The resultant of two vectors having magnitudes $2$ and $3$ is $1$. What is their cross product?

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

Consider a vector $\overrightarrow{F} = 4\hat{i} - 3\hat{j}$. Another vector that is perpendicular to $\overrightarrow{F}$ is

Find the component of vector $\vec{P} = 2 \hat{i} + 3 \hat{j}$ along the direction of vector $\vec{Q} = \hat{i} + \hat{j}$.

$\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors given by $\overrightarrow{A} = 2\widehat{i} + 3\widehat{j}$ and $\overrightarrow{B} = \widehat{i} + \widehat{j}$. The magnitude of the component (projection) of $\overrightarrow{A}$ on $\overrightarrow{B}$ is

The angle between the vectors $(\hat{i} + \hat{j})$ and $(\hat{i} - \hat{k})$ is ........ $^\circ$.