The two vectors have magnitudes $3$ and $5$. If the angle between them is $60^o$,then the dot product of the two vectors will be:

What is the product of two vectors if they are parallel or antiparallel?

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

Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.

If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors having unit magnitude. If $\vec{A} + \vec{B} + \vec{C} = \vec{0}$,then $\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}$ will be:

Define the vector product of two vectors.