For a homogeneous system to have a non-trivial solution , the determinant of its coefficient matrix must be zero.
The system is
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+(\sqrt2\sin\alpha)y+(\sqrt2\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 So the coefficient matrix is
A = ( 1 2 sin α 2 cos α 1 cos α sin α 1 sin α − cos α ) . A=\begin{pmatrix}
1 & \sqrt2\sin\alpha & \sqrt2\cos\alpha\\
1 & \cos\alpha & \sin\alpha\\
1 & \sin\alpha & -\cos\alpha
\end{pmatrix}. A = 1 1 1 2 sin α cos α sin α 2 cos α sin α − cos α . We need
det ( A ) = 0. \det(A)=0. det ( A ) = 0.
Apply row operations: replace
R 2 → R 2 − R 1 , R 3 → R 3 − R 1 . R_2\to R_2-R_1, \qquad R_3\to R_3-R_1. R 2 → R 2 − R 1 , R 3 → R 3 − R 1 . Then
det ( A ) = ∣ 1 2 sin α 2 cos α 0 cos α − 2 sin α sin α − 2 cos α 0 sin α − 2 sin α − cos α − 2 cos α ∣ . \det(A)=
\begin{vmatrix}
1 & \sqrt2\sin\alpha & \sqrt2\cos\alpha\\
0 & \cos\alpha-\sqrt2\sin\alpha & \sin\alpha-\sqrt2\cos\alpha\\
0 & \sin\alpha-\sqrt2\sin\alpha & -\cos\alpha-\sqrt2\cos\alpha
\end{vmatrix}. det ( A ) = 1 0 0 2 sin α cos α − 2 sin α sin α − 2 sin α 2 cos α sin α − 2 cos α − cos α − 2 cos α . Expanding along the first column,
det ( A ) = ∣ cos α − 2 sin α sin α − 2 cos α ( 1 − 2 ) sin α − ( 1 + 2 ) cos α ∣ . \det(A)=
\begin{vmatrix}
\cos\alpha-\sqrt2\sin\alpha & \sin\alpha-\sqrt2\cos\alpha\\
(1-\sqrt2)\sin\alpha & -(1+\sqrt2)\cos\alpha
\end{vmatrix}. det ( A ) = cos α − 2 sin α ( 1 − 2 ) sin α sin α − 2 cos α − ( 1 + 2 ) cos α .
Compute this determinant:
det ( A ) = ( cos α − 2 sin α ) ( − ( 1 + 2 ) cos α ) − ( sin α − 2 cos α ) ( 1 − 2 ) sin α . \det(A)=\big(\cos\alpha-\sqrt2\sin\alpha\big)\big(-(1+\sqrt2)\cos\alpha\big)
-\big(\sin\alpha-\sqrt2\cos\alpha\big)(1-\sqrt2)\sin\alpha. det ( A ) = ( cos α − 2 sin α ) ( − ( 1 + 2 ) cos α ) − ( sin α − 2 cos α ) ( 1 − 2 ) sin α . Now simplify term by term.
First term:
− ( 1 + 2 ) cos α ( cos α − 2 sin α ) = − ( 1 + 2 ) cos 2 α + 2 ( 1 + 2 ) sin α cos α . -(1+\sqrt2)\cos\alpha(\cos\alpha-\sqrt2\sin\alpha)
=-(1+\sqrt2)\cos^2\alpha+\sqrt2(1+\sqrt2)\sin\alpha\cos\alpha. − ( 1 + 2 ) cos α ( cos α − 2 sin α ) = − ( 1 + 2 ) cos 2 α + 2 ( 1 + 2 ) sin α cos α . Since
2 ( 1 + 2 ) = 2 + 2 , \sqrt2(1+\sqrt2)=2+\sqrt2, 2 ( 1 + 2 ) = 2 + 2 , this becomes
− ( 1 + 2 ) cos 2 α + ( 2 + 2 ) sin α cos α . -(1+\sqrt2)\cos^2\alpha+(2+\sqrt2)\sin\alpha\cos\alpha. − ( 1 + 2 ) cos 2 α + ( 2 + 2 ) sin α cos α . Second term:
( sin α − 2 cos α ) ( 1 − 2 ) sin α = ( 1 − 2 ) sin 2 α − 2 ( 1 − 2 ) sin α cos α . (\sin\alpha-\sqrt2\cos\alpha)(1-\sqrt2)\sin\alpha
=(1-\sqrt2)\sin^2\alpha-\sqrt2(1-\sqrt2)\sin\alpha\cos\alpha. ( sin α − 2 cos α ) ( 1 − 2 ) sin α = ( 1 − 2 ) sin 2 α − 2 ( 1 − 2 ) sin α cos α . Since
2 ( 1 − 2 ) = 2 − 2 , \sqrt2(1-\sqrt2)=\sqrt2-2, 2 ( 1 − 2 ) = 2 − 2 , we get
( 1 − 2 ) sin 2 α − ( 2 − 2 ) sin α cos α . (1-\sqrt2)\sin^2\alpha-(\sqrt2-2)\sin\alpha\cos\alpha. ( 1 − 2 ) sin 2 α − ( 2 − 2 ) sin α cos α . Therefore,
\det(A)=-(1+\sqrt2)\cos^2\alpha-(1-\sqrt2)\sin^2\alpha+ig[(2+\sqrt2)+(\sqrt2-2)\big]\sin\alpha\cos\alpha.
