azimuthal angle and θ is the angle from the positive z axis. But I've seen different notations and conventions, so make sure you understand what your book is using. Thus, the (1, pi/2, 0) point in spherical coordinates would be (0, 0, 1) in Cartesian coordinates (i.e., the north pole on a unit sphere).

    Note: for doing these "easy" type of conversions, I never bother with the formulas -- I just trace the points out in my head. Thus, I think "OK, it's unit radius; the phi means it's aligned on the y axis; the 0 for θ means it has to lie on the z axis".
     
  9. t_n_k

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    Mar 6, 2009
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    I'm surprised that one would consider solving the problem by translation to the orthogonal (x,y,z) system to do the vector additions and then back to the spherical system to state the solution.

    It certainly provides some conceptual insights regarding the link between the two systems but overlooks the obvious value & practicality of doing the whole thing in spherical co-ordinates. Presumably the expectation is that the problem is to be solved in that manner - although having an alternative approach might help sometimes.

    Conceptually we shouldn't have too many issues visualizing the spherical system anyway. As someonesdad comments imply, there is an obvious connection with our own spatial perspective living on planet Earth and the spherical co-ordinate system.
     
    Last edited: Sep 5, 2011
  10. someonesdad

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    Try an example: add the Cartesian i and j vectors. In spherical coordinates, i is (1, 0, pi/2) and j is (1, pi/2, pi/2). If you add the spherical components, you get (2, pi/2, pi), which is a vector pointing in the -z direction -- clearly wrong, as the answer must lie in the xy plane at an azimuth of pi/4 (in addition, the radial component has to be sqrt(2)).

    The problem is that the spherical unit vectors don't have constant direction -- hence, you can't add them like you can Cartesian unit vectors.

    You can also see the same type of problem in trying to add planar vectors using polar coordinates in the plane.
     
  11. tquiva

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    Oct 19, 2010
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    I'm still confused. So after I convert the origins of (1,pi/2,0) to (0,0,1) and (3,pi/2,pi/2) to (0,3,0).... what do I do with A and B? Do I convert these two vectors both to Cartesian form also?
     
  12. t_n_k

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    Yes I see my mistake - thanks.
     
  13. someonesdad

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    The way I'd solve the problem is:

    1. Get the Cartesian coordinates of the two vectors.

    2. Add the corresponding Cartesian coordinates.

    Did you make a wire model of things so you could see the two arrows in space? Then vector addition is that you bring the starting points of the vectors together and combine them in the plane that contains both vectors with the familiar parallelogram method (this is a coordinate-free definition of vector addition). Of course, for subtraction, you take the negative of one vector, then add them together. What's nice is that this mental picture works for vectors in any number of dimensions. In 3 dimensional space, the easiest way to do this is to use the Cartesian coordinates as indicated.

    I've attached how I'd work the problem on paper. Of course, check what I've done and make sure I haven't made any errors in assumptions...
     
  14. t_n_k

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    This is an interesting problem ...

    The transformation between unit vectors from Cartesian to spherical is related by ..

     \left [\bf \hat{\rho} \\ \bf \hat{\theta} \\  \bf \hat{\phi}  \right ]=\left[ \bf M \right ]\left  [ \bf \hat{x} \\ \bf \hat{y} \\  \bf \hat{z} \right ]=\left [ \begin{array}{cc} sin(\theta)cos(\phi) & sin(\theta)sin(\phi) & cos(\phi) \\ cos(\theta)cos(\phi) & cos(\theta)sin(\phi) & -sin(\theta) \\ -sin(\phi) & cos(\phi) & 0 \end{array}\right ] \ \left  [ \bf \hat{x} \\ \bf \hat{y} \\  \bf \hat{z} \right ]

    Consider the case of converting a Cartesian form

    A'=a_x\bf{\hat{x}}+a_y\bf{\hat{y}}+a_z\bf{\hat{z}}

    to the spherical form

    A=a_{\rho} \bf{\hat{\rho}}+a_{\theta} \bf{\hat{\theta}}+a_{\phi}\bf{\hat{\phi}}

    Note: There's a reason I'm going this direction rather than spherical to Cartesian

    Firstly, one needs to deduce the values

    \left [  \rho \\ \theta \\ \phi  \right ]=\left [ \sqr {(a_x^2+a_y^2+a_z^2)} \\ arccos(\frac{a_z}{\rho}) \\ arctan(\frac{a_y}{a_x})\right ]

    and hence form the matrix M using the values of

    \theta \ and \ \phi

    Taking your values for the transformed vector A in the Cartesian system

    xA=3sin(-7)cos(2)=0.8202
    yA=3sin(-7)sin(2)=-1.792
    zA=3cos(-7)=2.262

    gives

    \left [  \rho \\ \theta \\ \phi  \right ]=\left [ \sqr  {(x_A^2+y_A^2+z_A^2)} \\ arccos(\frac{z_A}{\rho}) \\  arctan(\frac{y_A}{x_A})\right ]=\left [ 3 \\ 0.717 \\ -1.14 \right]

    So the matrix operation

    \left [ \bf{M}\right ]\left [x_A \\ y_A \\ z_A \right]

    with M deduced using values

    \theta=0.717 \ and  \ \phi= -1.14

    should give back the original co-efficients of the spherical unit vectors.

    But this isn't the case. One just gets the result ...

