Finding the Rotation Matrix for Vector Alignment

[Athenian]

Homework Statement

Find the rotation matrix that will line up the orthogonal vectors,

$A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$

along the $y$- and $z$-axes, respectively.

Afterward, find a new basis in which the vectors $A$ and $B$ are aligned with the $\hat{x}_2'$ and $\hat{x}_3'$ basis vectors, respectively.

Relevant Equations

Refer Below $\longrightarrow$

Below is the attempted solution of a tutor. However, I do question his solution method. Therefore, I would sincerely appreciate it if anyone could tell me what is going on with the below solution.

First off, the rotation of the matrix could be expressed as below:
$$G = \begin{pmatrix} AB & -||A \times B|| & 0 \\ ||A \times B|| & AB & 0 \\ 0 & 0 & 1\end{pmatrix} $$

Solving for $AB$:
$$AB = 1-2+1=0$$

Solving for $A \times B$:
$$A \times B = -i$$
$$||A \times B|| = 1$$

Therefore, the solution to the first part of the question is:
$$G = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix} $$

My Questions:
First off, what kind of rotation matrix is $G$? Wasn't the rotation matrix supposed to consist of sines and cosines (as well as 0's and 1's)?

In addition, the wording of the question really confuses me. Essentially, the question wants me to find the rotation matrix that will allow vector $A$ to align itself along the $y$-axis whereas vector $B$ should align itself along the $z$-axis. Am I interpreting the question correctly?

Finally, how am I supposed to find $\hat{x}_2'$ and $\hat{x}_3'$ basis vectors? I watched through the given tutorial several times and I feel like the definition for the "basis vectors" (as well as how to use or find for it) are somewhat ill-defined.
However, I am going to simply assume the tutorial is fine and that I may have some knowledge gap that is preventing me from properly understanding the given material.
Therefore, any assistance in helping me out with this question would be greatly appreciated!

Side Note: I'll get back to working on this thread (i.e. https://www.physicsforums.com/threa...steady-state-temperature-distribution.987906/) soon enough. Thank you for your patience for those working with me there.

 

[Orodruin]

The given answer is obviously wrong as it does not change the z-component. Since A has a non-zero z-component the given matrix will not line it up along the y-axis.

First off, what kind of rotation matrix is GGG? Wasn't the rotation matrix supposed to consist of sines and cosines (as well as 0's and 1's)?

It does. It is just that the angle is 90 degrees so sines and cosines are 0, 1, or -1.

 

[Athenian]

Thank you very much for the comment, @Orodruin! I didn't know that the 0, 1, and -1 were all along taken from sines and consines. It makes a lot more sense that way. Thank you for the information!

Beyond that, I finally figured out the answer to the first part of the question. And, yes, you were right. The tutor was just flat-out wrong.

After two rotations along different axes, I got $A'' = \begin{pmatrix} 0 \\ \sqrt{3} \\ 0 \end{pmatrix}$ as well as $B'' = \begin{pmatrix} 0 \\ 0 \\ \sqrt{6} \end{pmatrix}$ respectively.

The problem wasn't really hard at all. I suppose the wording of the problem kind of threw me off a bit. But, once I understood what the problem was trying to ask (and yes, I was slow to the "uptake"), I was able to calculate the problem pretty straightforwardly.

However, how do I go about calculating the below follow-up problem?

Afterward, find a new basis in which the vectors $A$ and $B$ are aligned with the $\hat{x}_2'$ and $\hat{x}_3'$ basis vectors, respectively.

First off, I do know that:
$$\begin{pmatrix} \hat{x}_1 ' \\ \hat{x}_2 ' \\ \hat{x}_3 ' \end{pmatrix} = T_{(3)} \begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix}$$

where $T_{(3)}$ is the orthogonal matrix.

However, first off, how should I go about determining $\hat{x}_2'$ and $\hat{x}_3'$ basis vectors?

 

[Orodruin]

You do not need to do anything like that. You know that A and B are proportional to the basis vectors. Just normalize them.

 

[Athenian]

My apologies for the late reply as I was working on other - and more urgent - problems. Nonetheless, thank you very much for your kind assistance, @Orodruin .

Below is my attempted solution to solving the problem. What do you think?

Since $A$ is orthogonal to $B$ and $B$ is orthogonal to $A$, I will then continue by normalizing the two vectors as shown below:

Normalizing $A$ gives $\begin{pmatrix} \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix}$.

Normalizing $B$ gives $\begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix}$.

Note that the original values of $A$ and $B$ are $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -2 \\ 1\end{pmatrix}$ respectively.

With this in mind, below is the expression for my new basis:
$$\begin{pmatrix} \hat{x}_1 ' \\ \hat{x}_2 ' \\ \hat{x}_3 ' \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix} \begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix} $$

Would this be correct?

 

[Orodruin]

With this relation your third (or rather first) basis vector would be zero.

 

[Athenian]

So, it should be this instead then? Is this what you are suggesting?

$$\begin{pmatrix} \hat{x}_1 ' \\ \hat{x}_2 ' \\ \hat{x}_3 ' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix} $$

With this relation your third (or rather first) basis vector would be zero.

I don't quite get what you mean by the third basis vector is actually the first basis vector though. Do you mind explaining that?

 

[Orodruin]

Now your $\hat x_3’$ is zero. None of your basis vectors should be zero.

 

[Athenian]

I think I get what the question is trying to ask finally. For starters, though, does the below equation look accurate now?

$$\begin{pmatrix} \hat{x}_1 ' \\ \hat{x}_2 ' \\ \hat{x}_3 ' \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}}\end{pmatrix} \begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix} $$

If I plug vector $B$ into $\begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix}$, I get the value for vector B lying along the $z$-axis which I found in the first part of the problem. Similarly, doing the same thing with vector $A$ gives the value of vector $A$ lying on the $y$-axis instead (also dictated in the first part of the problem). Please refer to post #3 for more information.

So, was this what the question was trying to ask all along with the phrase "Afterward, find a new basis in which the vectors $A$ and $B$ are aligned with the $\hat{x}_2 '$ and $\hat{x}_3 '$ basis vectors, respectively"?

Thank you so much for your help and please do let me know what you think.

 

[Orodruin]

I think I get what the question is trying to ask finally. For starters, though, does the below equation look accurate now?

$$\begin{pmatrix} \hat{x}_1 ' \\ \hat{x}_2 ' \\ \hat{x}_3 ' \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}}\end{pmatrix} \begin{pmatrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{pmatrix} $$

No. This would still give $\hat x_1' = 0$. You cannot just put the first row of the matrix arbitrarily to zero.

 

[Athenian]

None of your basis vectors should be zero.

You're right. I wasn't looking carefully enough there.

But, in that case, I am getting pretty lost on what I am trying to do. What should be placed instead of three 0's instead then? I placed three 0's there precisely because I only have two vectors (i.e. $A$ and $B$) which I can find for their orthonormal vectors shown in the previous post. Any suggestions on what should be placed instead of the triple 0's?