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 $\triangle ABC$. The length of the median through $A$ is:

If $\bar{a} = 2\bar{i} - 3\bar{j} + 5\bar{k}$ and $\bar{b} = -\bar{i} + 3\bar{j} + 3\bar{k}$ are two vectors,then the vector of magnitude $28$ units in the direction of the vector $\bar{a} - \bar{b}$ is

$\bar{a}$ 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 $x =$

$A$ vector $\vec{r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{r}$ is $2 \sqrt{3}$ units,find $\vec{r}$.

The magnitude of the sum of vectors $(1, -\sqrt{2})$ and $(2, \sqrt{2})$ is $\ldots$

$A$ point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors) divides the line segment joining $A$ and $B$ in the ratio $2:1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$,then find the position vector of $B$.