Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $‘n’$ times the magnitude of $(\vec A - \vec B)$. The angle between $ \vec A$ and $\vec B$ is
${\cos ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]$
${\cos ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]$
${\sin ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]$
${\sin ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]$
The sum of two forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$ such that $|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{P}}| .$ The angle $\theta$ (in degrees) that the resultant of $2 \overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ will make with $\overrightarrow{\mathrm{Q}}$ is
$ABC$ is an equilateral triangle. Length of each side is $a$ and centroid is point $O$. then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=.......$
The resultant of two vectors $\overrightarrow P $ and $\overrightarrow Q $ is $\overrightarrow R .$ If $Q$ is doubled, the new resultant is perpendicular to $P$. Then $R $ equals
Given that $\overrightarrow A + \overrightarrow B = \overrightarrow C $and that $\overrightarrow C $ is $ \bot $ to $\overrightarrow A $. Further if $|\overrightarrow A |\, = \,|\overrightarrow C |,$then what is the angle between $\overrightarrow A $ and $\overrightarrow B $