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My professor assigned us this homework for a separate text book from our own. I am very lost on how to approach this problem. While our text covers homogeneous matrices none are the examples are like anything in the context of this problem. I was hoping someone could point me to a resource that would be helpful in attacking this problem.

The first transformation I got: $$ A^0_1 = \left[ {\begin{array}{cc} -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & C+E \\ 0 & -1 & 0 & A-D \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] $$ But for the second I get confused, I get: $$ A^1_2 = \left[ {\begin{array}{cc} 0 & -1 & 0 & B \\ 0 & 0 & -1 & A-D \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] $$

Thanks!

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Are those $x_i,y_i,z_i$ supposed to be unit vectors parallel to the $x,y,z$ axes respectively? (Plus, I assume all transformations are from the point indexed by $0$ to all the other indexed points.) Also, it might be helpful if you explained why this problem's context is different from what you've done before, and how you computed $A_1^0$, $A_2^1$. (Not strictly necessary, but it'd be helpful.) –  anon Sep 22 '11 at 19:06
    
Yes, your thought are correct. Why this is difficult was before we were given the two matrices and we would transform them mathematically, but now we are asked to do this visually by looking at the two points. –  Nick Sep 22 '11 at 19:19
    
To be honest I am a little lost. For the first matrix I approached it that the $$Z_0,Z_1 and Y_0,Y_1 $$ made a plane. So my logic was x is pointing out of the plane so its 0, y was C+E (length of plane), and z was A-d (height of plane). I applied the same logic to point 1 wrt point 2 x,y make a plane and z points out of the plane. I don't think this is right but its how my brain make sense of the problem. Anon are you talking about point 0 wrt to point 2? –  Nick Sep 22 '11 at 19:45
    
My comment was indeed talking about 0 wrt 2, which was why I deleted it. I've given a demonstration of how to compute $A_2^1$, tell me if it helps or if you don't grasp it fully. –  anon Sep 22 '11 at 20:01

1 Answer 1

up vote 1 down vote accepted

So apparently my initial assumption that the coordinate system remains the same for each matrix is false, and your matrix for $A_2^1$ is correct after all. Let's look at $A_3^2$ this time. The transformation maps $$e_1\to-e_3,$$ $$e_2\to e_2,$$ $$e_3\to e_2.$$ Mapping $e_i$ to $e_j$ means the $i$-th column of the $3\times3$ part is $e_j$ (so that $T(e_i)=e_j$). Any negative signs get multiplied for the final column. So we get $$\begin{pmatrix}-e_3&e_2&e_1\end{pmatrix}=\begin{pmatrix}0&0&1\\0&1&0\\-1&0&0\end{pmatrix}.$$ Now the displacement vector from point 2 to point 3 is $(e,0,a)$, so the final matrix is $$T=\begin{pmatrix}0&0&1&e\\0&1&0&0\\-1&0&0&a\\0&0&0&1\end{pmatrix}.$$

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Your logic makes perfect sense. I understand the 3x3 part (I did it from the wrong point before) the part where I continue to struggle is the translations of the last column. For instance with the same logic $$ A^3_2$$ 3x3 $$ \left[ {\begin{array}{cc} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} } \right]$$ I think is right? Fingers crossed :) but the last column still alludes me –  Nick Sep 22 '11 at 20:24
    
@Nick: Nope, wrong again. :) Write out the maps like I did and take it slowly. For point $2$ to point $3$, you should have got $-e_2\to-e_3$, $-e_1\to-e_1$ and $e_3\to-e_2$ to begin with. As for the last column, you need to ignore the $x_i$'s and look at just the vertex points: in other words the top three components of the last column will be a displacement vector that takes the original point to the destination point. Does that make sense? –  anon Sep 22 '11 at 20:36
    
I'm sorry, I'm stumped, I've been staring at this for 45mins. I can't figure out why that matrix isn't right. From point 2 to 3 all you have to do is rotate 90degrees about the y axis to get the same point and the displacement would be E in the x and just a in the z? So confused! Sorry it takes me a little bit to pick up on stuff :( –  Nick Sep 22 '11 at 21:34
    
@Nick: Those y vectors are pointed in the negative x direction, according to our original coordinate system inherited from point 0. Or is it the case that we're supposed to change the basis for each and every transformation matrix? That was never indicated in either the picture or your question, but if it's true and you're aware of that fact you should make it clear so we can help you. (Also, you're asking questions on a Q&A site, that's what this place is for so there's no need to apologize.) –  anon Sep 22 '11 at 21:45
    
To be honest, I'm uncertain as well. I went into my professor earlier today and the matrix in the original post $A^1_2$ he said was correct. I'm not sure if that helps clarify anything but I also assumed he could of looked at it wrong. His words were "Z is 0 because it coming out of the plane"(for that case.) I have all of the 3x3 parts done and he seem to think they were right when he briefly glanced at it. But now looking more closely at the question should I have 10 answers for this question? The 5 matrices for $A^{i-1}_i$ and 5 for $A^0_i$. –  Nick Sep 22 '11 at 22:09

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