MHB Determine the matrices that represent the following rotations of R^3

[kalish1]

I need to determine the matrix that represents the following rotation of $R^3$.

(a) angle $\theta$, the axis $e_2$

(b) angle $2\pi/3$, axis contains the vector $(1,1,1)^t$

(c) angle $\pi/2$, axis contains the vector $(1,1,0)^t$

Now, I would like to check if I got the right answers because this problem has been quite difficult for me. Any help is greatly appreciated.

Please forgive me for skipping the work because formatting matrices is a real pain. Especially when I have a lot of them.

For part $(a)$, I got that $(e_2,e_3,e_1)$ is an orthonormal basis of $R^3$. Then after simplification, the matrix is

$$
\begin{matrix}
\cos(\theta) & 0 & \sin(\theta) \\
0 & 1 & 0 \\
-\sin(\theta) & 0 & \cos(\theta) \\
\end{matrix}
$$

For part $(b)$, I got an orthonormal basis as $\{[1/\sqrt(3), 1/\sqrt(3), 1/\sqrt(3)]^t, [1/\sqrt(2),-1/\sqrt(2),0]^t,[1/\sqrt(6),1/\sqrt(6),-2/\sqrt(6)]^t\}$.

Then after simplification, the matrix is $$
\begin{matrix}
-\sqrt(3)/2 & 0 & -1/2 \\
0 & 1 & 0 \\
1/2 & 0 & -\sqrt(3)/2 \\
\end{matrix}
$$

Is what I have done so far correct such that I can proceed with part $(c)$?

 

[Ackbach]

Part a is correct. I always point people to this web site for rotations about arbitrary axes. In your case, since your axis for part b and your axis for part c go through the origin, you won't need Steps 1 or 7, and you can also lop off the fourth column and fourth row of the remaining necessary matrices.

 

[I like Serena]

I need to determine the matrix that represents the following rotation of $R^3$.

(a) angle $\theta$, the axis $e_2$

For part $(a)$, I got that $(e_2,e_3,e_1)$ is an orthonormal basis of $R^3$. Then after simplification, the matrix is

$$
\begin{bmatrix}
\cos(\theta) & 0 & \sin(\theta) \\
0 & 1 & 0 \\
-\sin(\theta) & 0 & \cos(\theta) \\
\end{bmatrix}
$$

Correct. :)

(b) angle $2\pi/3$, axis contains the vector $(1,1,1)^t$

For part $(b)$, I got an orthonormal basis as $\{[1/\sqrt(3), 1/\sqrt(3), 1/\sqrt(3)]^t, [1/\sqrt(2),-1/\sqrt(2),0]^t,[1/\sqrt(6),1/\sqrt(6),-2/\sqrt(6)]^t\}$.

Then after simplification, the matrix is $$
\begin{bmatrix}
-\sqrt(3)/2 & 0 & -1/2 \\
0 & 1 & 0 \\
1/2 & 0 & -\sqrt(3)/2 \\
\end{bmatrix}
$$

Is what I have done so far correct such that I can proceed with part $(c)$?

Your orthonormal basis is correct.

But let's see...
The axis is invariant.
What do you get if you multiply your matrix by the axis (1,1,1)?
Is it (1,1,1)?

Furthermore, (1,-1,0) is perpendicular to the axis.
So if I multiply the matrix with it, I should get a vector that makes an angle of $2\pi/3$.
Put otherwise, the dot product should be $\sqrt 2 \cdot \sqrt 2 \cdot \cos(2\pi/3) = -1$.
Is it?
Moreover, that vector should also be perpendicular to the axis (1,1,1). Is it?