If the system of equations aligned & x+( 2 α) y+( 2 α) z=0 & x+( α) y+( α) z=0 & x+( α) y-( α) z=0 aligned has a non-trivial solution, then α ∈(0, (π)/(2)) is equal to :

Correct answer: B. (5 π)/(24) aligned & | array ccc 1 & 2 α & 2 α 1 & α & - α 1 & α & α array |=0 & 1- 2 α( α+ α)+ 2 α( α- α)=0 & 1+ 2 2 α- 2 2 α=0 & 2 α- 2 α=- 1 2 & (2 α+(π)/(4))=-(1)/(2) & 2 α+(π)/(4)=2 n π ± (2 π)/(3) & α+(π)/(8)=n π ± (π)/(3) & n=0 & x=(π)/(3)-(π)/(8)=(5 π)/(24) aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Determinants

If the system of equations aligned & x+( 2 α) y+( 2 α) z=0 & x+( α)...

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

A $\frac{11 \pi}{24}$
B $\frac{5 \pi}{24}$
C $\frac{7 \pi}{24}$
D $\frac{3 \pi}{4}$

Correct answer

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

Step-by-step solution

\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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