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Two forces $\vec{F}_1$ and $\vec{F}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\overrightarrow{\mathrm{F}}_2$ is $\cos ^{-1}\left(\frac{1}{\mathrm{n}}\right)$. The value of $|\mathrm{n}|$ is__________.
$6$
$7$
$8$
$9$
Solution
$\left|\overrightarrow{\mathrm{F}}_1\right|=\mathrm{F}$
$\left|\overrightarrow{\mathrm{F}}_{\mathrm{R}}\right|=\left|\overrightarrow{\mathrm{F}}_2\right|=3 \mathrm{~F}$
$\mathrm{~F}_{\mathrm{R}}^2=\mathrm{F}_1^2+\mathrm{F}_2^2+2 \mathrm{~F}_1 \mathrm{~F}_2 \cos \theta$
$9 \mathrm{~F}^2=\mathrm{F}^2+9 \mathrm{~F}^2+6 \mathrm{~F}^2 \cos \theta$
$\cos \theta=-\frac{1}{6}$
$\theta=\cos ^{-1}\left(\frac{1}{-6}\right)$
$\mathrm{n}=-6$
$|\mathrm{n}|=6$
