Question: While studying Rotation of Rigid Bodies, I came across the following situation:

> 
> ![1](https://i.stack.imgur.com/8tjtD.png)
> 
> 
> Consider a rigid body in pure rotational motion about a fixed axis (for example the z-axis). For any particle in the object, its linear velocity is given by $$ v=r \omega=R \sin \theta \omega $$ where $\mathbf{R}$ is the position vector of the particle from the origin (see Fig. 7.9) and $\theta$ is the angle between the position vector and the $z$-axis. As shown in Fig. 7.9, the direction of $y$ is perpendicular to the plane formed by $\boldsymbol{\omega}$ and $\mathbf{R}$ where it can be verified using the right-hand rule.
> 
> 
> 

**Problem:**

Why is the direction of $y$ perpendicular to the plane formed by $\boldsymbol{\omega}$ and $\mathbf{R}$? It is stated that this relation can be obtained by using the right hand rule. However, I am having trouble using the right hand rule in this scenario. Furthermore, I just cannot properly visualize the plane formed by $\boldsymbol{\omega}$ and $\mathbf{R}$. I would appreciate if someone could further elaborate an explanation regarding the above statements.

Answer: 
![1](https://i.stack.imgur.com/eUS1L.png)

Look at this figure. The plane formed by $\boldsymbol{\omega}$ and $\mathbf{R}$ is perpendicular to y’ axis, thus $\vec e\_{y’}\cdot \vec e\_R=0$.
The vector representing (ω) is defined as being along the axis of rotation (z). The angle (θ) is measured in the plane defined by (ω) and (R). Looking at the sketch, I would say that the (x) axis is perpendicular to that plane. In any case, such a statement requires that both the (x) and (y) axes are rotating with the rigid body.