Finding the distance from a point to a curve.

Question

How can I find the distance from $(0,0,0)$ to the curve $$C=\{(x,y,z)\in\mathbb{R^3};\frac{x^2}{4}+y^2+\frac{z^2}{4}=1,x+y+z=1\}$$

Answer 1

Hints: Consider these two functions of the three coordinates.

$$f=\frac{x^2}{4}+y^2+\frac{z^2}{4}-1$$ $$g=x+y+z-1$$

Then the tangent $\hat{t}$ to the curve $C$ must be perpendicular to both $\nabla f$ and $\nabla g$ (why?). In other words $$\hat{t} \propto \nabla f \times \nabla g$$

At the point $p$ on the curve C that is closest to the origin, you get that this tangent vector must be perpendicular to the vector $(x,y,z)$ (why?)

Solve this simultaneously with $f=g=0$ and you get two points on the ellipsoid where these are all met. One is the nearest to the origin and one is the farthest. Pick the smaller distance of these (smallest norm). [Edited, thanks to comments below]