Find the points of the ellipsoid $x^2+2y^2+3z^2 = 1$ which are closest to and furthest from the plane $x+y+z=10$
Hi am i going in the right direction?
I wan to use the fact that the distance formula is….
$$d^2 = (x-u)^2+(y-v)^2+(z-w)^2 $$
So i want to maximise and minimise u,v,w. And i want to use two constraints
$$f(x,y,z,u,v,w) = (x-u)^2+(y-v)^2+(z-w)^2 $$
$$h(x….w) = u+v+w-10 = 0 \quad constraint \, (1) $$
$$g(x….w) = x^2 + 2y^2 +3z^2 – 1 = 0 \quad constraint \, (2) $$
And then using Lagrange multipliers i want to say that
$$\nabla f = \lambda \nabla h + \mu \nabla g $$
For which i found that
$$f_u: -2 (x-u) = \lambda $$
$$f_v: -2 (y-v) = \lambda $$
$$f_w: -2 (z-w) = \lambda $$
$$f_x: (x-u) = \mu x $$
$$f_x: (y-v) = 2\mu y $$
$$f_x: (z-w) = 3\mu z $$
Saying that $f_u = f_v = f_w $
$$(x-u) = (y-v) = (z-w) = \frac{ \lambda }{-2} $$
Then subbing this this into g
$$ g(x … w) = \lambda^2 + \lambda^2 + \lambda^2 = 0 $$ $$ \lambda^2 = 0 $$ $$ \lambda = 0 $$
If this is true then from f_x
$$(x-u) = \mu x $$
$$\lambda = \mu x \quad \text{from $f_u = f_v = f_w $ } $$
$$0 = \mu x $$
$$\mu = 0 $$
$$x = 0 $$
not sure where to go after here
Best Answer
at the points that are closest and the farthest from the plane $x+y+z = 10$ should have the normal $(2x, 4y, 6z)$ of the surface $x^2 + 2y^2 + 3z^2 = 1$ be parallel to the normal $(1,1,1)$ of the plane. therefore we can take $x = 6t, y = 3t, z= 2t.$ making this point on the ellipsoid requires $$1=(6t)^2+2(3t)^2 + 3(2t)^2 = 66t^2 \to t = \pm 1/\sqrt{66}.$$ these values of $t$ give you the required points.