If $ABCDEF$ is a regular hexagon,then $\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = .....$

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\bar{a}|=\sqrt{31}, 4|\bar{b}|=|\bar{c}|=2$ and $2(\bar{a} \times \bar{b})=3(\bar{c} \times \bar{a})$ and if the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2\pi}{3}$,then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{d}=\vec{a} \times \vec{b}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|$,$|\vec{c}-2\vec{a}|^2=8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$,then $|10-3\vec{b} \cdot \vec{c}|+|\vec{d} \times \vec{c}|^2$ is equal to . . . . . .

$A, B, P, Q, R$ are five points in a plane. If the forces acting at point $A$ are $\overline{AP}, \overline{AQ}, \overline{AR}$ and the forces acting at point $B$ are $\overline{PB}, \overline{QB}, \overline{RB}$,find the resultant of all these forces.

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.

The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is