If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$

$A$ point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors) divides the line segment joining $A$ and $B$ in the ratio $2:1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$,then find the position vector of $B$.

$A$ unit vector in $XY$-plane making an angle $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle $60^{\circ}$ with $3\hat{i}-4\hat{j}$ is

Let $A, B, C, D$ be the points in the plane with position vectors $\vec{a} = -2\hat{i} - \hat{j}$, $\vec{b} = 4\hat{i}$, $\vec{c} = 3\hat{i} + 3\hat{j}$, and $\vec{d} = -3\hat{i} + 2\hat{j}$ respectively. Then $ABCD$ is:

If the position vectors of two points $A$ and $B$ are $\vec{a} - 3\vec{b}$ and $6\vec{b} - 2\vec{a}$ respectively,then the position vector of the point dividing $AB$ in the ratio $1 : 2$ is:

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$