let $f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2},a>b>c>0$, what is the maximum of $f(x,y,z)$ under the restriction $x^2+y^2+z^2=1,x\cos\alpha +y\cos\beta+z\cos\gamma=0,\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1?$
I noticed that the valid $(x,y,z)$ is on a circle where the normal vector of the surface of the circle is $(\cos\alpha,\cos\beta,\cos\gamma)$, but I don't know how to do then.
Another method I've tried is the Lagrange Multiplier method, but I still don't work it out.
I'll be grateful if there's any help :) .
Let $u=\cos\alpha$, $v=\cos\beta$, and $w=\cos\gamma$. Then the Lagrange multiplier system of equations is
$$ \begin{align} 2x/a^2 - 2\lambda_1 x - \lambda_2 u &= 0 \nonumber \\ 2y/b^2 - 2\lambda_1 y - \lambda_2 v &= 0 \nonumber \\ 2z/c^2 - 2\lambda_1 z - \lambda_2 w &= 0 \nonumber \\ x^2+y^2+z^2-1 &= 0 \nonumber \\ xu + yv + zw &= 0. \end{align} $$
The equation $$ \begin{align} \lambda_1^2 &- \left[ u^2(1/b^2+1/c^2) + v^2(1/a^2+1/c^2) + w^2(1/a^2+1/b^2)\right]\lambda_1 \nonumber \\ &+ (u^2/(b^2 c^2) + v^2/(a^2 c^2) + w^2/(a^2 b^2)) \nonumber \\ &= 0 \end{align} $$ follows from the previous system of equations. The maximum and minimum values of $f$ are the maximum and minimum solutions for $\lambda_1$ in this quadratic equation. I numerically verified this, but did not prove it algebraically.