The points with position vectors $\bar{a}+\bar{b}$,$\bar{a}-\bar{b}$,and $\bar{a}+k\bar{b}$ are collinear:

$R$ divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively in the ratio $2: 1$ externally. $S$ divides $PQ$ internally in the ratio $2: 1$. Then,the position vector of the midpoint of the line joining $R$ and $S$ is

The vectors $a$ and $b$ are non-collinear. The value of $x$ for which the vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,is

The angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of the $X$-axis is:

Five points given by $A, B, C, D, E$ are in a plane. Three forces $\overrightarrow{AC}, \overrightarrow{AD},$ and $\overrightarrow{AE}$ act at $A$,and three forces $\overrightarrow{CB}, \overrightarrow{DB},$ and $\overrightarrow{EB}$ act at $B$. Then their resultant is:

Find the position vector of the point which divides the join of the points with position vectors $2\vec{a} - 3\vec{b}$ and $3\vec{a} - 2\vec{b}$ externally in the ratio $2:3$.