I've been trying to solve this problem but I got really confused with it.
Assume C1 is intersection of sphere $x^2 + y^2 +z^2 =1$ and plane $x+y+z=\frac{3}{2}$ in $\mathbb{R^3}$ and C2 is intersection of sphere $x^2 + y^2 +z^2 =9$ and plane $z=0$ in $\mathbb{R^3}$. What is the minimum distance of sets C1 and C2 using Lagrange multipliers.
I tried using Lagrange double constraints(g and h) but I'm not sure what my constraints should be.I used circles equations derived from intersections as constraints and $f=(x_{1}-x_{0})^2+(y_{1}-y_{0})^2+(z_{1}-z_{0})^2$ and tried to use Lagrange method as below but I couldn't find the points: $$ \nabla(f)=\lambda_{1} \nabla(g)+\lambda_{2} \nabla(h) $$
I'd appreciate if anyone can guide me through this problem.
You need to minimize the distance between two generic points, one in each curve. So, let $(x,y,z)$ be a point in $C_1$ and $(u,v,w)$ a point in $C_2$. What you are looking for in $$ \min f(x,y,z,u,v,w) = (x-u)^2 + (y-v)^2+(z-w)^2$$
subject to: \begin{align*} x^2+y^2+z^2=1\\ x+y+z = \frac 32\\ u^2+v^2+w^2 = 9\\ w = 0 \end{align*}
So, you "just" need solve a system with 10 equations. You can make it a little simpler by substituting $w=0$ and $z = \frac 32 -x-y$ right from the start. You would need to determine $$ \min f(x,y,u,v)=(x-u)^2+(y-v)^2+(\frac 32-x-y)^2 $$
subject to: \begin{align*} x^2+y^2+(\frac 32 -x-y)^2=1\\ u^2+v^2=9 \end{align*}
which is more manageable, but still requires the solution of a system with 6 equations.
You will find 4 critical points of the Lagrangian, from which you can choose the one leading to the minimal distance.