Question:

Given three unit vectors $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$ and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Step 1: Calculate $\lambda$. The given condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ implies that $$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{0}|^2$$ Expanding the left side: $$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0$$ Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, $|\vec{a}|^2=1$, $|\vec{b}|^2=1$, and $|\vec{c}|^2=1$. The expression for $\lambda$ is defined as $\lambda = \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$. Substituting these values into the equation: $$1+1+1+2\lambda=0$$ $$3+2\lambda=0$$ $$2\lambda=-3$$ $$\lambda=-\frac{3}{2}$$ Step 2: Calculate $\mu$. The condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ for three unit vectors implies that these vectors are coplanar and form an equilateral triangle when placed head-to-tail. Consequently, the angle between any pair of these vectors is $120^\circ$. Let $\theta$ be the angle between any two vectors, for example, $\vec{a}$ and $\vec{b}$. Then $\theta = 120^\circ$. The magnitude of the cross product of two unit vectors is given by $|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta$. For $\vec{a}$ and $\vec{b}$: $$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin(120^\circ) = (1)(1)\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ Similarly, for $\vec{b}$ and $\vec{c}$, and for $\vec{c}$ and $\vec{a}$: $$|\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\sin(120^\circ) = (1)(1)\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ $$|\vec{c}\times\vec{a}| = |\vec{c}||\vec{a}|\sin(120^\circ) = (1)(1)\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ The expression for $\mu$ is defined as $\mu=|\vec{a}\times\vec{b}|+|\vec{b}\times\vec{c}|+|\vec{c}\times\vec{a}|$. Substituting the calculated magnitudes: $$\mu = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$$ Thus, the pair $(\lambda, \mu)$ is $\left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$.