The sum of two forces is $18 \ N$ and the resultant,whose direction is at right angles to the smaller force,is $12 \ N$. The magnitudes of the two forces are:

If $|\bar{a}|=3, |\bar{b}|=4, |\bar{a}-\bar{b}|=5$,then $|\bar{a}+\bar{b}|=$

Suppose that $\bar{p}, \bar{q}$ and $\bar{r}$ are three non-coplanar vectors in $\mathbb{R}^3$. Let the components of a vector $\bar{s}$ along $\bar{p}, \bar{q}$ and $\bar{r}$ be $4, 3$ and $5$ respectively. If the components of this vector $\bar{s}$ along $(-\bar{p}+\bar{q}+\bar{r}), (\bar{p}-\bar{q}+\bar{r})$ and $(-\bar{p}-\bar{q}+\bar{r})$ are $x, y$ and $z$ respectively,then the value of $2x+y+z$ is

The position vector of point $C$ relative to $B$ is $(\hat{i} + \hat{j})$ and the position vector of $B$ relative to $A$ is $(\hat{i} - \hat{j})$. The position vector of $C$ relative to $A$ is:

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$,$-2\bar{i}+\bar{j}+3\bar{k}$,and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron,then $GD=$

$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to: