$\bar{a}$ 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 $x =$

If $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + \hat{j} - \hat{k}$,and $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{p} = 2\vec{a} + \vec{b}$ and $\vec{q} = \vec{a} + 2\vec{b}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $O$ is the origin and $C$ is the midpoint of $A(2, -1)$ and $B(-4, 3)$,then the value of $\overrightarrow{OC}$ is:

Let $a = \hat{i} + x \hat{j} + \hat{k}$,$b = \hat{i} + \hat{j} + \hat{k}$ and $|a + b| = |a| + |b|$,then

Let $PQRS$ be a quadrilateral. If $M$ and $N$ are midpoints of the sides $PQ$ and $RS$ respectively,then $\overrightarrow{PS} + \overrightarrow{QR} =$

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 $2x-z=$