Recent content by xsever

I guess I really needed a break. It's \pi ! Thank you for your help. Much appreciated.

Well, I can see that the signs are opposite, but playing with the sign or even\pi/2 does not do the job. Wow! I am really not seeing this !

I know that \phi(radians) is negative and that would mean it is on the negative side. Doing 2\pi-0.982 will result in the same trigonometric values. I am sorry but I do not see what you are trying to explain to me.

In my book, \phi=tan-1 (y/x) = tan-1 (3/-2) = -0.982. Where is my mistake ?

Could you please upload a scan of your solution. I am trying my method and it's giving me wrong answers. I got R=\sqrt{14} \theta=1.3 \phi=-0.982 Therefore, my matrix looks like this: [AR ] [.535 -.801 .267 ] [2] [A\theta] [ .148 -.222 -.963] [-5] [A\phi] [.831 .555...

By using the method I mentioned, I am not getting the right answer. The right answer is: A= -R4.276 - (theta)4.299 + (phi)1.1094. If anyone can get that answer, please let me know of the method used. Thanks.

The matrix will take a vector from cartesian to spherical. Here it is:

Here is what I was thinking: The point (-2,3,1) should be transformed into spherical form. Then, vector A would be transformed using the matrix. As the matrix contains cos(theta) and cos(phi), those theta and phi values would be used from the transformation of the point. Does this make any sense?

I scanned the question just to confirm that I did not miss anything: The course is Engineering Electromagnetics and this question is in chapter 1. The chapter covers coordinate systems and transformations.

I have written the problem exactly as it was written in the book. I modified nothing and this is why I could not solve it and came here. Any thoughts/questions ?

Homework Statement Transform the vector A = x2 - y5 + z3 into spherical coordinates at (x= -2, y= 3, z=1 ).Homework Equations http://www.equationsheet.com/eqninfo/Equation-0348.html http://en.wikipedia.org/wiki/Spherical_coordinate_system The Attempt at a Solution I know the transformation...