QJEEM24Ma7f93: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7f93 Solution: If the system of equations has a non-trivial solution, then is…

$\begin{aligned}& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\& x+(\cos \alpha) y+(\sin \alpha) z=0 \\& x+(\sin \alpha) y-(\cos \alpha) z=0\end{aligned}$

$\because$ Non-trivial solution

$\Rightarrow D=0$

$\begin{aligned}& \left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\1 & \cos \alpha & \sin \alpha \\1 & \sin \alpha & -\cos \alpha\end{array}\right|=0 \\& 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\\end{aligned}$

$\begin{aligned}& -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\& \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\& \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\& \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\& \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\& \Rightarrow \alpha=\frac{5 \pi}{24}\end{aligned}$