$ABCDEF$ is a regular hexagon. Find the sum of the vectors $\vec{BE} + \vec{BC} + \vec{EF} + \vec{BA} + \vec{CF} + \vec{AF}$.

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and the points represented by position vectors $\bar{a}-2 \bar{b}+3 \bar{c}$, $-4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $x \bar{a}-9 \bar{b}+z \bar{c}$ are collinear, then $2x-z=$

If $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$,then $\vec{CE}$ equals to

Let $\overline{OA}=\overline{a}, \overline{OB}=\overline{b}$. If the vector along the angle bisector of $\angle AOB$ is given by $x \frac{\overline{a}}{|\overline{a}|}+y \frac{\overline{b}}{|\overline{b}|}$,then which of the following is true?

Find $|\vec{a}-\vec{b}|$,if two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2, |\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$.

If $2 \hat{i}+4 \hat{j}-5 \hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{j}+2 \hat{k}$ are the position vectors of the vertices $A$, $B$, and $C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is