At what angle must the two forces $(x + y)$ and $(x -y)$ act so that the resultant may be $\sqrt {({x^2} + {y^2})} $
${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{2({x^2} - {y^2})}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{2({x^2} - {y^2})}}{{{x^2} + {y^2}}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right)$
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
A vector $\vec A $ is rotated by a small angle $\Delta \theta$ radian $( \Delta \theta << 1)$ to get a new vector $\vec B$ In that case $\left| {\vec B - \vec A} \right|$ is
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 :
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
Three vectors $\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}}$ and $\overrightarrow{\mathrm{OR}}$ each of magnitude $A$ are acting as shown in figure. The resultant of the three vectors is $A \sqrt{x}$. The value of $x$ is. . . . . . . . .