If $M$ and $N$ are the mid-points of the sides $BC$ and $CD$ respectively of a parallelogram $ABCD$, then $AM + AN$ equals

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 $\overline{i}-2 \overline{j}+\overline{k}, \overline{i}+\overline{j}-2 \overline{k}, 2 \overline{i}-\overline{j}-\overline{k}$ and $\overline{i}+\overline{j}+\overline{k}$ be the position vectors of four points $A, B, C$ and $D$ respectively. If a point $P$ divides $AB$ in the ratio $2:1$ internally and a point $Q$ divides $CD$ in the ratio $1:2$ externally,then the ratio in which the point with position vector $5 \overline{i}-6 \overline{j}-5 \overline{k}$ divides $PQ$ is

Consider two points $P$ and $Q$ with position vectors $\overrightarrow{OP} = 3\vec{a} - 2\vec{b}$ and $\overrightarrow{OQ} = \vec{a} + \vec{b}$. Find the position vector of a point $R$ which divides the line segment joining $P$ and $Q$ in the ratio $2:1$ internally.

The value of $x$ if $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is:

If $|\overline{u}|=2$ and $\overline{u}$ makes angles of $60^{\circ}$ and $120^{\circ}$ with axes $OX$ and $OY$ respectively,then $\overline{u}=$