    \left [ 3 \\ 0 \\ 0 \right ]

    rather than the original coefficients

    \left [ 3 \\ -7 \\ 2 \right ]

    I suspect finding the equivalent coefficients of the unit Cartesian vectors corresponding to spherical vector A (given in the question) would be found from

    \left [ Inverse(\bf{M}) \right ]\left [ 3 \\ -7 \\ 2 \right ]

    Where M is deduced using

    \theta=\frac{\pi}{2} \ and  \ \phi= 0

    as stated in the question for the origin of A in the spherical system

    The result for the Cartesian "equivalent" of A would then be

    A'=3\bf{\hat{x}}+2\bf{\hat{y}}+7\bf{\hat{z}}

    In the same way the B vector would translate to

    B'=-2\bf{\hat{x}}+-2\bf{\hat{y}}+4\bf{\hat{z}}

    In this case using

    \theta=\frac{\pi}{2} \ and  \ \phi= \frac{\pi}{2}

    to deduce M

    in ...

    \left [ Inverse(\bf{M}) \right ]\left [-2 \\ -4 \\ 2 \right ]
     
    Last edited: Sep 6, 2011
  15. tquiva

    Thread Starter Member

    Oct 19, 2010
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    Thank you both for the given solutions. But I have a question for t_n_k:

    What do you mean by "M deduced" ?

    Is M a vector you're multiplying to the vector next to it?
     
  16. t_n_k

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    Clearly the two methods presented are different solutions with different answers - so take your pick as to which you think is correct.


     \left[ \bf M \right ] =\left [ \begin{array}{cc} sin(\theta)cos(\phi)  & sin(\theta)sin(\phi) & cos(\phi) \\ cos(\theta)cos(\phi) &  cos(\theta)sin(\phi) & -sin(\theta) \\ -sin(\phi) & cos(\phi)  & 0 \end{array}\right ]

    You substitute the appropriate values of

    \theta \  and \ \phi

    into the elements of M to come up with a 3x3 matrix in this case.

    So if

    \theta=\frac{\pi}{2} \ and \ \phi=0

    then the top left element of matrix M is given by

    M(1,1)=sin(\theta)cos(\phi)=sin(\frac{\pi}{2})cos(0)=1

    and so forth
     
  17. t_n_k

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    Returning to the original question. The required answer is the new vector A-B.

    The problem arises when translating the Cartesian difference vector back to the spherical system is deciding where the resultant vector is to originate.

    Presumably this would most likely be the same origin specified for spherical vector A in the question.

    What are the options?

    Firstly, if I use the same method outlined in post #14 for performing the translation I start with the Cartesian difference vector

    A'-B'=5\bf{\hat{x}}+4\bf{\hat{y}}+3\bf{\hat{z}}

    which translates to

    A-B=7.071 \bf{\hat{\rho}}+0 \bf{\hat{\theta}}+0\bf{\hat{\phi}}

    in the spherical system.

    This tells me that the solution is non unique, with the resulting vector mapping out a sphere of radius 7.071. Not particularly useful!

    If I place the same vector at the stated origin of vector A

    NB: Using only the \theta and \phi values of that origin

    the Cartesian difference vector translates to the spherical equivalent.

    A-B=5\bf{\hat{\rho}}-3 \bf{\hat{\theta}}+4\bf{\hat{\phi}}

    I'm not sure if this would be the required answer.
     
  18. someonesdad

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    Jul 7, 2009
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    One other problem here is that the spherical coordinate system isn't defined here, so there's some ambiguity -- what exactly are the angles θ and phi and their domains? These need to be defined because there are different conventions. The stated problem gives values for the spherical coordinates that are outside the typical domains -- for example, the conventions I've always used are the radial coordinate is >= 0, the azimuth is between 0 and 2*pi, and the angle off the z axis is between 0 and pi.

    To summarize how I think this problem needs to be solved, the two vectors are assumed to be free vectors. Then you'd effectively use parallel transport to move one of them to the other vector so that their starting points coincide. Negate the vector being subtracted (B). Then add them as usual with the parallelogram law.

    When you're studying such stuff for the first time, it helps to create some physical models -- that's why I encouraged the use of some wire to represent the vectors. You can get some square blocks of styrofoam or foam and use them to hold a wire of the requisite length at the desired orientation in space. Then place things on a piece of graph paper on a table top and you can have a coordinate system. Then it's easier to do things like the parallel transport, negating, etc. and see what's going on in space. This helps develop your ability to make mental pictures from the given conditions.
     
  19. t_n_k

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    The stated origins for the vectors A & B fall within the typical domains you mention.

    In your posted solution you seem to interpret the unit vectors \hat{\bf \theta} and \hat{\bf \phi} as angular rotations. Are the stated co-efficients of these unit vectors what you considered as being outside of the domain ranges? I believe the co-efficients of those unit vectors are normally interpreted as linear displacements.

    In regard to the latter point I found this to be a useful link http://web.mit.edu/6.013_book/www/appendices/app1.html
     
    Last edited: Sep 8, 2011
  20. someonesdad

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    Hi, t_n_k: If I understand the problem, the origins of the vectors are irrelevant; I'm interpreting them as free vectors in space. Then I interpreted the coordinates of the vectors as spherical coordinates. Thus, A = (3, -7, 2) = (r, θ, phi) where -7 and 2 are angles in radian measure. It never occurred to me to interpret them as anything different -- maybe that explains why we're doing things differently.

    The OP needs to chime in and remove some of the ambiguity of the problem. :p
     
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