$a$ and $b$ are two non-collinear vectors,then $xa + yb$ (where $x$ and $y$ are scalars) represents a vector which is

The position vector of a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})= k \overrightarrow{FE}$,then $k$ is equal to

The position vectors of four points $P, Q, R, S$ are $2a + 4c$,$5a + 3\sqrt{3}b + 4c$,$-2\sqrt{3}b + c$,and $2a + c$ respectively. Then:

In a quadrilateral $PQRS$,$A$ divides $SR$ in the ratio $1:3$ and $B$ is the mid-point of $PR$. If $3SR - QR - 3PS - PQ = kAB$,then $k=$

$\bar{a}$ and $\bar{b}$ are non-collinear vectors. If $\bar{p} = (2x + 1)\bar{a} - \bar{b}$ and $\bar{q} = (x - 2)\bar{a} + \bar{b}$ are collinear vectors,then $x =$