Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = a, \overrightarrow{OB} = b, \overrightarrow{OC} = 2a + 3b$ and $\overrightarrow{OD} = a - 2b$. If $|a| = 3|b|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is

If $|a| = 3$,$|b| = 4$ and the angle between $a$ and $b$ is $120^\circ$,then $|4a + 3b| = $

If in a $\Delta ABC$,$O$ and $O^{\prime}$ are the incentre and orthocentre respectively,then $\vec{O^{\prime}A} + \vec{O^{\prime}B} + \vec{O^{\prime}C}$ is equal to

If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b}$,then $\theta = $