The angle between vector $(\overrightarrow{{A}})$ and $(\overrightarrow{{A}}-\overrightarrow{{B}})$ is :
$\tan ^{-1}\left(\frac{-\frac{{B}}{2}}{{A}-{B} \frac{\sqrt{3}}{2}}\right)$
$\tan ^{-1}\left(\frac{{A}}{0.7 {B}}\right)$
$\tan ^{-1}\left(\frac{\sqrt{3} {B}}{2 {A}-{B}}\right)$
$\tan ^{-1}\left(\frac{{B} \cos \theta}{{A}-{B} \sin \theta}\right)$
Two vectors $\dot{A}$ and $\dot{B}$ are defined as $\dot{A}=a \hat{i}$ and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega t \hat{\mathrm{i}}+\sin \omega t \hat{j}$ ), where a is a constant and $\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1}$. If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $t=\tau$ for the first time, the value of $\tau$, in, seconds, is. . . . . .
$\overrightarrow A \, = \,3\widehat i\, + \,2\widehat j$ , $\overrightarrow B \, = \widehat {\,i} + \widehat j - 2\widehat k$ then find their addition by algebric method.
The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is ....... $^o$