The area of the parallelogram having diagonals $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$ is:

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

Find the scalar and vector products of two vectors $\vec{a} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k})$ and $\vec{b} = (-2 \hat{i} + \hat{j} - 3 \hat{k})$.

The diagonals of a parallelogram are $2\,\hat{i}$ and $2\,\hat{j}$. What is the area of the parallelogram in square units?

If $\vec A$ and $\vec B$ are perpendicular vectors,where $\vec A = 5\hat i + 7\hat j - 3\hat k$ and $\vec B = 2\hat i + 2\hat j - a\hat k$,then the value of $a$ is:

If $|\overrightarrow A \times \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$,then the angle between $\overrightarrow A$ and $\overrightarrow B$ will be ........ $^o$.