$ABC$ is an isosceles triangle right-angled at $A$. Forces of magnitude $2\sqrt{2}$,$5$,and $6$ act along $\overrightarrow{BC}$,$\overrightarrow{CA}$,and $\overrightarrow{AB}$ respectively. The magnitude of their resultant force is

If $\vec{a}$ and $\vec{b}$ are the position vectors $(P.V.)$ of two points $A$ and $B$ respectively,and $C$ divides the line segment $AB$ in the ratio $2 : 1$,then the position vector of $C$ is:

If the position vector of a point $A$ is $a + 2b$ and a point $P$ divides $AB$ in the ratio $2:3$,where the position vector of $P$ is $a$,then the position vector of $B$ is:

If $\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$,and $\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ are three vectors such that $\vec{a}=x \vec{b}+y \vec{c}$,then $x+y=$

If the position vectors of the points $A, B, C, D$ are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 5\hat{j}$,$\vec{c} = 3\hat{i} + 2\hat{j} - 3\hat{k}$,and $\vec{d} = \hat{i} - 6\hat{j} - \hat{k}$,then the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is:

If $\vec{r}_1 = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{r}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$,find the unit vector parallel to the resultant vector of $\vec{r}_1$ and $\vec{r}_2$.