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Homework Help: Vectors in the form r<θ

  1. Jan 14, 2012 #1
    Hi Guys,

    At school a few of our classes this semester have begun using vectors in the form of r<θ to solve problems but I never learn this method of solving problems and I would really appriciate it if you guys would help me out with this! (:

    I know that r<θ = r cosθ i + r sin j but what happens when you have a 3D based coordinate (i.e. vecotrs with i, j and k components)?

    Also, I know that when you dot two vectors in the form of r<θ together, you get (r + r) cos (θ+θ) similarly when you cross you get (r + r) sin (θ+θ) right?

    Now when you cross the vectors, how do you know what component you end up with? I know that if you cross 3i + 2j with 6i + 2j you end up with a vector with only a k component but how do you see this? How if I have like 4k x r<θ how do I solve it then?

    I know it'a mouthful but I'm really hoping that someone can guide me on this or if possible direct me to a site or (preferbly) a youtube video that teaches this. I haven't been able to find anything at all. ):

    Thanks guys!
  2. jcsd
  3. Jan 14, 2012 #2

    Simon Bridge

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    I've not seen the notation r<θ before, however I've see the RHS before:

    You understand that this is just r = xi + yj - which you should have seen before. All they've done is put x=rcosθ and y=rsinθ where r = |r| = √(x2+y2). This is a polar representation of Cartesian coordinates.

    For 3D, you can just keep the z axis coordinate, which would be a cylindrical representation, or you can introduce a second angle (usually taken from the +z axis) for a spherical representation.

    IRL: you are more likely to use polar, cylindrical, and spherical coordinate systems directly.
  4. Jan 14, 2012 #3


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    TS probably meant this [itex]r\angle\theta[/itex] which is common shorthand for polar notation.
  5. Jan 14, 2012 #4

    Simon Bridge

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    Yeah - makes sense, thanks.
    Limitations of text only representations.
  6. Jan 14, 2012 #5

    Simon Bridge

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    u=(x,y) and v=(w,z) then [itex]\vec{u}\cdot \vec{v}=uv\cos(A)[/itex] where A is the angle between them.

    Continuing, using cap letters for angles:

    in polar notation, u=(r,B), v=(s,C) then surely A=|B-C| is the angle between them?

    So [itex]\vec{u} \cdot \vec{v} = rs\cos(B-C)[/itex]

    the cross product would be [itex]\vec{u} \times \vec{v} = rs\sin(C-B)[/itex] (final minus initial).

    eg. when B=0, the angle between them is just C. As B increases, the angle between them decreases.
  7. Jan 15, 2012 #6
    Thank you so much for the clarifications Simon! It's really very much appriciated. Also can I ask how you calculate the cross products when you have something like lets say:

    1. r<210 x r<150 - I know you get a vector that only has a k component vector but how do you know this?

    2. r<210 x 4k? - Is there a method to do this or does r<210 have to be converted to cartician coordinates first?

    Thank you guys!
  8. Jan 15, 2012 #7

    Simon Bridge

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    1. the cross product of two vectors is perpendicular to both and both example starting vectors are in the x-y plane ... the only perpendicular direction is in the k direction.

    If you use the i-j notation, and multiply them out, you get terms in ixi, jxj, ixj, and jxi ... since the unit vectors are orthogonal ixi=jxj=0 and ixj=-jxi=k But to know "why the cross product is always perpendicular", you have to look into what the term means.

    Wikipedia associates the cross product with "perpendicularness" but I tend to relate it to rotations ... momentum is a vector, angular momentum is also ... except it is a curly one. We use the right-hand-screw rule to unambiguously represent the rotation as a vector perpendicular to it.

    2. It would be easiest to do directly by converting r<120 to cartesian ... or you can use cylindrical coordinates. But it is more likely that you will change coordinates so the vectors both lie in the x-y plane. For any two vectors, this is always possible - which will be partly why you are being taught it. (In general, the orientation of axis is arbitrary - so we pick the orientation that makes the math easiest.) There are many ways to skin a cat - pick the easy one for the situation.

    Eventually you'll be understanding these in terms of matrix calculations. example
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