Find the points of the ellipsoid $x^2+2y^2+3z^2 = 1$ which are closest to and furthest from the plane $x+y+z=10$

Question

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

Answer 1

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.

Answer 2

I got pretty far..... But the two sets of points i got are equidistant at $d = \frac{\sqrt{2}3}{11}=1.2792$

I want 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

$$f_u: -2 (x-u) = \lambda......(1) $$ $$f_v: -2 (y-v) = \lambda .......(2)$$ $$f_w: -2 (z-w) = \lambda.....(3) $$ $$f_x: (x-u) = \mu x .......(4) $$ $$f_x: (y-v) = 2\mu y ..,,..(5) $$ $$f_x: (z-w) = 3\mu z .......(6)$$

Saying that $f_u = f_v = f_w $

$$(x-u) = (y-v) = (z-w) = \frac{\lambda}{2} ...(a) $$

This means that $(4)=(5)=(6)= \frac{\lambda}{2}$ $$\mu x = 2\mu y = 3\mu z = \frac{\lambda}{2}$$ $$x = 2y = 3z = \frac{\lambda}{2\mu} ... (b)$$

From b $x=2y$ $y=1/2x$ $z=1/3x#

Sub this in to g $$x^2+ 2\frac{1}{4}x^2+3\frac{1}{9}x^2 = 1 $$ $$6x^2+ 3x^2+2x^2 = 6 $$ $$11x^2=6$$ $$x= \pm \sqrt{\frac{6}{11}}$$

meaning... $$y=\pm \frac{1}{2} \sqrt{\frac{6}{11}}, \quad z= \pm \frac{1}{3} \sqrt{\frac{6}{11}} $$

(x,y,z) points are $$point \, A = \left(\sqrt{\frac{6}{11}} ,\frac{1}{2} \sqrt{\frac{6}{11}}, \frac{1}{3} \sqrt{\frac{6}{11}}\right)$$ $$point \, B = \left(-\sqrt{\frac{6}{11}} , -\frac{1}{2} \sqrt{\frac{6}{11}}, -\frac{1}{3} \sqrt{\frac{6}{11}}\right)$$

From (b) consider that $$ x = \frac{\lambda}{-2\mu}$$ $$\sqrt{\frac{6}{11}} = \frac{\lambda}{-2\mu} \quad x= \sqrt{\frac{6}{11}}$$ $$\lambda = -2\mu \sqrt{\frac{6}{11}}$$

Now from (1) $$ -(x-u) = \lambda$$ $$-2x = \lambda \quad from \, (c)$$ $$ -2 \frac{\lambda}{-2\mu} = \lambda$$ $$ \frac{1}{\mu} = 1$$ $$ hence \, \mu = 1$$

$$\therefore \lambda = -2 \sqrt{\frac{6}{11}}$$ so... $$ \frac{\lambda}{-2} = \frac{-2 \sqrt{\frac{6}{11}}}{-2}= \sqrt{\frac{6}{11}}$$

$$ \therefore (x-y) = (y-v) = (z-w) = \sqrt{\frac{6}{11}}.... (c) $$

Then i got that $$ u = x - \sqrt{\frac{6}{11}}$$ $$ v = y - \sqrt{\frac{6}{11}}$$ $$ w = z - \sqrt{\frac{6}{11}}$$

Using thid i found that At "point A" $$G(x,y,z,u,v,w) = \left( \sqrt{\frac{6}{11}}, 1/2 \sqrt{\frac{6}{11}}, 1/3 \sqrt{\frac{6}{11}}, 0, -1/2 \sqrt{\frac{6}{11}}, -2/3 \sqrt{\frac{6}{11}} \right)$$

At point "Point B" $$G(x,y,z,u,v,w) = \left( -\sqrt{\frac{6}{11}}, -1/2 \sqrt{\frac{6}{11}}, -1/3 \sqrt{\frac{6}{11}}, -2\sqrt{\frac{6}{11}}, -3/2 \sqrt{\frac{6}{11}}, -1/3 \sqrt{\frac{6}{11}} \right)$$

Which when i plug in to d both get $d = \frac{\sqrt{2}3}{11}=1.2792$