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Question:
Answered & Verified by Expert
If the system of equations
$\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
Options:
$\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
Solution:
1301 Upvotes
Verified Answer
The correct answer is:
$\frac{5 \pi}{24}$
$\begin{aligned} & \left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \sin \alpha & -\cos \alpha \\ 1 & \cos \alpha & \sin \alpha\end{array}\right|=0 \\ & \Rightarrow 1-\sqrt{2} \sin \alpha(\sin \alpha+\cos \alpha)+\sqrt{2} \cos \alpha(\cos \alpha-\sin \alpha)=0 \\ & \Rightarrow 1+\sqrt{2} \cos 2 \alpha-\sqrt{2} \sin 2 \alpha=0 \\ & \cos 2 \alpha-\sin 2 \alpha=-\frac{1}{\sqrt{2}} \\ & \cos \left(2 \alpha+\frac{\pi}{4}\right)=-\frac{1}{2} \\ & 2 \alpha+\frac{\pi}{4}=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} \\ & \alpha+\frac{\pi}{8}=\mathrm{n} \pi \pm \frac{\pi}{3} \\ & \mathrm{n}=0 \\ & \mathrm{x}=\frac{\pi}{3}-\frac{\pi}{8}=\frac{5 \pi}{24}\end{aligned}$
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