The magnitude of vectors $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$ in the given figure are equal. The direction of $\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }$ with $x$-axis will be
$\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$
$\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1+\sqrt{3}-\sqrt{2})}$
$\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1-\sqrt{3}+\sqrt{2})}$
$\tan ^{-1} \frac{(1+\sqrt{3}-\sqrt{2})}{(1-\sqrt{3}-\sqrt{2})}$
Two forces each numerically equal to $10$ $dynes$ are acting as shown in the adjoining figure, then the magnitude of resultant is.........$dyne$
If vectors $P, Q$ and $R$ have magnitude $5, 12$ and $13 $ units and $\overrightarrow P + \overrightarrow Q = \overrightarrow R ,$ the angle between $Q$ and $R$ is
There are two force vectors, one of $5\, N$ and other of $12\, N $ at what angle the two vectors be added to get resultant vector of $17\, N, 7\, N $ and $13 \,N$ respectively
Two forces ${F_1} = 1\,N$ and ${F_2} = 2\,N$ act along the lines $x = 0$ and $y = 0$ respectively. Then the resultant of forces would be
Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^o$, then which of the following vectors may be equal to $\frac{{\vec A}}{2} - \vec B$