Question

If the system of equations
\( x + \left( \sqrt{2} \sin \alpha \right) y + \left( \sqrt{2} \cos \alpha \right) z = 0 \)
\( x + \left( \cos \alpha \right) y + \left( \sin \alpha \right) z = 0 \)
\( x + \left( \sin \alpha \right) y - \left( \cos \alpha \right) z = 0 \)
has a non-trivial solution, then \( \alpha \in \left( 0, \frac{\pi}{2} \right) \) is equal to:

• A. \( \frac{3\pi}{4} \)
• B. \( \frac{7\pi}{24} \)
• C. \( \frac{5\pi}{24} \)
• D. \( \frac{11\pi}{24} \)

Answer 1

The Correct Option is C

System Representation: The system of equations is expressed in matrix form as: \[ \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

Condition for Non-Trivial Solution: A non-trivial solution exists if the determinant of the coefficient matrix equals zero: \[ \text{det} \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix} = 0 \]

Determinant Calculation: Evaluating the determinant yields: \[ \text{det} = 1 \times (\cos \alpha \times (-\cos \alpha) - \sin \alpha \times \sin \alpha) - \sqrt{2} \sin \alpha \times (1 \times -\cos \alpha - 1 \times \sin \alpha) + \sqrt{2} \cos \alpha \times (1 \times \sin \alpha - 1 \times \cos \alpha) \] The simplification of this determinant results in a trigonometric equation for \(\alpha\).

Solving for \(\alpha\): The solution to the derived trigonometric equation is found to be \(\alpha = \frac{5\pi}{24}\), based on the relation \[ \alpha + \frac{\pi}{8} = n\pi \pm \frac{\pi}{3} \] For \(n = 0\), the specific solution is \[ x = \frac{\pi}{3} - \frac{\pi}{8} = \frac{5\pi}{24}. \]