If the system of equations gathered x+(√2 )y+(√2 )z=0 x+( )y+( )z=0 x+( )y-( )z=0 gathered…

Question

If the system of equations

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

has a non-trivial solution, then $\alpha \in (0,\frac{\pi }{2})$ is equal to :

Accepted Answer

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

$\because $ Non-trivial solution

$\Rightarrow D=0$

$\begin{gathered} |\begin{array}{ccc}1 \quad  \sqrt{2}\sin \alpha  \quad  \sqrt{2}\cos \ 1 \quad  \cos \alpha  \quad  \sin \alpha \ 1 \quad  \sin \alpha  \quad  -\cos \alpha \end{array}|=0 \ 1{[-{\cos }^{2}\alpha -\sin \alpha ]}^{-1}[-\sqrt{2}\sin \alpha \cos \alpha -\sqrt{2}\sin \alpha \cos \alpha ]+1[\sqrt{2}{\sin }^{2}\alpha -\sqrt{2}{\cos }^{2}\alpha ]=0 \end{gathered}$

$\begin{gathered} -1+2\sqrt{2}\sin \alpha \cos \alpha +\sqrt{2}({\sin }^{2}\alpha -{\cos }^{2}\alpha )=0 \ \sqrt{2}\sin 2\alpha -\sqrt{2}\cos 2\alpha =1 \ \frac{\sin 2\alpha }{\sqrt{2}}-\frac{\cos 2\alpha }{\sqrt{2}}=\frac{1}{2} \ \sin (2\alpha -\frac{\pi }{4})=\sin \frac{\pi }{6} \ \Rightarrow 2\alpha -\frac{\pi }{4}=n\pi +(-1{)}^{n}\frac{\pi }{6}n=0 \ \Rightarrow \alpha =\frac{5\pi }{24} \end{gathered}$

Solutions · 1