Find the unit vector perpendicular to the two vectors $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.

$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})$ is equal to :

If $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$,then what is the value of $|\vec{A} + \vec{B}|$?

If $|\vec{A}| = 2$ and $|\vec{B}| = 4$,then match the relation in Column-$I$ with the angle $\theta$ between $\vec{A}$ and $\vec{B}$ in Column-$II$.

Column-$I$	Column-$II$
$(a) |\vec{A} \times \vec{B}| = 0$	$(i) \theta = 30^{\circ}$
$(b) |\vec{A} \times \vec{B}| = 8$	$(ii) \theta = 45^{\circ}$
$(c) |\vec{A} \times \vec{B}| = 4$	$(iii) \theta = 90^{\circ}$
$(d) |\vec{A} \times \vec{B}| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$

If $|\vec A| = 2$ and $|\vec B| = 4$,then match the relation in Column $-I$ with the angle $\theta$ between $\vec A$ and $\vec B$ in Column $-II$.

Column $-I$	Column $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

If $\overrightarrow{ P } \times \overrightarrow{ Q } = \overrightarrow{ Q } \times \overrightarrow{ P }$,the angle between $\overrightarrow{ P }$ and $\overrightarrow{ Q }$ is $\theta$ $(0^{\circ} < \theta < 360^{\circ})$. The value of $\theta$ will be ........ (in $^{\circ}$)