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

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be non-coplanar vectors. If $P, Q, R$ and $S$ are four points with position vectors $-\bar{a}+4\bar{b}-3\bar{c}$,$3\bar{a}+2\bar{b}-5\bar{c}$,$-3\bar{a}+8\bar{b}-5\bar{c}$ and $-3\bar{a}+2\bar{b}+\bar{c}$ respectively,then the ordered pair $(x, y)$ of real numbers such that $\overline{PQ} = x \cdot \overline{PR} + y \cdot \overline{PS}$ is

The projections of a vector on the three coordinate axes are $6, -3, 2$ respectively. The direction cosines of the vector are:

Which of the following is not a property of vectors?

In $\triangle ABC$,if $D$ and $E$ are the mid-points of the sides $BC$ and $CA$ respectively,then $2(\vec{AD}+\vec{EB})=$

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear. If the vector $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$,then $\vec{a} + 2\vec{b} + 6\vec{c} = \dots$