# Homework Help: Am I doing this right? Doesn't feel right. (find axis of rotation)

1. Mar 17, 2012

### skyturnred

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

So here is the question:

Matrix A corresponds to the linear transformation T obtained by first rotating a vector in R3 through angle ∏/3 about the z axis and then through angle ∏/4 about the x-axis. Find the parametric equation for the axis of rotation.

2. Relevant equations

3. The attempt at a solution

Finding matrix A: First I write down the two standard rotations with the first one on the right and multiply them:

This gives me matrix A. I then take the result and subtract the 3x3 identity matrix (so Mat(A) - I3). I augment this by the 3x1 zero vector and rref. So the following is what I am rref-ing. (so I am solving this system (A-I)[w]=0, and the axis parallel to [w] is the axis of rotation)

But when I rref this, I get the following:

W3=t where t is in the reals
W2=-0.4142t
W1=0.7174t

This doesn't seem right to me.. so the parametric form of the axis of rotation is this:

x=0.7174t
y=-0.4142t
z=t

2. Mar 17, 2012

### sunjin09

The rotation axis must be perpendicular to both the starting vector and the ending vector, how do you find such a vector that is perpendicular to both?

3. Mar 18, 2012

### skyturnred

I don't quite understand, what are the "starting" and "ending" vectors? Does that mean that my method above is wrong?

I know that the cross product is how you find a vector that is perpendicular to two other vectors.

4. Mar 18, 2012

### sunjin09

I'm sorry for providing wrong information. Please ignore my last post.
Regarding the problem. If you wrote down the rotation matrix correctly, you should get the right answer. The rotation axis is nothing but the eigenvector of the rotation matrix with eigenvalue 1, which, after I solved for it, is exactly [0.7174, -0.4142, 1]'. If you get a different answer, try to check your calculation. FYI, the rotation matrix I got was
[0.5,-0.866,0
0.6124,0.3536,-0.7071
0.6124,0.3536,0.7071]