The sum of three forces ${\vec F_1} = 100\,N,{\vec F_2} = 80\,N$ and ${\vec F_3} = 60\,N$ acting on a particle is zero. The angle between $\vec F_1$ and $\vec F_2$ is nearly .......... $^o$

Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved  part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :

Two vectors $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ have equal magnitude. The magnitude of $(\overrightarrow{{X}}-\overrightarrow{{Y}})$ is ${n}$ times the magnitude of $(\overrightarrow{{X}}+\overrightarrow{{Y}})$. The angle between $\overrightarrow{{X}}$ and $\overrightarrow{{Y}}$ is -

The position vectors of points $A, B, C$ and $D$ are $\vec A = 3\hat i + 4\hat j + 5\hat k,\,\vec B = 4\hat i + 5\hat j + 6\hat k,\,\vec C = 7\hat i + 9\hat j + 3\hat k$ and $\vec D = 4\hat i + 6\hat j$ then the displacement vectors $\overrightarrow {AB} $ and $\overrightarrow {CD} $ are

How many minimum number of non-zero vectors in different planes can be added to give zero resultant

If $\overrightarrow A = 4\hat i - 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$ then magnitude and direction of $\overrightarrow A \, + \overrightarrow B $ will be