Question

If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that $\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}$ and deduce that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$

Answer 1

$$\begin{gathered} \text{We have} \\ \overrightarrow{BC}=\overrightarrow{a} \\ \overrightarrow{CA}=\overrightarrow{b} \\ \overrightarrow{AB}=\overrightarrow{c} \\ \left|\overrightarrow{a}\right|\text{=}a \\ \left|\overrightarrow{b}\right|=b(\because \mathrm{Length}\mathrm{is}\mathrm{always}\mathrm{positive}) \\ \overrightarrow{c}\mathrm{=c} \\ \mathrm{Now}, \\ \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}\left(\mathrm{Given}\right) \\ \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0} \\ \Rightarrow \overrightarrow{a}\times \left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{a}\times \overrightarrow{0} \\ \Rightarrow \overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{0} \\ \Rightarrow \overrightarrow{0}+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0} \\ \Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a} \\ \Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin C=\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|\sin B \\ \Rightarrow ab\sin C=ca\sin B \\ \text{Dividing both sides by}\mathit{\text{abc,}}\text{we get} \\ \mathit{\Rightarrow }\frac{\sin \mathit{}C}{c}\mathit{=}\frac{\sin \mathit{}B}{b}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}...\left(1\right) \\ \text{Again,} \\ \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}\mathit{=}\overrightarrow{\mathit{0}} \\ \mathit{\Rightarrow }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\mathit{=}\overrightarrow{\mathit{0}} \\ \mathit{\Rightarrow }\overrightarrow{b}\mathit{\times }\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{b}\mathit{\times }\overrightarrow{0} \\ \mathit{\Rightarrow }\overrightarrow{b}\mathit{\times }\overrightarrow{a}+\overrightarrow{b}\mathit{\times }\overrightarrow{b}+\overrightarrow{b}\mathit{\times }\overrightarrow{c}\mathit{=}\overrightarrow{0} \\ \mathit{\Rightarrow }\mathit{-}\overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{+}\overrightarrow{0}\mathit{+}\overrightarrow{b}\mathit{\times }\overrightarrow{c}\mathit{=}\overrightarrow{0} \\ \mathit{\Rightarrow }\overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{=}\overrightarrow{b}\mathit{\times }\overrightarrow{c} \\ \mathit{\Rightarrow }\left|\overrightarrow{a}\right|\mathit{}\left|\overrightarrow{b}\right|\sin C\mathit{=}\left|\overrightarrow{b}\right|\mathit{}\left|\overrightarrow{c}\right|\mathit{}\sin \mathit{}A \\ \mathit{\Rightarrow }ab\mathit{}\sin \mathit{}C\mathit{=}bc\mathit{}\sin A \\ \text{Dividing both sides by}\mathit{\text{abc,}}\text{we get} \\ \mathit{\Rightarrow }\frac{\sin \mathit{}C}{c}\mathit{=}\frac{\sin \mathit{}A}{a}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}...\left(2\right) \\ \text{From (1) and (2), we get} \\ \frac{\sin \mathit{}\mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}} \\ \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \end{gathered}$$