Two forces $3\,N$ and $2\,N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $6\,N$ and the resultant becomes $2R$. The value of $\theta$ is ....... $^\circ$.

The vectors $5\hat{i} + 8\hat{j}$ and $2\hat{i} + 7\hat{j}$ are added. The magnitude of the sum of these vectors is

The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ has a magnitude $R_{1}$. If the direction of $\overrightarrow{Q}$ is reversed,the resultant has a magnitude $R_{2}$. The value of $(R_{1}^{2} + R_{2}^{2})$ is:

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

The vector sum of the two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence, the forces $\vec{A}$ and $\vec{B}$ are

The sum of two forces acting at a point is $16\, N.$ If the resultant force is $8\, N$ and its direction is perpendicular to the minimum force,then the forces are: