How Do You Handle 3D Vector Operations with Polar Coordinates?

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Homework Help Overview

The discussion revolves around the application of 3D vector operations using polar coordinates, specifically the notation r<θ. Participants are exploring how to handle vector operations such as dot and cross products in this context, particularly when transitioning from 2D to 3D representations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the conversion of polar coordinates to Cartesian coordinates and how this affects vector operations in three dimensions. Questions arise about the nature of the cross product and its resulting direction, especially when involving vectors in the k direction.

Discussion Status

There is an ongoing exploration of the concepts, with some participants providing clarifications about the notation and the underlying principles of vector operations. Multiple interpretations of the notation and methods for calculating cross products are being discussed, indicating a productive exchange of ideas.

Contextual Notes

Some participants express uncertainty about the notation r<θ and its implications for 3D vector operations. There are mentions of potential limitations in text-based explanations and the need for visual aids to better understand the concepts.

Icetray
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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 appreciate 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!
 
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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.
 
Simon Bridge said:
I've not seen the notation r<θ before.

TS probably meant this [itex]r\angle\theta[/itex] which is common shorthand for polar notation.
 
Yeah - makes sense, thanks.
Limitations of text only representations.
 
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.
 
Simon Bridge said:
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.

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 let's 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!
 
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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