$\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j}) = $

Show that the area of the triangle formed by the vectors $\vec{a}$ and $\vec{b}$ is one-half of the magnitude of $\vec{a} \times \vec{b}$.

If $\hat{A}$ is a unit vector in a given direction,then the value of $\hat{A} \cdot \frac{d\hat{A}}{dt}$ is

The dot product of two mutually perpendicular vectors is:

The diagonals of a parallelogram are represented by vectors $\vec{A} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} - \hat{k}$. What is the area of the parallelogram (in $\sqrt{3}$)?

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