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Let me first fix some notation to avoid confusion. Assume I have two frames $A$ and $B$, having basis vector $i_{A}, j_{A}, k_{A}$ and $i_{B}, j_{B}, k_{B}$, respectively, and a rotation matrix $R_{A/B}$ that converts a three dimensional vector represented in the $B$ frame, say $v|_{B}$, to the same vector represented in the $A$ frame, that is

$$ v|_{A} = R_{A/B}v|_{B}$$

Also, assume that the last row of $R_{A/B}$, represented by $R_{A/B}(3)$, is

$$R_{A/B}(3) = \begin{bmatrix}r_{1} &r_{2}&r_{3}\end{bmatrix}$$

Now, if $v|_{A}=\begin{bmatrix}v_{1} &v_{2}&v_{3}\end{bmatrix}$, and $v|_{B}=\begin{bmatrix}w_{1} &w_{2}&w_{3}\end{bmatrix}$, then

$$v_{3} = r_{1}w_{1}+r_{2}w_{2}+r_{3}w_{3}$$

Now my question is: If I convert $v|_{B}$ to a two dimensional vector my taking a norm, say $v_{n}|_{B} = \begin{bmatrix}||w_{1},w_{2}||&w_{3}\end{bmatrix}$, where this new two dimensional vector is now in the 2D frame $B'$, where $k_{B'} = k_{B}$ (similarly for A'), how can I get a rotation matrix from $B'$ to $A'$ that gives me the same $v_{3}$ as before.

By brute force calculation, I can set the third row of the new 2x2 rotation matrix, say $\begin{bmatrix}a & b\end{bmatrix}$, to give me the desired result:

$$ v_{3} = r_{1}w_{1}+r_{2}w_{2}+r_{3}w_{3} = a||w_{1},w_{2}|| + bw_{3} $$

This gives me $a = (r_{1}w_{1}+r_{2}w_{2})/||w_{1},w_{2}||$ and $b = r_{3}$. However, you can notice that my new 2x2 rotation matrix is a function of the vector I am trying to rotate. I want to avoid this, but I don't know how.

My brain keeps going towards axis angle and taking advantage of that somehow, but I have been thinking about this for some hours and my head is spinning. Any advice or direction would be greatly appreciated.

Mr. Fegur
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  • Is a frame a basis? – copper.hat Sep 22 '13 at 21:26
  • Yes. Each frame has associated with it three basis unit vectors. So you can say FA has associated with it iA, jA, and kA, and FB has associated with it iB, jB, and kB. I am trying to compress two of them by usage of the norm. – Mr. Fegur Sep 22 '13 at 21:31
  • I can't figure out what $v_n \mid_B$ has to do with a rotation matrix from $B$ to $A$? – copper.hat Sep 22 '13 at 21:52
  • Oh I see what you are asking now. Say $F_{B}$ has basis vectors $i_{B}$, $j_{B}$, $k_{B}$, and we write $v = w_{1}i_{B} + w_{2}j_{B}+w_{3}k_{B}$. Now when I compress $w_{1}$ and $w_{2}$ to $||w_{1},w_{2}||$, the resulting vector $\begin{bmatrix}||w_{1},w_{2}|| & w_{3}\end{bmatrix}$ is not in the $B$ frame. Well, can I just assume it is in some new frame $B'$ with basis $i_{B'}$, $k_{B'}$, $k_{B'}=k_{B}$ and similarly for $A'$, and so $v|{n} = ||w{1},w_{2}||i_{B'} + w_{3}k_{B'}$? I am mostly interested in having $w_{3}$ unchanged after transformation from $B'$ to $A'$.Thanks! – Mr. Fegur Sep 22 '13 at 22:39
  • Perhaps I am missing something but can't you just use a rotation that leaves the 3rd component unchanged, and rotates $(w_1,w_2)$ to $(0, |(w_1,w_2)|)$? – copper.hat Sep 22 '13 at 22:57
  • Ok clearly I need to pause before I hit the enter button. In my last comment, I said "I am mostly interested in having $w_{3}$ unchanged after transformation from $B′$ to $A′$.", but what I actually meant was that I am interested in having $w_{3}$ equal to $v_{3}$ after transformation from $B'$ to $A'$, where $v_{3}$ as defined above is $v_{3}=r_{1}w_{1}+r_{2}w_{2}+r_{3}w_{3}$. Sorry for the confusion and thanks for bearing with me. – Mr. Fegur Sep 22 '13 at 23:11

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