Question

Find the points on the curve of intersection of the ellipsoid x^2 + 4y^2 + 4z^2 = 4 and the plane x-4y-z=0 that are closest to the origin.

Answer 1

looks like we could equate the 2 structures to define the shape

x^2 + 4y^2 + 4z^2 - 4 = x - 4y - z

(x^2-x) + (4y^2+4y) + (4z^2-z) - 4 = 0

(x-1/2)^2 -1/4 + 4(y+1/2)^2 - 1 + 4(z-1/8) - 1/16 - 4 = 0

(x-1/2)^2 + 4(y+1/2)^2 + 4(z-1/8) - 85/16 = 0

\[(x-1/2)^2 + 4(y+1/2)^2 + 4(z-1/8)^2 = \frac{85}{16}\]
this at least puts all the points in a single equation to test

Answer 2

without remembering alot of the thrms or formulas, i would say that we are looking for a sphere centered at the origin that just fits inside this

Answer 3

lagrange multipliers come to mind

Answer 4

i already tried using lagrange multipliers
\[x=x \lambda+\mu/2\]
\[y=4y \lambda-2\mu\]
\[z=4z lambd-\mu/2\]

Answer 5

just a thought, but it might help if we dont complete the squares:

x^2 + 4y^2 + 4z^2 - 4 = x - 4y - z

x^2 + 4y^2 + 4z^2 - 4 -x + 4y + z = x^2 + y^2 +z^2 - r^2

3y^2 + 3z^2 - 4 -x + 4y + z = - r^2

r^2 = x - (3y^2 + 4y) - (z^2+z) + 4, and minimize r^2

Answer 6

i tried solving those 5 eqns(including the two constraints), but i always end up with 2 unknowns..

Answer 7

btw, why is it a sphere? i thought i am looking for a curve.

Answer 8

the closest point(s) to the origin in 3d space will be the result of inflating a "spherical" balloon centered at the origin. when the radius of the sphere is minimized to accomodate the "closest point" we would have our solution

Answer 9

that and the distance formula is also the equation of a sphere/circle

Answer 10

\[d=\sqrt{x^2+y^2+z^2}\]
\[d^2=x^2+y^2+z^2\]

Answer 11

thanks for your help! i think i already got it
(0, 1/sqrt17, +-4/sqrt.17) :)

Answer 12

i wish i could verify that result, but im too rusty at it :)