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

Answer: (C)

$\frac{5\pi}{24}$

Answer 2

**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 $n = 0$, $x = \frac{\pi}{3} - \frac{\pi}{8} = \frac{5\pi}{24}.$