Solution

Rigid body with constant angular velocity 

$\omega=3\ \frac{\text{rad}}{\text{s}}$ rotate around fixed axis. That axis passes through point O as shown on picture below. Vector $\bold{r}$ connects point O with point P. Our particle of interest of this body is positioned in point P on edge of rigid body. Distance from axis to particle in point P is $d$. Angle between axis of rotation and vector $\bold{r}$ is $\theta$.

Because this rigid body is rotating, we know that particle of our interest in point P has circumferential velocity $\bold{u}$, which is important for us to find in this task. This velocity is easy to find if we have cross product of vector of angular velocity $\bm{\omega}$ and vector of distance $\bold{r}$ between points O and P:

u=ω×r

Also, it is correct to say that scalar or magnitude of circumferential velocity is:

∣u∣=d∣ω∣=∣r∣sin(θ)∣ω∣=∣ω×r

We know from text of this task that axis of rotation is parallel to vector $\bold{A}=\bold{a_x}-2\bold{a_y}+2\bold{a_z}$ and it is passing through point $\text{O}=(2,-3,1)$. This point must be point O because axis of rotation passes through point O as shown on picture. Now we can find vector of angular velocity because we have magnitude given from task $\omega=3\ \frac{\text{rad}}{\text{s}}$ and direction that must be parallel to vector $\bold{A}$.

Firstly, we need to transform vector $\bold{A}$ to unit vector that has magnitude of $1$. To do that we need to find magnitude of vector $\bold{A}$ and than divide this vector $\bold{A}$ with it:

$$\begin{aligned} |\bold{A}|&=\sqrt{1^2+(-2)^2+2^2}\\ &=\sqrt{1+4+4}\\ &=\sqrt{9}=3 \end{aligned}$$

Now we can divide vector $\bold{A}$ with its magnitude $|\bold{A}|=3$ to get unit vector $\bold{a}$which only contains information about direction:

Now we subtract two vectors $\bold{O}$, $\bold{P}$ to get vector of distance $\bold{r}$:

$$\begin{aligned} \bold{r}&=\bold{P}-\bold{O}\\ &=\bold{a_x} + 3\bold{a_y} + 4\bold{a_z}-(2\bold{a_x} -3\bold{a_y} + \bold{a_z})\\ &=-\bold{a_x} +6\bold{a_y} + 3\bold{a_z} \end{aligned}$$

Subtraction of two vectors must be $\bold{P}-\bold{O}$ and not $\bold{O}-\bold{P}$ because vector $\bold{r}$must be pointing from point O to point P.

Finally, we can calculate circumferential velocity $\bold{u}$ at point P where particle is located:

$$\begin{aligned} \bold{u}&=\bm{\omega}\times\bold{r}\\ &=(\bold{a_x}-2\bold{a_y}+2\bold{a_z})\times(-\bold{a_x} +6\bold{a_y} + 3\bold{a_z}) \end{aligned}$$

Cross product can be simply calculated:

$$\begin{aligned} \bm{A}\times\bold{B}&=(A_x\bold{a_x}+A_y\bold{a_y}+A_z\bold{a_z})\times(B_x\bold{a_x} +B_y\bold{a_y} + B_z\bold{a_z})\\ &= \begin{vmatrix} \bold{a_{x}}&\bold{a_{y}}&\bold{a_{z}} \\ \\ A_x&A_y&A_z \\ \\ B_x&B_y&B_z \end{vmatrix}\\ &=(A_yB_z-A_zB_y)\bold{a_{x}}-(A_xB_z-A_zB_x)\bold{a_{y}}+(A_xB_y-A_yB_x)\bold{a_{z}} \end{aligned}$$

We will insert cross product back to expression for circumferential velocity $\bold{u}$:

$$\begin{aligned} \boxed{\bold{u}=\bm{\omega}\times\bold{r}= -18\bold{a_{x}}-5\bold{a_{y}}+4\bold{a_{z}}} \end{aligned}$$

Also, magnitude of circumferential velocity $\bold{u}$ is:

$$\begin{aligned} |\bold{u}&|=\sqrt{(-18)^2+(-5)^2+4^2}\\ &=\sqrt{324+25+14}=\sqrt{365} \end{aligned}$$

Answer

u=−18ax​−5ay​+4az​​