Find the angle between the vectors $A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$. (in $^{\circ}$)

The value of $(\overrightarrow A + \overrightarrow B ) \times (\overrightarrow A - \overrightarrow B )$ is

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

If a vector $A=2i+2j+3k$ and $B=3i+6j+nk$ are perpendicular to each other,then the value of $n$ is

If the component of the vector $\vec{A}$ along the vector $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$, then the ratio of magnitudes of vectors $\vec{A}$ and $\vec{B}$ 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-$II$
$(a) |\vec{A} \times \vec{B}| = 0$	$(i) \theta = 30^{\circ}$
$(b) |\vec{A} \times \vec{B}| = 8$	$(ii) \theta = 45^{\circ}$
$(c) |\vec{A} \times \vec{B}| = 4$	$(iii) \theta = 90^{\circ}$
$(d) |\vec{A} \times \vec{B}| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$