If the collinear points $A, B$ and $C$ have position vectors respectively $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$, then $x+y=$

$A$ unit vector $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis. If $\vec{a} + \hat{i} + \hat{j}$ is a unit vector,then $\vec{a}$ is equal to

$ABC$ 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

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 $M_1, M_2, M_3$ and $M_4$ are respectively the magnitudes of the vectors $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -3\hat{i} - 4\hat{j} - 4\hat{k}$,$\vec{a}_3 = -\hat{i} + \hat{j} - \hat{k}$,and $\vec{a}_4 = -\hat{i} + 3\hat{j} + \hat{k}$,then the correct order of $M_1, M_2, M_3$ and $M_4$ is:

$A$ unit vector in $XY$-plane making an angle $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle $60^{\circ}$ with $3\hat{i}-4\hat{j}$ is