If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

If the vectors $\bar{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\bar{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\bar{c}=p \hat{i}+\hat{j}+q \hat{k}$ are mutually orthogonal,then $(p, q)$ is equal to

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

If $a \neq 0, b \neq 0$ and $|a + b| = |a - b|$,then the vectors $a$ and $b$ are . . . .

If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$, then the angle between $a$ and $b$ is

If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is: