3D rotation of coordinates axis

Question

I am struggling to understand the 3D rotation.

The initial object position is $A(x_1, y_1, z_1).$ We need rotate it into $A'(x_2,y_2,z_2)$ about $z$-axis by angle $\angle A'PA=\theta.$ And we take $P$ be the any point in $z$-axis. It makes $\angle APS=\phi.$

Now we find $cos \phi= \frac{PS}{OP}=\frac{x_1}{r}\Rightarrow x_1=rcos \phi.$ But we know $OQ =x_1$ but here slides take $PS =x_1$ and they forcefully match the answer which is wrong.

My question is slides has wrong or my concepts is wrong? If both are wrong so how we find 3D rotation about any coordinate axis?

Answer 1

Suppose $P = (x, y, z) $ a point in $3D$.

We want to rotate $P$ about the $z$ axis.

Note that this rotation does not change the $z$ coordinate of $P$, and that the image of the $x$ and $y$ coordinates follow that of the usual $2D$ rotation.

Thus if $P' = (x', y', z')$ is the image of $P$ under a rotation about the $z$ axis by an angle $\theta$ then

$z' = z$

$x' = x \cos \theta - y \sin \theta $

$y' = x \sin \theta + y \cos \theta $

In matrix form

$\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} x \\ y \\ z \end{bmatrix} $

Answer 2

Following the notation in the question, we have the original point as $A = (x_1, y_1, z_1)$ and its image is $A' = (x_2, y_2, z_2) $

Define $r = \sqrt{x_1^2 + y_1^2} $

Then there is a unique angle $\phi \in [0, 2\pi) $

such that

$x_1 = r \cos \phi$

$y_1 = r \sin \phi$

$\phi$ is the angle that the vector extending from $(0, 0, z_1) $ to $(x_1, y_1, z_1)$ makes with the direction of the positive $x$ axis.

If we now rotate $A$ about the $z$ axis by an angle $\theta$ then we are effectively keeping the $z$ coordinate of $A$ constant, and change the angle $\phi$ to the angle $\phi + \theta$

Hence, the new coordinates of $A'(x_2, y_2, z_2)$ are:

$z_2 = z_1 $

$x_2 = r \cos(\phi + \theta)$

$y_2 = r \sin( \phi + \theta)$

Now we note that:

$\cos(\phi + \theta) = \cos \phi \cos \theta - \sin \phi \sin \theta $

$\sin (\phi + \theta) =\sin \phi \cos \theta + \cos \phi \sin \theta $

therefore,

$x_2 = (r \cos \phi) \cos \theta - (r \sin \phi) \sin \theta = x_1 \cos \theta - y_1 \sin \theta $

and

$y_2 = (r \cos \phi) \sin \theta + (r \sin \phi) \cos \theta = x_1 \sin \theta + y_1 \cos \theta$

Answer 3

This is a short note on $3D$ rotation that I wrote more than $6$ years ago.

Answer 4

The following is a note on $3D$ rotation that I wrote a while ago. Click on the image and ZOOM +.