Write down a unit vector in the $XY$-plane,making an angle of $30^{\circ}$ with the positive direction of the $x$-axis.

If $\overline{a}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k}$,$\overline{b}=4 \hat{\imath}+5 \hat{\jmath}+3 \hat{k}$ and $\overline{c}=6 \hat{\imath}+\hat{\jmath}+5 \hat{k}$ are the position vectors of the vertices of a triangle $ABC$ respectively,then the position vector of the intersection of the medians (centroid) of the triangle $ABC$ is:

Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position vectors $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$:

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is a point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is a point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to:

$ABCD$ is a parallelogram and $P$ is the mid-point of the side $AD$. The line $BP$ meets the diagonal $AC$ in $Q$. Then, the ratio of $AQ:QC$ is equal to

The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $