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 :
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
Mark the correct statement :-
If the sum of two unit vectors is a unit vector, then magnitude of difference is
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
The resultant force of $5 \,N$ and $10 \,N$ can not be ........ $N$
$\overrightarrow A \, = \,2\widehat i\, + \,3\widehat j + 4\widehat k$ , $\overrightarrow B \, = \widehat {\,i} - \widehat j + \widehat k$, then find their substraction by algebric method.