If in a triangle $ABC$, $\overrightarrow{BC} = \frac{\vec{u}}{|\vec{u}|\vec{v}|}$ and $\overrightarrow{AC} = \frac{2\vec{u}}{|\vec{u}|}$ where $|\vec{u}||\vec{v}|$, then $1 + \cos 2A + \cos 2B + \cos 2C = _______.
Step-by-Step Solution
Key Concept: Use dot products to enforce orthogonality conditions; express unit vectors and manipulate them to find relationships between angles.
Given $\vec{BC} = \frac{\vec{u}}{|\vec{u}|} - \frac{\vec{v}}{|\vec{v}|}$ and $\vec{AC} = \frac{2\vec{u}}{|\vec{u}|}$, we compute $\vec{AB} = \vec{AC} - \vec{BC} = \frac{\vec{u}}{|\vec{u}|} + \frac{\vec{v}}{|\vec{v}|}$. From $\vec{AB} \cdot \vec{BC} = 0$, the orthogonality condition simplifies to $1 + \cos 2A + \cos 2B + \cos 2C - 1 - 4\cos A \cos B \cos C = 0$ with $B = \frac{\pi}{2}$.
Correct Answer: I need to find the value of $1 + \cos 2A + \cos 2B + \cos 2C$.
From the solution, it states that $B = \frac{\pi}{2}$, which means the triangle is right-angled at B.
For a right-angled triangle with $B = \frac{\pi}{