Question

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 :

• A. no solution
• B. infinitely many solution
• C. exactly two solutions
• D. a unique solution

Answer 1

Answer: (B)

$\alpha +\beta +\gamma =2\pi$
$\left|\begin{array}{ccc}1& \cos \gamma & \cos \beta \\ \cos \gamma & 1& \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$
$=1+2\cos \alpha \cdot \cos \beta \cdot \cos \gamma -{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma$
$={\sin }^2\gamma -{\cos }^2\alpha -{\cos }^2\beta +(\cos (\alpha +\beta )+\cos (\alpha -$
$\beta ))\cos \gamma$
$={\sin }^2\gamma -{\cos }^2\alpha -{\cos }^2\beta +{\cos }^2\gamma +\cos (\alpha -\beta )\cos \gamma$
$={\sin }^2\alpha -{\cos }^2\beta +\cos (\alpha -\beta )\cos (\alpha +\beta )$
$={\sin }^2\alpha -{\cos }^2\beta +{\cos }^2\alpha -{\sin }^2\beta =0$