Fill in the blanks:
$(1)$ If $|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}$,then the angle between $\vec{A}$ and $\vec{B}$ is ............ .
$(2)$ The angle between the angular momentum and linear momentum of a particle in rotational motion is ............ .
$(3)$ $A$ force $F\hat{k}$ acts on a particle having position vector $(2\hat{i} + \hat{j})$. The torque acting on the particle is ............ .

(N/A) $(1)$ Given $|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}$.
$AB \sin \theta = AB \cos \theta$.
$\tan \theta = 1$,so $\theta = 45^{\circ}$.
$(2)$ By definition,angular momentum $\vec{L} = \vec{r} \times \vec{p}$. Since $\vec{L}$ is perpendicular to both $\vec{r}$ and $\vec{p}$,the angle between angular momentum $\vec{L}$ and linear momentum $\vec{p}$ is $90^{\circ}$.
$(3)$ Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Given $\vec{r} = (2\hat{i} + \hat{j})$ and $\vec{F} = F\hat{k}$.
$\vec{\tau} = (2\hat{i} + \hat{j}) \times F\hat{k} = 2F(\hat{i} \times \hat{k}) + F(\hat{j} \times \hat{k})$.
Using cross products $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$,we get $\vec{\tau} = -2F\hat{j} + F\hat{i} = F(\hat{i} - 2\hat{j})$.