The resultant of two forces acting at an angle of $120^{\circ}$ is $10 \,kg$-wt and is perpendicular to one of the forces. That force is

If the magnitude of the sum of two vectors is equal to the magnitude of the difference of two vectors,then the angle between them is: (in $^{\circ}$)

If two forces of equal magnitudes act simultaneously on a body in the east and the north directions,then:

If vectors $\overrightarrow P, \overrightarrow Q$ and $\overrightarrow R$ have magnitudes $5, 12$ and $13$ units respectively and $\overrightarrow P + \overrightarrow Q = \overrightarrow R$,then the angle between $\overrightarrow Q$ and $\overrightarrow R$ is

Two vectors of magnitude $3$ and $4$ have a resultant which makes angles $\alpha$ and $\beta$ respectively with them (given $\alpha + \beta \neq 90^o$).

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?