Let $ABCDEF$ be a regular hexagon with the vertices $A, B, C, D, E, F$ in counterclockwise order. Then the vector $\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF}$ is equal to

$A$ vector of magnitude $14$ lies in the $xy-$ plane and makes an angle of $60^\circ$ with the $x-$ axis. The components of the vector in the direction of the $x-$ axis and $y-$ axis are:

If $r \cdot i = r \cdot j = r \cdot k$ and $|r| = 3$, then $r = $

The position vector of the point of intersection of the medians (centroid) of a triangle,whose vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$ is

The magnitude of the sum of vectors $(1, -\sqrt{2})$ and $(2, \sqrt{2})$ is $\ldots$

Let $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. Let $D$ divide $BC$ in the ratio $3:1$ internally and $E$ divide $AD$ in the ratio $4:1$ internally. Let $BE$ meet $AC$ in $F$. If $E$ divides $BF$ in the ratio $3:2$ internally, then the position vector of $F$ is