# 3d matrix rotation

I've been reading up on rotation transformation and following this PowerPoint.

The assignment I'm working on is asking to perform a rotation and find the image $$Q$$ of the point P = (1, 2, -1) after a $$45$$ degree $$y$$-roll. I was under the impression that a roll was on the $$z$$-axis; which is where I'm confused.

I'm trying to set up the problem like such (from WikiPedia):

But I'm lost as to translate the 3 points into a matrix. Is this correct? The way I'm trying to visualize it as below without the translation.

• I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $\frac{\pi}{4}$. – copper.hat Apr 4 '13 at 20:16
• @copper.hat Would this be an accurate representation? – Kermit Apr 4 '13 at 20:19
• Looks good... ${}{}{}$ – copper.hat Apr 4 '13 at 20:23
• I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis. – rschwieb Apr 4 '13 at 20:23
• Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation. – rschwieb Apr 4 '13 at 20:32

To roll $\theta=\pi/4$ around the $y$ axis, the matrix becomes:
$$\begin{bmatrix}\sqrt{2}/2&0&\sqrt{2}/2\\0&1&0\\-\sqrt{2}/2&0&\sqrt{2}/2\end{bmatrix}$$
$$\begin{bmatrix}\sqrt{2}/2&0&\sqrt{2}/2\\0&1&0\\-\sqrt{2}/2&0&\sqrt{2}/2\end{bmatrix}\begin{bmatrix}1\\2\\-1\end{bmatrix}=\begin{bmatrix}0\\2\\-\sqrt{2}\end{bmatrix}$$
So, $Q=(0,2,-\sqrt{2})$