1. **Problem Statement:** Prove the projection law in any triangle $\triangle ABC$ using vector methods: $$a = b \cos C + c \cos B$$ where $a$, $b$, and $c$ are the sides opposite vertices $A$, $B$, and $C$ respectively. 2. **Given Vector Relation:** In vector form, the sides satisfy $$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$ which implies $$\vec{a} = -\vec{b} - \vec{c}$$. 3. **Dot Product Step:** Take the dot product of both sides with $\vec{a}$: $$\vec{a} \cdot \vec{a} = \vec{a} \cdot (-\vec{b} - \vec{c}) = -\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$$ 4. **Express Dot Products Using Cosines:** Recall that for vectors $\vec{u}$ and $\vec{v}$, $$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$$ where $\theta$ is the angle between them. Here, $|\vec{a}| = a$, $|\vec{b}| = b$, $|\vec{c}| = c$. The angles between $\vec{a}$ and $\vec{b}$, and $\vec{a}$ and $\vec{c}$ correspond to $180^\circ - C$ and $180^\circ - B$ respectively because vectors are oriented along sides of the triangle but in opposite directions. Thus, $$\vec{a} \cdot \vec{b} = ab \cos(180^\circ - C) = -ab \cos C$$ $$\vec{a} \cdot \vec{c} = ac \cos(180^\circ - B) = -ac \cos B$$ 5. **Substitute Back:** $$a^2 = -(-ab \cos C) - (-ac \cos B) = ab \cos C + ac \cos B$$ 6. **Divide Both Sides by $a$:** $$a = b \cos C + c \cos B$$ **Conclusion:** The projection law is proven using vector dot product properties and the given vector relation in the triangle.