The speed of a particle changes from $\sqrt{5} \ m/s$ to $2\sqrt{5} \ m/s$ in a time $t$. If the magnitude of change in its velocity is $5 \ m/s$,the angle between the initial and final velocities of the particle is (in $^{\circ}$)

What should be the angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ so that the magnitude of the resultant vector is minimum (in $^{\circ}$)?

Given $A = 2 \hat{i} + \hat{j}$,$B = 3 \hat{j} - \hat{k}$,and $C = 6 \hat{i} - 2 \hat{k}$. The value of $A - 2B + 3C$ is:

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

The resultant $\vec{R}$ of $\vec{P}$ and $\vec{Q}$ is perpendicular to $\vec{P}$. Also,$|\vec{P}| = |\vec{R}|$. Find the angle between $\vec{P}$ and $\vec{Q}$.

The sum of two forces $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$ such that $|\overrightarrow{R}| = |\overrightarrow{P}|$. The angle $\alpha$ (in degrees) that the resultant of $2\overrightarrow{P}$ and $\overrightarrow{Q}$ will make with $\overrightarrow{Q}$ is