Let the angle with the positive x-axis be \( \alpha \).
\(\text{Given, } \beta = \frac{\alpha}{2} \text{ and } \gamma = \frac{\alpha}{2}. \)
\(\text{We know that } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \)
\(\text{Substituting the values of } \beta \text{ and } \gamma: \)
\(\cos^2 \alpha + \cos^2 \left( \frac{\alpha}{2} \right) + \cos^2 \left( \frac{\alpha}{2} \right) = 1 \)
\(\Rightarrow \cos^2 \alpha + 2 \cos^2 \left( \frac{\alpha}{2} \right) = 1 \)
\(\text{Using the identity } \cos \alpha = 2 \cos^2 \left( \frac{\alpha}{2} \right) - 1, \text{ we get } \)
\(2 \cos^2 \left( \frac{\alpha}{2} \right) = \cos \alpha + 1 \)
\(\text{So,} \)
\(\cos^2 \alpha + \cos \alpha + 1 = 1 \)
\(\Rightarrow \cos^2 \alpha + \cos \alpha = 0 \)
\(\Rightarrow \cos \alpha (\cos \alpha + 1) = 0 \)
\(\text{This gives } \cos \alpha = 0 \text{ or } \cos \alpha = -1. \)
Case 1:
\(\cos \alpha = 0 \)
\(\Rightarrow \alpha = \frac{\pi}{2} \text{ or } \alpha = \frac{3\pi}{2} \)
\(\text{Since the angles are with the positive axes, } 0 \le \alpha, \beta, \gamma \le \pi. \)
\(\text{If } \alpha = \frac{\pi}{2}, \text{ then } \beta = \frac{\pi}{4} \)
\(\text{If } \alpha = \frac{3\pi}{2}, \text{ this is not possible as } \beta = \frac{3\pi}{4} \text{ and } \gamma = \frac{3\pi}{4}, \)
\(\text{leading to } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 0 + \frac{1}{2} + \frac{1}{2} = 1. \)
Case 2:
\(\cos \alpha = -1 \)
\(\Rightarrow \alpha = \pi \)
\(\Rightarrow \beta = \frac{\pi}{2}, \quad \gamma = \frac{\pi}{2} \)
\(\Rightarrow \cos^2 \pi + \cos^2 \frac{\pi}{2} + \cos^2 \frac{\pi}{2} = 1 + 0 + 0 = 1 \)
\(\text{Possible values of } \beta \text{ are } \frac{\pi}{4} \text{ and } \frac{\pi}{2}. \)
\(\text{Sum of possible values of } \beta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}. \)
Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:
Let the shortest distance between the lines $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha + \beta$ is
Match List-I with List-II: List-I