- Home
- Standard 12
- Mathematics
If $\alpha+\beta+\gamma=2 \pi$, then the system of equations
$x+(\cos \gamma) y+(\cos \beta) z=0$
$(\cos \gamma) x+y+(\cos \alpha) z=0$
$(\cos \beta) x+(\cos \alpha) y+z=0$
has :
no solution
infinitely many solution
exactly two solutions
a unique solution
Solution
$\alpha+\beta+\gamma=2 \pi$
$\left|\begin{array}{ccc}1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$
$=1+2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma$
$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+(\cos (\alpha+\beta)+\cos (\alpha-\beta)) \cos \gamma$
$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \gamma+\cos (\alpha-\beta) \cos \gamma$
$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos (\alpha-\beta) \cos (\alpha+\beta)$
$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \alpha-\sin ^{2} \beta=0$
