Question

If a vector makes angles \( \frac{\pi}{3}, \frac{\pi}{4} \) and \( \gamma \) with \( \hat{i}, \hat{j} \), and \( \hat{k} \), respectively, where \( \gamma \in \left( \frac{\pi}{2}, \pi \right) \), then the angle \( \gamma \) is:

• A. \( \frac{3\pi}{4} \)
• B. \( \frac{7\pi}{12} \)
• C. \( \frac{11\pi}{12} \)
• D. \( \frac{5\pi}{6} \)
• E. \( \frac{2\pi}{3} \)

Hint:

The sum of the squares of the directional cosines of a normalized vector is always 1, a fundamental property derived from the Pythagorean theorem in the context of Euclidean space.

Answer 1

Answer: (E)

The cosine of the angle a vector \( \vec{v} \) makes with the standard basis vectors \( \hat{i}, \hat{j}, \) and \( \hat{k} \) is related to its components. For \( \vec{v} \) normalized, the directional cosines are given by: \[ \cos(\frac{\pi}{3}) = \frac{1}{2}, \quad \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}, \quad \cos(\gamma) \] The sum of the squares of these cosines for a normalized vector equals 1: \[ \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2(\gamma) = 1 \] \[ \frac{1}{4} + \frac{1}{2} + \cos^2(\gamma) = 1 \] \[ \cos^2(\gamma) = 1 - \frac{3}{4} = \frac{1}{4} \] \[ \cos(\gamma) = \pm \frac{1}{2} \] Given that \( \gamma \) is in the interval \( \left( \frac{\pi}{2}, \pi \right) \), the cosine function is negative in this range, so: \[ \cos(\gamma) = -\frac{1}{2} \] Thus, \( \gamma = \frac{2\pi}{3} \).