2
$\begingroup$

I'm trying to calculate the rotating matrix around the Z axis in a counter clockwise direction.

In-order to do that I've decided on a vector A [1 0 0] and I wish to rotate it around the z axis so it'll end up as At [0 1 0] (90 degrees rotation around the z axis)

Let's consider our matrix R. Now, what I've done is set At coefficients to the product of the matrix R by the vector A as follows:

At.x = A.x * R00 + A.y * R10 + A.z * R20
At.y = A.x * R01 + A.y * R11 + A.z * R21
At.z = A.x * R02 + A.y * R12 + A.z * R22

Now, replacing the A and At coefficients we get:

0 = 1 * R00 + 0 * R10 + 0 * R20
1 = 1 * R01 + 0 * R11 + 0 * R21
0 = 1 * R02 + 0 * R12 + 0 * R22

Next, to make these equations right I've deducted the following:

R00 must be 0
R01 must be 1
R02 must be 0
others R coefficients are irrelevant to me

Creating the matrix so far:

0     1     0
0     0     0
0     0     0

Since we rotated around the z axis by 90 degrees we can say the following about the earlier deduction of the matrix coefficients:

R00 = cos(θ)
R01 = sin(θ)
R02 = cos(θ)

And finally, applying this to the matrix:

cos(θ)     sin(θ)     cos(θ)
  0          0          0
  0          0          0

After building my matrix, I've tested it here: Matrix Calculation Online Website

With the following same vector A [1 0 0] and matrix R I ended up with And got the desired Result of At [0 1 0]

So, summing it all the process I've done, my question is why for every resource I looked for the answer the rotating around the z axis matrix should look like this: Z axis rotation matrix

What have I done wrong / missed calculating to get the same matrix ?

$\endgroup$
  • $\begingroup$ You should test some different cases (maybe you have). It is very easy to get the right result accidentally. $\endgroup$ – Carser Jul 30 '16 at 16:06
  • $\begingroup$ @Carser I suppose you're right, but that would only make me realize that the way I found this matrix is wrong I wish to know how to be able to compute it $\endgroup$ – Jorayen Jul 30 '16 at 16:12
  • $\begingroup$ Well it seems you've designed your matrix specifically for the case of $90^\circ$, whereas the rotation matrix you link to is the more general case for any angle $\theta$. Basically, you're case is the specific instance for the general matrix where $\theta = 90^\circ$. $\endgroup$ – Carser Jul 30 '16 at 16:13
  • $\begingroup$ @Carser I see, could you tell me on how could I design it for any value of θ $\endgroup$ – Jorayen Jul 30 '16 at 16:17
  • $\begingroup$ As a quick sanity check of your solution, remember that the columns of a transformation matrix are the images of the basis vectors. The matrix you’ve ended up with collapses everything onto the $x$-axis! This can’t possibly be right, since, among other things, the $z$-axis shouldn’t be affected at all by this rotation. $\endgroup$ – amd Jul 30 '16 at 21:02
0
$\begingroup$

It seems like you've designed a set of equations specifically for the case of $\theta=90^\circ$. If you want to derive the more general rotation matrix about axis $z$, then you have to allow for any rotation $\theta$.

Consider rotating some vector like your vector $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ or slightly more generally $\begin{bmatrix} r \\ 0 \\ 0 \end{bmatrix}$ around the $z$ axis by an arbitrary angle $\theta$.
enter image description here (https://en.wikipedia.org/wiki/Rotation_matrix#/media/File:Counterclockwise_rotation.png)

The position of the head of the new vector can be calculated using trigonometry. It's really the same mathematics as polar coordinates where here you're rotating a radius of $r$.

So the new $x$ and $y$ coordinates of the vector head position can be found with $$ x = r \cos(\theta) $$ $$ y = r \sin(\theta) $$ Writing this with matrices gives $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos(\theta) & 0 \\ 0 & \sin(\theta) \end{bmatrix} \begin{bmatrix} r \\ 0 \end{bmatrix} $$ This gives a slightly more general formula but requires the vector start parallel to the $x$ axis. We can generalize further by considering rotating any vector $\begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix}$ in 3D about the $z$ axis. Using the same trig we can deduce that the new $x$ position of the vector head is $$ x = v_x \cos(\theta) - v_y\sin(\theta) $$ the new $y$ position is $$ y = v_x \sin(\theta) + v_y\cos(\theta) $$ and since we're rotating around $z$ the $v_z$ value does not change $$ z = v_z $$ which writing with matrices gives you the rotation matrix you'll recognize $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix} $$

$\endgroup$
  • $\begingroup$ Could you explain what you said for a 3D vector "x=vxcos(θ)−vysin(θ)" and "y=vxsin(θ)+vycos(θ)" how did you got to this equation, comparing it to the method I used to design the matrix I don't quite understand what you did here $\endgroup$ – Jorayen Jul 30 '16 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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