So
det ( A ) = − ( 1 + 2 ) cos 2 α − ( 1 − 2 ) sin 2 α + 2 2 sin α cos α . \det(A)=-(1+\sqrt2)\cos^2\alpha-(1-\sqrt2)\sin^2\alpha+2\sqrt2\sin\alpha\cos\alpha. det ( A ) = − ( 1 + 2 ) cos 2 α − ( 1 − 2 ) sin 2 α + 2 2 sin α cos α .
Set determinant equal to zero:
− ( 1 + 2 ) cos 2 α − ( 1 − 2 ) sin 2 α + 2 2 sin α cos α = 0. -(1+\sqrt2)\cos^2\alpha-(1-\sqrt2)\sin^2\alpha+2\sqrt2\sin\alpha\cos\alpha=0. − ( 1 + 2 ) cos 2 α − ( 1 − 2 ) sin 2 α + 2 2 sin α cos α = 0. Multiply by − 1 -1 − 1
( 1 + 2 ) cos 2 α + ( 1 − 2 ) sin 2 α − 2 2 sin α cos α = 0. (1+\sqrt2)\cos^2\alpha+(1-\sqrt2)\sin^2\alpha-2\sqrt2\sin\alpha\cos\alpha=0. ( 1 + 2 ) cos 2 α + ( 1 − 2 ) sin 2 α − 2 2 sin α cos α = 0. Divide by cos 2 α \cos^2\alpha cos 2 α α ∈ ( 0 , π / 2 ) \alpha\in(0,\pi/2) α ∈ ( 0 , π /2 ) cos α ≠ 0 \cos\alpha\neq 0 cos α = 0
( 1 + 2 ) + ( 1 − 2 ) tan 2 α − 2 2 tan α = 0. (1+\sqrt2)+(1-\sqrt2)\tan^2\alpha-2\sqrt2\tan\alpha=0. ( 1 + 2 ) + ( 1 − 2 ) tan 2 α − 2 2 tan α = 0. Let t = tan α t=\tan\alpha t = tan α
( 1 − 2 ) t 2 − 2 2 t + ( 1 + 2 ) = 0. (1-\sqrt2)t^2-2\sqrt2 t+(1+\sqrt2)=0. ( 1 − 2 ) t 2 − 2 2 t + ( 1 + 2 ) = 0.
Solve the quadratic:
( 1 − 2 ) t 2 − 2 2 t + ( 1 + 2 ) = 0. (1-\sqrt2)t^2-2\sqrt2 t+(1+\sqrt2)=0. ( 1 − 2 ) t 2 − 2 2 t + ( 1 + 2 ) = 0. Notice that t = 1 + 2 t=1+\sqrt2 t = 1 + 2
( 1 − 2 ) ( 1 + 2 ) 2 − 2 2 ( 1 + 2 ) + ( 1 + 2 ) = 0. (1-\sqrt2)(1+\sqrt2)^2-2\sqrt2(1+\sqrt2)+(1+\sqrt2)=0. ( 1 − 2 ) ( 1 + 2 ) 2 − 2 2 ( 1 + 2 ) + ( 1 + 2 ) = 0. Also, using product of roots,
t 1 t 2 = 1 + 2 1 − 2 = − ( 1 + 2 ) 2 . t_1t_2=\frac{1+\sqrt2}{1-\sqrt2}=-(1+\sqrt2)^2. t 1 t 2 = 1 − 2 1 + 2 = − ( 1 + 2 ) 2 . So the other root is negative. Since α ∈ ( 0 , π / 2 ) \alpha\in(0,\pi/2) α ∈ ( 0 , π /2 ) t = tan α > 0 t=\tan\alpha>0 t = tan α > 0
tan α = 1 + 2 . \tan\alpha=1+\sqrt2. tan α = 1 + 2 .
Use the standard value
tan ( π 8 ) = 2 − 1. \tan\left(\frac{\pi}{8}\right)=\sqrt2-1. tan ( 8 π ) = 2 − 1. Therefore,
tan ( 3 π 8 ) = cot ( π 8 ) = 1 2 − 1 = 1 + 2 . \tan\left(\frac{3\pi}{8}\right)=\cot\left(\frac{\pi}{8}\right)=\frac{1}{\sqrt2-1}=1+\sqrt2. tan ( 8 3 π ) = cot ( 8 π ) = 2 − 1 1 = 1 + 2 . Thus,
α = 3 π 8 = 9 π 24 . \alpha=\frac{3\pi}{8}=\frac{9\pi}{24}. α = 8 3 π = 24 9 π .
Compare with options:
A: 5 π 24 \frac{5\pi}{24} 24 5 π
B: 11 π 24 \frac{11\pi}{24} 24 11 π
C: 7 π 24 \frac{7\pi}{24} 24 7 π
D: 3 π 4 \frac{3\pi}{4} 4 3 π
None equals 3 π 8 \frac{3\pi}{8} 8 3 π
So the mathematically derived answer is
α = 3 π 8 . \boxed{\alpha=\frac{3\pi}{8}}. α = 8 3 π . The stored correct answer A = 5 π 24 A=\frac{5\pi}{24} A = 24 5 π not match.