Two forces $\overline{\mathrm{F}}_1$ and $\overline{\mathrm{F}}_2$ 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 $\vec{F}_2$ is $\cos ^{-1}\left(\frac{1}{n}\right)$. The value of $|n|$ is _______.

<p>Let's denote the magnitude of the smaller force as $F$, hence the magnitude of the larger force is $3F$. The resultant force $\vec{R}$ is equal in magnitude to the larger force, which means $|\vec{R}| = 3F$. When two forces $\vec{F}_1$ and $\vec{F}_2$ act on a body, the magnitude of their resultant $\vec{R}$ can be found using the law of vector addition:</p> <p>$$|\vec{R}| = \sqrt{|\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2|\cos\theta}$$,</p> <p>where $\theta$ is the angle between $\vec{F}_1$ and $\vec{F}_2$. Given that in our case $|\vec{R}| = 3F$, $|\vec{F}_1| = F$ and $|\vec{F}_2| = 3F$, by substituting these values into the equation, we get:</p> <p>$3F = \sqrt{F^2 + (3F)^2 + 2(F)(3F)\cos\theta}$</p> <p>$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta$</p> <p>Simplifying this equation by subtracting $10F^2$ from both sides gives:</p> <p>$-F^2 = 6F^2\cos\theta$</p> <p>Dividing both sides by $-F^2$ gives:</p> <p>$-1 = -6\cos\theta$</p> <p>Therefore, $\cos\theta = \frac{1}{6}$.</p> <p>It is given that the angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$, hence comparing this with the above result, we find that $n = 6$. Therefore, $|n| = 6$.</p>