The position vectors of $A$ and $B$ are $(\hat{i}+\hat{j}+\hat{k})$ and $(\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k})$. If $B$ divides the line segment $AC$ in the ratio $2:1$,then the position vector of $C$ is

Write all the unit vectors in the $XY$-plane.

If the vectors $\vec{AB} = -3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$,then the length of the median through $A$ is

The direction cosines of the resultant of the vectors $(i + j + k)$,$(-i + j + k)$,$(i - j + k)$ and $(i + j - k)$ are

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

If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$