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 $^{\circ}$)

$A$ vector $P$ has $X$ and $Y$ components of magnitude $2$ units and $4$ units respectively. $A$ vector $Q$ along the negative $X$-axis has a magnitude of $6$ units. The vector $(Q - P)$ will be

If the angle between two vectors $A$ and $B$ is $120^{\circ}$,then their resultant $C$ will be:

If the resultant of two forces has a magnitude smaller than the magnitude of the larger force,the two forces must be

Two forces are such that the sum of their magnitudes is $18 \; N$ and their resultant is $12 \; N$,which is perpendicular to the smaller force. Then the magnitudes of the forces are:

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?