Let \theta be the angle between two unit vectors \vec{a} and \vec{b}. If \vec{a}+2\vec{b} is perpendicular to 5\vec{a}-4\vec{b}, then what is \cos~\theta+\cos~2\theta equal to?

  1. A. 0 ✓
  2. B. 1/2
  3. C. 1
  4. D. \frac{\sqrt{3}+1}{2}

Correct Answer: A. 0

Explanation

Since the vectors are perpendicular, their dot product is zero: (\vec{a}+2\vec{b}) \cdot (5\vec{a}-4\vec{b}) = 0. Expanding gives 5|\vec{a}|^2 - 4\vec{a}\cdot\vec{b} + 10\vec{b}\cdot\vec{a} - 8|\vec{b}|^2 = 0. Since they are unit vectors, |\vec{a}|=|\vec{b}|=1, so 5 + 6\cos\theta - 8 = 0, giving \cos\theta = 1/2. We evaluate \cos\theta + \cos 2\theta = \cos\theta + (2\cos^2\theta - 1) = \frac{1}{2} + 2(\frac{1}{4}) - 1 = 0.

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