The $x$ and $y$ components of vector $\vec{P}$ have magnitudes $1$ and $3$,and the $x$ and $y$ components of the resultant of $\vec{P}$ and $\vec{Q}$ have magnitudes $5$ and $6$ respectively. What is the magnitude of $\vec{Q}$?

The vector sum of two forces is perpendicular to their vector difference. In that case,the forces

Two forces of magnitude $3\;N$ and $4\;N$ respectively are acting on a body. Calculate the resultant force if the angle between them is $0^{\circ}$. (in $;N$)

If $\sqrt{A^2+B^2}$ represents the magnitude of the resultant of two vectors $(\vec{A}+\vec{B})$ and $(\vec{A}-\vec{B})$,then the angle between the two vectors $\vec{A}$ and $\vec{B}$ is

Let the two forces have equal magnitude $A$. If the magnitude of the resultant is $\frac{2A}{3}$,then the angle between those two forces is:

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?