If the system of equations x + ( 2 sin ) y + ( 2 cos ) z = 0 x + ( cos ) y + ( sin ) z = 0 x + ( sin ) y - ( cos ) z = 0 has a non-trivial solution, then ? ( 0 , 2 ) is equal to [2024]

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Solution

Q.
If the system of equations

$x + (\sqrt{2} \sin α) y + (\sqrt{2} \cos α) z = 0$

$x + (\cos α) y + (\sin α) z = 0$

$x + (\sin α) y - (\cos α) z = 0$

has a non-trivial solution, then $α \in (0, \frac{π}{2})$ is equal to [2024]

1 $\frac{7 π}{24}$

2 $\frac{3 π}{4}$

3 $\frac{11 π}{24}$

4 $\frac{5 π}{24}$

Ans.
(4)

Given, $x + (\sqrt{2} \sin α) y + (\sqrt{2} \cos α) z = 0$

$x + (\cos α) y + (\sin α) z = 0$

$x + (\sin α) y - (\cos α) z = 0$

For non-trivial solution,

$| \begin{matrix} 1 & \sqrt{2} \sin α & \sqrt{2} \cos α \\ 1 & \cos α & \sin α \\ 1 & \sin α & - \cos α \end{matrix} |$ $= 0$

$\Rightarrow 1 [- \cos^{2} α - \sin^{2} α] - 1 [- \sqrt{2} \sin α \cos α - \sqrt{2} \sin α \cos α] + 1 [\sqrt{2} \sin^{2} α - \sqrt{2} \cos^{2} α] = 0$

$\Rightarrow - 1 + 2 \sqrt{2} \sin α \cos α + \sqrt{2} (\sin^{2} α - \cos^{2} α) = 0$

$\Rightarrow \sqrt{2} \sin 2 α - \sqrt{2} \cos 2 α = 1$

$\Rightarrow \sin (2 α - \frac{π}{4}) = \sin \frac{π}{6} \Rightarrow 2 α - \frac{π}{4} = n π + {(- 1)}^{n} \frac{π}{6}$

for $n = 0 \Rightarrow α = \frac{5 π}{24}$

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