Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following problem. Does anyone know how to start with? Any links on what to read in order to understand what I have to do?

Which of the following transformations map the points $t_0=(-1,-1,-1)^T,t_1=(1,1,-1)^T$, $t_2=(-1,1,1)^T$, and $t_3=(1,-1,1)^T$ of the tetrahedron $T$ to the points $u_0=(2,2,-1)^T$, $u_1=(2,2,3)^T,u_2=(0,0,1)^T$, and $u_3=(4,0,1)^T$ (i.e. map $t_i$ to $u_i$ for $i=0,1,2,3$)?

  1. Rotation about the $z$-axis by $45^\circ$, then scaling with $s_x=\sqrt2,s_y=\sqrt2,s_z=1$, then translation by $(2,1,-1)$, and then rotation about the $x$-axis by $90^\circ$.
  2. Rotation about the $x$-axis by $90^\circ$, then scaling with $s_x=\sqrt2,s_y=1,s_z=\sqrt2$, then rotation about the $y$-axis by $45^\circ$, and then translation by $(2,1,1)$.
  3. Translation by $(1,-1,2)$, then rotation about the $z$-axis by $45^\circ$, then scaling with $s_x=\sqrt2,s_y=\sqrt2,s_z=1$, and then rotation about the $x$-axis by $90^\circ$.
  4. Scaling with $s_x=\sqrt2,s_y=\sqrt2,s_z=1$, then rotation about the $z$-axis by $45^\circ$, then translation by $(2,1,-1)$, and then rotation about the $x$-axis by $90^\circ$.
  5. Rotation about the $x$-axis by $90^\circ$, then rotation about the $y$-axis by $45^\circ$, then translation by $(1,1,\sqrt{1/2})$, and then scaling with $s_x=2,s_y=1,s_z=\sqrt2$.

Any help would be grateful.

share|improve this question
    
The problem should be copied here rather than given as a link. Links to remote sites go down and then we have no access to your question. People are also more likely to answer a question here than on a remote site. –  robjohn Nov 2 '12 at 22:43
    
yes you are right but you should have more than 10 reputation and I don't so it didn't let me to post an image. If any moderator can edit my post it would be great. –  kokos Nov 2 '12 at 22:46
    
Please check my copying of the image. –  Brian M. Scott Nov 2 '12 at 23:09
    
@BrianM.Scott thank you very much! Ok now that is displayed in a nice way, I would like some help to start with it if anyone knows. –  kokos Nov 2 '12 at 23:16
add comment

1 Answer

up vote 1 down vote accepted

Rotation about a coordinate axis is a linear transformation, so it has a correponding matrix. For instance, there is a matrix $A$ such that $A(x,y,z)^T$ is the vector that you get when you rotate $(x,y,z)^T$ $45^\circ$ about the $z$-axis. To find the matrix of a linear transformation, just work out where the transformation sends the basis vectors $(1,0,0)^T,(0,1,0)^T$, and $(0,0,1)^T$, and make those images the columns of the matrix.

For instance, the matrix that performs a $45^\circ$ rotation about the $z$-axis is $$A=\begin{bmatrix}1/\sqrt2&-1/\sqrt2&0\\1/\sqrt2&1/\sqrt2&0\\0&0&1\end{bmatrix}\;,$$ because this rotation takes $(1,0,0)^T$ to $(1/\sqrt2,1/\sqrt2,0)^T$, $(0,1,0)^T$ to $(-1/\sqrt2,1/\sqrt2,0)^T$, and $(0,0,1)^T$ to $(0,0,1)^T$.

Similarly, the matrix that performs a $90^\circ$ rotation about the $x$-axis is

$$B=\begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}\;,$$ because that rotation takes the basis vectors $(1,0,0)^T,(0,1,0)^T$, and $(0,0,1)^T$ to $(1,0,0)^T$, $(0,0,1)^T$, and $(0,-1,0)^T$, respectively.

Scaling by $s_x,s_y,s_z$ means transforming $(x,y,z)^T$ to $(s_xx,s_yy,s_zz)^T$; this operation is performed by the matrix $$\begin{bmatrix}s_x&0&0\\0&s_y&0\\0&0&s_z\end{bmatrix}\;;$$ if you don’t immediately see why, calculate the vector $$\begin{bmatrix}s_x&0&0\\0&s_y&0\\0&0&s_z\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\;.$$

Translation by by $(2,1,-1)^T$ is not a linear transformation; to translate $(x,y,z)^T$ by $(2,1,-1)^T$, just add to get $(x+2,y+1,z-1)^T$.

Thus, if you start with a vector $(x,y,z)^T$ and perform the operations described in (1), you get the vector

$$B\left(SA\begin{bmatrix}x\\y\\z\end{bmatrix}+\begin{bmatrix}2\\1\\-1\end{bmatrix}\right)\;,\tag{1}$$

where $S$ is the appropriate scaling matrix.

Now you can simply substitute the vectors $t_0,t_1,t_2$, and $t_3$ for $(x,y,z)^T$, calculate their images under this sequence of transformations, and see whether these images are $u_0,u_1,u_2$, and $u_3$, respectively.

At worst you can do the same thing with each of (2)-(5); this is tedious, but it works. If you’ve already learned something about how rotations, scaling, and translations interact with one another, you may be able to shorten the work considerably. For example, if you know that rotation commutes with scaling, you’ll see that the operations in (1) and (4) have exactly the same result; in terms of $(1)$ this is because $SA=AS$, since $S$ is a diagonal matrix.

share|improve this answer
    
it's more clear for me now because of your detailed explanation! thank you. –  kokos Nov 2 '12 at 23:57
    
@kokos: You’re welcome. –  Brian M. Scott Nov 2 '12 at 23:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.