Given $$x^2+y^2+z^2=121$$ $$x\sqrt{11} + 4y + z\sqrt{22}=77$$ Find $$ \frac{\sqrt{11} + 4 + \sqrt{22}}{x+y+z} $$
I tried to plug in something for z at first, since x and y should have unique values for every value of z, but that didn't seem to work.
The answer is 7/11, which is clearly the second equation divided by the first but I don't understand how or why that would lead the final expression.
$x^2+y^2+z^2=121$ Is the equation of a sphere centered at the origin of radius 11.
$x\sqrt{11} + 4y + z\sqrt{22}=77$ is the equation of a plane
Origin is $\frac {77}{\sqrt {11 + 4^2 +22}} = 11$ units from the plane!
The plane is tangent at to the sphere.
the point of tangency $(x,y,z) = \frac {11}{7}\cdot(\sqrt{11} , 4, \sqrt {22})$
$x+y+z = \frac {11}{7} (\sqrt {11} + 4 + \sqrt{22})$
$\frac {\sqrt {11} + 4 + \sqrt{22}}{x+y+z} = \frac {7}{11}$
HINT:
WLOG $x=11\cos u\cos v,y=11\cos u\sin v,z=11\sin u$
$$77=\sqrt{11}\cdot11\cos u\cos v+4\cdot11\cos u\sin v+\sqrt{22}\cdot11\sin u$$
$$\iff7=S=\sqrt{11}\cos u\cos v+4\cos u\sin v+\sqrt{22}\sin u$$ $$=3\sqrt3\cos u\cos\left(v-\arcsin\dfrac4{3\sqrt3}\right)+\sqrt{22}\sin u$$
For $\cos u\ge0,$
$$S\le3\sqrt3\cos u+\sqrt{22}\sin u$$
Now $3\sqrt3\cos u+\sqrt{22}\sin u=7\cos\left(u-\arccos\dfrac{3\sqrt3}7\right)\le7$
So, we need $\cos\left(u-\arccos\dfrac{3\sqrt3}7\right)=\cos\left(v-\arcsin\dfrac4{3\sqrt3}\right)=1$
$\implies u\equiv\arccos\dfrac{3\sqrt3}7, v\equiv\arcsin\dfrac4{3\sqrt3}=\arccos\dfrac{\sqrt{11}}{3\sqrt3}\pmod{2\pi}$
HINT:
$$49\cdot121-77^2=(x\sqrt{11}+4y+z\sqrt{22})^2-(x^2+y^2+z^2)(11+16+22)$$
$$\iff0=(4x-\sqrt{11}y)^2+(\sqrt{22}x-\sqrt{11}z)^2+(\sqrt{22}y-4z)^2$$