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 line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .