Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 20 maths

To have    $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Hint:

To solve this we use $|\vec{a} \times \vec{b}| \text { and }|\vec{a}||\vec{b}|$ formula

Given:

Solution  : $\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B}=0$

\begin{aligned} &\overrightarrow{A B} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{A B} \times 0 \\\\ &a c(\sin B-\sin A)=0 \end{aligned}

Divide by abc

\begin{aligned} &\frac{\sin B}{b}-\frac{\sin A}{a}=0 \\\\ &\frac{\sin B}{b}=\frac{\sin A}{a} \end{aligned}                                   .......(i)

\begin{aligned} &\overrightarrow{B C} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin B}{b}=\frac{\sin C}{c} \end{aligned}.......(ii)

\begin{aligned} &\overrightarrow{C A} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin C}{c}=\frac{\sin A}{a} \end{aligned}.......(iii)

From (i),(ii) and (iii)

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$