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

Set up the system in matrix form: 
The system of equations can be represented 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 a Non-Trivial Solution: 
For the system to have a non-trivial solution, the determinant of the matrix must be zero: \[ \text{det} \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix} = 0 \] 

Calculate the Determinant: 
Expanding the determinant: \[ \text{det} = 1 \times (\cos \alpha \times (-\cos \alpha) - \sin \alpha \times \sin \alpha) - \sqrt{2} \sin \alpha \times (1 \times -\cos \alpha - 1 \times \sin \alpha) + \sqrt{2} \cos \alpha \times (1 \times \sin \alpha - 1 \times \cos \alpha) \] 
Simplifying this determinant leads to an equation in terms of \(\alpha\) that must be solved for \(\alpha\). 

Solve for \(\alpha\): 
Solving the resulting trigonometric equation, we find that \(\alpha = \frac{5\pi}{24}\). 
\[ \alpha + \frac{\pi}{8} = n\pi \pm \frac{\pi}{3} \] For \(n = 0\), \[ x = \frac{\pi}{3} - \frac{\pi}{8} = \frac{5\pi}{24}. \]

Answer 2

Step 1: Write the system of equations as a matrix.
The given system of equations is: \[ x + \sqrt{2} \sin \alpha \, y + \sqrt{2} \cos \alpha \, z = 0 \] \[ x + \cos \alpha \, y + \sin \alpha \, z = 0 \] \[ x + \sin \alpha \, y - \cos \alpha \, z = 0 \] This system can be written in matrix form as: \[ \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 \] For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

Step 2: Calculate the determinant of the coefficient matrix.
Let the coefficient matrix be: \[ A = \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix} \] The determinant of \( A \) is: \[ \text{det}(A) = 1 \cdot \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} - \sqrt{2} \sin \alpha \cdot \begin{vmatrix} 1 & \sin \alpha \\ 1 & -\cos \alpha \end{vmatrix} + \sqrt{2} \cos \alpha \cdot \begin{vmatrix} 1 & \cos \alpha \\ 1 & \sin \alpha \end{vmatrix} \] Now, calculate each of the 2x2 determinants: \[ \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -\cos^2 \alpha - \sin^2 \alpha = -1 \] \[ \begin{vmatrix} 1 & \sin \alpha \\ 1 & -\cos \alpha \end{vmatrix} = -\cos \alpha - \sin \alpha \] \[ \begin{vmatrix} 1 & \cos \alpha \\ 1 & \sin \alpha \end{vmatrix} = \sin \alpha - \cos \alpha \] Thus, the determinant becomes: \[ \text{det}(A) = 1 \cdot (-1) - \sqrt{2} \sin \alpha \cdot (-\cos \alpha - \sin \alpha) + \sqrt{2} \cos \alpha \cdot (\sin \alpha - \cos \alpha) \] Simplify: \[ \text{det}(A) = -1 + \sqrt{2} \sin \alpha (\cos \alpha + \sin \alpha) + \sqrt{2} \cos \alpha (\sin \alpha - \cos \alpha) \] \[ \text{det}(A) = -1 + \sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} \sin^2 \alpha + \sqrt{2} \cos \alpha \sin \alpha - \sqrt{2} \cos^2 \alpha \] \[ \text{det}(A) = -1 + 2\sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} (\sin^2 \alpha - \cos^2 \alpha) \] We can simplify this expression further, but for now, we will use the condition that the determinant must be zero for the non-trivial solution: \[ \text{det}(A) = 0 \] This leads to a transcendental equation in \( \alpha \). Solving this equation numerically or symbolically, we find that \( \alpha = \frac{5\pi}{24} \).

Step 3: Conclusion.
Thus, the value of \( \alpha \) that satisfies the equation is: \[ \boxed{\frac{5\pi}{24}} \]