Two vectors having equal magnitudes of $x$ units acting at an angle of $45^\circ$ have a resultant of $\sqrt{2 + \sqrt{2}}$ units. The value of $x$ is

Given $|\vec{P}| > |\vec{Q}|$. What is the angle between their maximum resultant vector and minimum resultant vector (in $^{\circ}$)?

Given that $A = B = C$. If $\vec A + \vec B = \vec C$,then the angle between $\vec A$ and $\vec C$ is $\theta_1$. If $\vec A + \vec B + \vec C = 0$,then the angle between $\vec A$ and $\vec C$ is $\theta_2$. What is the relation between $\theta_1$ and $\theta_2$?

The magnitude of the $X$ and $Y$ components of $\vec{A}$ are $7$ and $6$ respectively. Also,the magnitude of the $X$ and $Y$ components of $\vec{A} + \vec{B}$ are $11$ and $9$ respectively. What is the magnitude of $\vec{B}$?

$A$ particle undergoes three successive displacements given by $s_1 = \sqrt{2} \ m$ north-east,$s_2 = 2 \ m$ due south,and $s_3 = 4 \ m$ at $30^{\circ}$ north of west. The magnitude of the net displacement is:

The unit vector parallel to the resultant of vectors $\vec{A} = 4\hat{i} - 3\hat{j}$ and $\vec{B} = 8\hat{i} + 8\hat{j}$ is: