\cos\left( \theta - \left(\theta + \frac{2\pi}{7} \right) \right) = \cos\left( -\frac{2\pi}{7} \right) = \cos\left( \frac{2\pi}{7} \right).

February 22, 2026

$\cos\left( \theta - \left(\theta + \frac{2\pi}{7} \right) \right) = \cos\left( -\frac{2\pi}{7} \right) = \cos\left( \frac{2\pi}{7} \right).$

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Solution: The angle between two unit vectors $ \vec{v} $ and $ \vec{w} $ is given by the cosine of the difference of their angles:
\cos \phi = \vec{v} \cdot \vec{w} = \cos \theta \cos\left(\theta + \frac{2\pi}{7}\right) + \sin \theta \sin\left(\theta + \frac{2\pi}{7}\right).
Using the cosine of difference identity:
Therefore, the angle between $ \vec{v} $ and $ \vec{w} $ is $ \frac{2\pi}{7} $.
\boxed{\frac{2\pi}{7}}
Question: A pharmacologist is modeling the interaction of two molecules as vectors in 3D space. If molecule A is represented by $ \vec{a} = (1, 2, 3) $ and molecule B by $ \vec{b} = (4, 5, 6) $, find the area of the parallelogram formed by these vectors.

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