Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 450 Question 450 of 472 Medium MCQ
If the system of equations
\[ x + ( 2 sin α ) y + ( 2 cos α ) z = 0 x + ( cos α ) y + ( sin α ) z = 0 x + ( sin α ) y − ( cos α ) z = 0 \begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned} x + ( 2 sin α ) y + ( 2 cos α ) z = 0 x + ( cos α ) y + ( sin α ) z = 0 x + ( sin α ) y − ( cos α ) z = 0 \]
has a non-trivial solution, then \[\alpha \in\left(0, \frac{\pi}{2}\right)\] is equal to :
Solutions · 1 Post your solution Official solution
\[ x + ( 2 sin α ) y + ( 2 cos α ) z = 0 x + ( cos α ) y + ( sin α ) z = 0 x + ( sin α ) y − ( cos α ) z = 0 \begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned} x + ( 2 sin α ) y + ( 2 cos α ) z = 0 x + ( cos α ) y + ( sin α ) z = 0 x + ( sin α ) y − ( cos α ) z = 0 \]
\[\because\] Non-trivial solution
\[\Rightarrow D=0\]
\[ \begin{aligned}
& \left|\[\begin{array}{|ccccccc|}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha\end{array}\end{aligned} \]\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 \\
\]
\[ − 1 + 2 2 sin α cos α + 2 ( sin 2 α − cos 2 α ) = 0 2 sin 2 α − 2 cos 2 α = 1 sin 2 α 2 − cos 2 α 2 = 1 2 sin ( 2 α − π 4 ) = sin π 6 ⇒ 2 α − π 4 = n π + ( − 1 ) n π 6 for n = 0 ⇒ α = 5 π 24 \begin{aligned}
& -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\
& \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\
& \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\
& \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\
& \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\
& \Rightarrow \alpha=\frac{5 \pi}{24}
\end{aligned} − 1 + 2 2 sin α cos α + 2 ( sin 2 α − cos 2 α ) = 0 2 sin 2 α − 2 cos 2 α = 1 2 sin 2 α − 2 cos 2 α = 2 1 sin ( 2 α − 4 π ) = sin 6 π ⇒ 2 α − 4 π = nπ + ( − 1 ) n 6 π for n = 0 ⇒ α = 24 5 π \] Answer the question to unlock