How to Convert Vectors to Spherical Coordinates at Given Points?

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Homework Help Overview

The discussion revolves around converting vectors to spherical coordinates at specified points, specifically transforming the vector 10ax at the point P (x = -3, y = 2, z = 4). The subject area includes vector transformations and coordinate systems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationships between Cartesian and spherical coordinates, questioning how to derive the angles θ and φ, as well as the radius r. There are discussions about the validity of the vector notation used and the implications of the transformation equations provided.

Discussion Status

The conversation is ongoing, with participants offering insights into geometric interpretations and the nature of the vector notation. Some guidance has been provided regarding the relationships between the coordinates, but there is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants note potential confusion regarding the notation of the vector 10ax and its representation in Cartesian versus spherical coordinates. There is also mention of the complexity involved in the calculations required to find the necessary variables.

Dong Hoon Lee
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Homework Statement


transform the following vectors to spherical coordinates at the points given

10ax at P (x = -3 , y = 2, z=4)

Homework Equations


x y z can be chage into x = rsinθcosφ , y=rsinθsinφ , z=cosθ

The Attempt at a Solution


ax vector can be expressed ar,aθ,aφ so, I can change x , y, z through 2.

therefore we have to find θ, φ, r

also, we can know the sinθcosφ = x/r sinθsinφ = y/r z=cosθ



10ax = 10ax ar + 10ax aθ + 10ax aφ

= 10sinθcosφ + 10cosθcosφ - 10sinφ



>>>> I can fill out innerproduct between x and r but how to solve the others,

Is it right answer? I want to find more correctly one (it make lots of number, because find each variables through lots of calculation)



>>>> I want to more objective soultion !



>>>>> want to know how to chage between sehperical coordinates and cartesian coordinates
Thank you for your attention for me
 
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>>>> I can fill out innerproduct between x and r but how to solve the others,
You have three equations and three unknowns.
It is better to use geometry though ... ferinstance "r" is the magnitude of the vector ##\vec r##, which is given by ##r^2=x^2+y^2+z^2##
Try sketching the vector.

I notice that "10ax" is not a vector though.
 
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Of course you could always just look up the transformation from cartesian to spherical ...
 
Dong Hoon Lee said:

Homework Statement


transform the following vectors to spherical coordinates at the points given

10ax at P (x = -3 , y = 2, z=4)

Homework Equations


x y z can be chage into x = rsinθcosφ , y=rsinθsinφ , z=cosθ

The Attempt at a Solution


ax vector can be expressed ar,aθ,aφ so, I can change x , y, z through 2.

therefore we have to find θ, φ, r

also, we can know the sinθcosφ = x/r sinθsinφ = y/r z=cosθ
10ax = 10ax ar + 10ax aθ + 10ax aφ

= 10sinθcosφ + 10cosθcosφ - 10sinφ
>>>> I can fill out innerproduct between x and r but how to solve the others,

Is it right answer? I want to find more correctly one (it make lots of number, because find each variables through lots of calculation)
>>>> I want to more objective soultion !



>>>>> want to know how to chage between sehperical coordinates and cartesian coordinates
Thank you for your attention for me

When you write ##10ax## do you really mean ##10 a \vec{r} = 10 a \langle x,y,z \rangle?## One of these is a vector and the other is not.
 
Hmmm ... looking at the later notation: OP may be using "a" to indicate the unit vector ... see ar aφ etc later on, as in
10ax = 10ax ar + 10ax aθ + 10ax aφ
So 10ax would mean ##\vec r = 10\hat a_x = (10,0,0)## cartesian ... which is very easy to put in spherical coordinates.
OTOH: that does not fit so well with the rest of the problem statement: the vector does not depend on position for example.

OP has been back since I replied and "liked" the reply ... presumably got what was needed.
 

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