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

If $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$,then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

Three vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are such that $\vec{A} \cdot \vec{B} = 0$ and $\vec{A} \cdot \vec{C} = 0$. Then $\vec{A}$ is parallel to:

If $\vec{a}$ and $\vec{b}$ make an angle $\cos^{-1}\left(\frac{5}{9}\right)$ with each other,then $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$ for $|\vec{a}|=n|\vec{b}|$. The integer value of $n$ 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 $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

The dot product of unit vectors $\hat{n}_1$ and $\hat{n}_2$ that are parallel to $5 \hat{i}+12 \hat{j}$ and $3 \hat{i}+4 \hat{j}$ respectively is