# rotation matrices

• Sep 25th 2011, 06:24 AM
shelford
rotation matrices
Hi, I am trying to find some rotation matrices of $R^3$.

So for angle $\frac{\pi}{2}$ and axis $e_2$ I think the rotation matrix is

$R_{e_2} (\frac{\pi}{2}) = \left(\begin{matrix} cos(\frac{\pi}{2}) &0 &sin(\frac{\pi}{2})\\0 &1 &0\\-sin(\frac{\pi}{2}) &0 &cos(\frac{\pi}{2})\end{matrix}\right) = \left(\begin{matrix} 0 &0 &1\\0 &1 &0\\-1 &0 &0\end{matrix}\right)$
I just used the given rotation matrix in $R^3$ from the text with no real working out needed. Is this correct?

Need help on this one:

For the angle $\frac{\pi}{6}$ and axis containing the vector $(1,1,1)^t$.

Not sure how to use the $(1,1,1)^t$ in this question.

Thanks for any help.
• Sep 27th 2011, 06:30 PM
shelford
Re: rotation matrices
I think I have misunderstood what the axis e2 is and I think my working out is wrong.

Any help would be nice.

Shelford
• Sep 29th 2011, 03:38 AM
shelford
Re: rotation matrices
Anyone have any ideas as to how I should interpret the axis $e_2$ and use $(1,1,1)^t$?

Thanks

And sorry for making another post, wanted to edit a previous post but I couldn't.
• Sep 29th 2011, 07:33 AM
Ackbeet
Re: rotation matrices
Quote:

Originally Posted by shelford
Hi, I am trying to find some rotation matrices of $R^3$.

So for angle $\frac{\pi}{2}$ and axis $e_2$ I think the rotation matrix is

$R_{e_2} (\frac{\pi}{2}) = \left(\begin{matrix} cos(\frac{\pi}{2}) &0 &sin(\frac{\pi}{2})\\0 &1 &0\\-sin(\frac{\pi}{2}) &0 &cos(\frac{\pi}{2})\end{matrix}\right) = \left(\begin{matrix} 0 &0 &1\\0 &1 &0\\-1 &0 &0\end{matrix}\right)$
I just used the given rotation matrix in $R^3$ from the text with no real working out needed. Is this correct?

Yes.

Quote:

Need help on this one:

For the angle $\frac{\pi}{6}$ and axis containing the vector $(1,1,1)^t$.

Not sure how to use the $(1,1,1)^t$ in this question.

Thanks for any help.
This is a much more complicated question, and requires more advanced treatment. The best information I can give you is to go to this website and use that procedure. As you can see, it's a multi-step procedure.