$A$ force ${F_1} = 1\,N$ acts along the direction of $x = 0$,and a force ${F_2} = 2\,N$ acts along the direction of $y = 0$. Find the resultant force.

$A$ particle moves $21 \ m$ along the direction of the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$,then it moves $14 \ m$ along the direction of the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$. What is its total displacement (in meters)?

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 $^{\circ}$)?

The displacement of a particle from a point having position vector $2 \hat{i} + 4 \hat{j}$ to another point having position vector $5 \hat{i} + 1 \hat{j}$ is ........ units.

The magnitudes of vectors $\vec A, \vec B$ and $\vec C$ are $3, 4$ and $5$ units respectively. If $\vec A + \vec B = \vec C$,the angle between $\vec A$ and $\vec B$ is

Magnitudes of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are $4$ units and $3$ units respectively. If these vectors are $(i)$ in the same direction $(\theta = 0^{\circ})$ and $(ii)$ in the opposite direction $(\theta = 180^{\circ})$,find the magnitude of the resultant vector in each case.