The resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If $\vec{Q}$ is doubled,the new resultant vector is perpendicular to $\vec{P}$. What is the magnitude of $\vec{R}$?

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

The resultant of two vectors at an angle $150^{\circ}$ is $10$ units and is perpendicular to one vector. The magnitude of the smaller vector is ....... units

Four forces act on a stationary particle at the origin of a coordinate system: $\overrightarrow{F_1} = 3\hat{i} - \hat{j} + 9\hat{k}$,$\overrightarrow{F_2} = 2\hat{i} - 2\hat{j} + 16\hat{k}$,$\overrightarrow{F_3} = 9\hat{i} + \hat{j} + 18\hat{k}$,and $\overrightarrow{F_4} = \hat{i} + 2\hat{j} - 18\hat{k}$. In which plane will the particle move under the influence of these forces?

The magnitudes of vectors $\overrightarrow{A}$,$\overrightarrow{B}$,and $\overrightarrow{C}$ are $12$,$5$,and $13$ units respectively,and $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$. Then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is:

Let $\vec{P} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{Q} = -(\hat{i} + \hat{j} + \hat{k})$. The angle between $(\vec{P} - \vec{Q})$ and $\vec{P}$ is (in $^\circ$)