If in a triangle $\mathrm{ABC}, \overrightarrow{B C}=\frac{\vec{u}}{|\vec{u}|}-\frac{\vec{v}}{|\vec{

Question

If in a triangle $\mathrm{ABC}, \overrightarrow{B C}=\frac{\vec{u}}{|\vec{u}|}-\frac{\vec{v}}{|\vec{v}|}$ and $\overrightarrow{A C}=\frac{2 \vec{u}}{|\vec{u}|}$ where $|\vec{u}| 
eq|\vec{v}|$, then $1+\cos 2 A+\cos 2 B+\cos 2 C=$

Professor PC Thomas Classes · Answer

$\overrightarrow{B C}=\frac{\vec{u}}{|\vec{u}|}-\frac{\vec{v}}{|\vec{v}|}, \overrightarrow{A C}=\frac{2 \vec{u}}{|\vec{u}|} \Rightarrow \overrightarrow{A B}=\overrightarrow{A C}-\overrightarrow{B C}=\frac{\vec{u}}{|\vec{u}|}+\frac{\vec{v}}{|\vec{v}|}$\ $\Rightarrow \overrightarrow{A B} \cdot \overrightarrow{B C}=\left(\frac{\vec{u}}{|\vec{u}|}+\frac{\vec{v}}{|\vec{v}|}ight) \cdot\left(\frac{\vec{u}}{|\vec{u}|}-\frac{\vec{v}}{|\vec{v}|}ight)=\left(\frac{\vec{u} \cdot \vec{u}}{|\vec{u}|^2}-\frac{\vec{v} \cdot \vec{v}}{|\vec{v}|^2}ight)=\frac{|\vec{u}|^2}{|\vec{u}|^2}-\frac{|\vec{v}|^2}{|\vec{v}|^2}=1-1=0$\ $\Rightarrow \quad B=\frac{\pi}{2} \Rightarrow 1+\cos 2 A+\cos 2 B+\cos 2 C=1-1-4 \cos A \cos B \cos C=0$