# Converting a unit vector from cartesian to cylindrical

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1. May 19, 2015

### vector_problems

Hi, I was wondering if anyone could help with a vector question that I have.

If I have a unit vector defined in cartesian co-ordinates as p= (0,1,0) how would I go about converting this vector to a cylindrical geometry.

I understand that I will probably need to use p_r=sqrt(px^2+py^2) and p_theta=arctan(py/px) but also want to convert the vector to a cylindrical geometry and i don't think it can be as simple as this

thanks

2. May 19, 2015

### BiGyElLoWhAt

You need to use the sine and cosine definitions. I'll draw a picture.

3. May 19, 2015

### vector_problems

thank you

4. May 19, 2015

### BiGyElLoWhAt

Look at the definitions, and see if you can figure out what to do. If not, post back and we'll take it from there.

#### Attached Files:

• ###### unit_vector_definition.png
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5. May 19, 2015

### BiGyElLoWhAt

By the way, welcome to physics forums vector problems.

6. May 19, 2015

### vector_problems

thank you very much for the welcome :)

just to clarify in case this changes things. The vector i am working with is a direction vector rather than a position vector. i.e. it describes the orientation of a particle with preferred orientation (0,1,0).

Will the same theory apply, thanks for the help. This is really outside my wheel house

7. May 19, 2015

### BiGyElLoWhAt

Yea, it doesn't matter, as far as math is concerned, a vector is a vector. And you're welcome for the welcome. The definition I provided in the image holds true for any position on that circle.

8. May 19, 2015

### vector_problems

from the picture it would seem that p_polar=(r,theta,z)=(1,pi/2/0) but this just seems too simple.

I would have thought i would have needed to use some sort of transformation matrix to transform from cartesian to polar e.g. p_r=cos(theta)p_x+sin(theta)p_y, p_theta=-sin(theta)p_x+cos(theta)p_y but is this instead only for a flow with multiple components.

Thanks again

9. May 19, 2015

### BiGyElLoWhAt

You can use a transformation matrix, but it's also equivalent to a system of equations relating the coordinates. Since y is independant of x as well as for z in any order, the matrix is unnecessary. Also, what you have is correct. (1,pi/2,0) is the correct set of coordinates for (0,1,0).

10. May 19, 2015

### vector_problems

so if i had a more general point in cartesian co-ordinates; p_cart=(px,py,pz) to express this in a cylindrical geometry do I just do p_polar=(sqrt(px^2+py^2),arctan(py/px),pz) or will I need a transformation matrix in this case

thanks

11. May 19, 2015

### BiGyElLoWhAt

nope, that's good.

12. May 19, 2015

### vector_problems

ok thats great, thank you very much for your help

13. May 19, 2015