Magnetic force on a moving charge

$\vec{F} = q(\vec{v}\times\vec{B})$ with $q = -2\times10^{-6}$ C, $\vec{v} = (2\hat{i} + 3\hat{j})\times10^6$ m/s, $\vec{B} = 2\hat{j}$ T
$\vec{v}\times\vec{B} = (2\hat{i} + 3\hat{j})\times10^6\times2\hat{j} = 4\times10^6(\hat{i}\times\hat{j}) + 6\times10^6(\hat{j}\times\hat{j})$
$\hat{i}\times\hat{j} = \hat{k}$ and $\hat{j}\times\hat{j} = 0$, so $\vec{v}\times\vec{B} = 4\times10^6\hat{k}$
$\vec{F} = -2\times10^{-6}\times4\times10^6\hat{k} = -8\hat{k}$ N
The force is 8 N in the –z direction.
Note: the first two options are printed identically in the source ('8N in z – direction'); the minus sign of one has been lost. They are taken here as +z and –z, with –z as the keyed answer.

Magnetic force on a moving charge: $\vec{F} = q(\vec{V} \times \vec{B})$
Its magnitude is $F = qVB\sin\theta$, where $\theta$ is the angle between $\vec{V}$ and $\vec{B}$.
When $\theta = 0^\circ$ or $\theta = 180^\circ$, $\sin\theta = 0$ and therefore $F = 0$.
For every other value of $\theta$, $\sin\theta \neq 0$ and the force is non-zero.
The force is maximum at $\theta = 90^\circ$, but it is not necessary that the angle be $90^\circ$ for the force to be non-zero.
So a non-zero force implies that the angle between $\vec{V}$ and $\vec{B}$ can have any value other than zero and $180^\circ$.