1. Sep 24, 2011

SBoskovi

I would just like to start off by saying the problem comes from Intro to electrodynamics, 3rd edition, griffiths. the problem is number 1.9.

Question: Find the transformation matrix R that descries a rotation by 120 degrees about an axis from the origin through the point (1,1,1) the rotation is clockwise as you look down at the axis from the origin.

1. The problem statement, all variables and given/known data

A(bar)x=Az
A(bar)z=Ay
A(bar)y=Ax

2. Relevant equations

Transformation law =

A(bar)x Rxx Rxy Rxz Ax
A(bar)y = Ryx Ryy Ryz Ay
A(bar)z Rzx Rzy Rzz Az

3. The attempt at a solution

I know that the solution is

R= 0 0 1
1 0 0
0 1 0

I am confused on how you get this? I have been trying to figure this out for a while. I know that i am over looking something here because it should be a very simple problem to do...

I have been looking at this problem for a long period of time...

I DO HAVE THE ANSWER KEY TO THE BOOK AND I DO HAVE CRAMSTER; I JUST CANNOT FIGURE OUT HOW TO GET THIS ANSWER... SOMEONE PLEASE HELP ME OUT! The answers that the answer key and Cramster provide as missing information to the point where I cannot follow what has occurred.

2. Sep 24, 2011

vela

Staff Emeritus

Try breaking that rotation down into rotations for which you do know how to write the matrices down.

3. Sep 24, 2011

SBoskovi

i know how the matrix is suppose to work,

the

Az=Rxx multiplies the Ax + Rxy multiplies the Ay + Rxz multiplies the Az
Ax=Ryx multiplies the Ax + Ryy multiplies the Ay + Ryz multiplies the Az
Ay=Rzx multiplies the Ax + Rzy multiplies the Ay + Rzz multiplies the Az\

^^ I understand what is going on there, simple..^^

but this is where i am getting confused, how do you end up with these actual numbers?

this results in

Rxx= 0 Rxy= 1 Rxz=0
Ryx= 1 Ryy= 0 Ryz=0
Rzx= 0 Rzy= 1 Rzz=0

This is not a homework problem, this is me trying to learn from the book and reading through the problem. I have tried to ask my professor for help but he refuses to help me on such a "simple" problem...

If you dont feel comfortable explaining it to me on the actual forum bc someone might use it as the homework answer you can PM. Please i am really confused

4. Sep 24, 2011

steve233

There's a solution manual online.
I used it when I took this course to practise for the final and midterm (since we didn't have questions assigned from the text book). You might want to consider downloading it.

Try google-ing the name of the text book and solution manual.

5. Sep 24, 2011

SBoskovi

I have the solutions manual, and I also have Cramster.

I AM NOT INTERESTED IN JUST GETTING THE ANSWER, I AM INTERESTED IN FIGURING OUT HOW TO DO THE PROBLEM. I USE THE ANSWER KEY TO FOLLOW MY WORK AND ONCE I GET STUCK TO SEE WHAT HAPPENED BUT WHEN IT COMES TO THIS PROBLEM I AM JUST PLAIN OUT STUCK AND HAVE NO IDEA WHAT HAPPENED.

6. Sep 24, 2011

steve233

Right, I only suggested it because they have step by step solutions and I remember it being helpful in solving problems.

7. Sep 24, 2011

vela

Staff Emeritus

Like I said earlier, break it into several separate rotations.

8. Aug 26, 2012

cdonlan87

IMO, this problem's answer is HARD. How the hell did the book get [(001);(100);(010)]? Only thing I can think of is the vector is sweeping through two of those cubes from pg 6

Anyways,if you want to actually get a decent *generic* translation, use a rotation matrix for X, Y and Z axis. Generate them using the method dilineated in the book. Apply them in succession.

For instance, the actual answer to the problem would be:

A*Xrot(-120)*Xrot(45)*Yrot(45)*Zrot(45) = A-bar.

Last edited: Aug 26, 2012
9. Aug 26, 2012

TSny

Consider the vector A = (1, 0, 0). What are the components of this vector in the rotated frame (i.e., what are the components of A(bar))?

Substitute these into your transformation law and deduce the value of some of the matrix elements of R.

Construct a couple of other vectors A that will allow you to find the other components of R.

Last edited: Aug 26, 2012
10. Aug 26, 2012

cdonlan87

This problem was solved analytically. Meaning it has nothing to do with rotation matrices and math and everything to do with the fact that the vector was precisely (1,1,1), and the degree distance was precisely 120 degrees.

One is supposed to notice that the vector 1,1,1 is 60 degrees away from each axis. Looking down the Vector, 360 degrees appears to be divided into 3 parts instead of 4.

Each of the 3 parts is 120 degrees. Thus a 120 degree rotation will move a vector from one set of quadrants to another...no editing of values...no component dependencies...no deductions...just irrelevant trickiness....

(well maybe not completely irrelevant...)

Its much the same thing one would do when told to rotate something 180 degrees around one of the axis.

Last edited: Aug 26, 2012
11. Aug 27, 2012

bigerst

There's a clever trick to this problem as cdonlan87 said. Don't try to get the generalized transformation matrix. It is indeed possible, but it would take you several hours. Think about this problem closely. (1,1,1) is a very special location, and 120 degrees is 1/3 of 360. Instead of thinking of the axis as you normally would, think of yourself looking down at the origin from (1,1,1). If that image was projected on a plane, what are the angles between the x, y, and z axes?

Last edited by a moderator: Aug 28, 2012
12. May 10, 2013

Superman1271

I don't understand how the vector is 60 degrees to each axis.

cosx^2+cosy^2+cosz^2=1

All three angles are the same, so 3(cosx)^2=1
which gives an angle of 54.7 degrees.

13. May 10, 2013

TSny

You are right, the angle is 54.7o.