If a charged particle travels without acceleration in a region...

The Lorentz force on a charged particle is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.

For the particle to move without acceleration, the net force must be zero:

$q(\vec{E} + \vec{v} \times \vec{B}) = 0$

$\vec{E} = -(\vec{v} \times \vec{B}) = \vec{B} \times \vec{v}$

From the properties of the cross product, the vector $\vec{E}$ must be perpendicular to both $\vec{B}$ and $\vec{v}$.

Therefore, $\vec{E} \perp \vec{B}$ and $\vec{v} \perp \vec{E}$.

The magnitude of the electric field is $E = vB \sin \theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Thus, $E$ is not necessarily equal to $vB$, and $\vec{v}$ is not necessarily perpendicular to $\vec{B}$.