When $n$ vectors of different magnitudes are added, we get a null vector. Then the value of $n$ cannot be

Which of the four arrangements in the figure correctly shows the vector addition of two forces $\overrightarrow {{F_1}} $ and $\overrightarrow {{F_2}} $ to yield the third force $\overrightarrow {{F_3}} $

If $\vec A$ and $\vec B$ are two non-zero vectors such that $\left| {\vec A + \vec B} \right| = \frac{{\left| {\vec A - \vec B} \right|}}{2}$ and $\left| {\vec A} \right| = 2\left| {\vec B} \right|$ then the angle between $\vec A$ and $\vec B$ is

Unit vector parallel to the resultant of vectors $\vec A = 4\hat i - 3\hat j$and $\vec B = 8\hat i + 8\hat j$ will be

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

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