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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

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  • $\begingroup$ the distance of a point $(x,y,z)$ to a plane is:$\dfrac{|ax+by+cz+d|}{\sqrt{a^2+b^2+c^2}}$,so you need to rework. $\endgroup$ – chenbai Nov 3 '15 at 8:28
  • $\begingroup$ Nah this is find it's kind of like $ d = \sqrt{(x-u)^2 + (y-v)^2} $ also i was asked specifically to use it. Also it's distance of a vector space. $\endgroup$ – kingportable Nov 3 '15 at 8:40
  • $\begingroup$ OK. but $\lambda=0$ seems wrong. and $\mu $ have another case which force $x=2y=3z$ $\endgroup$ – chenbai Nov 3 '15 at 10:10
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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.

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    $\begingroup$ I was about to give this answer myself as fast as I woke up, but you beat me with an hour or so. It is a nice strategy in general, to try and restrict the problem as much as possible as early as possible and since the plane has constant normal everywhere that is the easiest starting point. $\endgroup$ – mathreadler Nov 4 '15 at 2:23
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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 $d = \frac{\sqrt{2}3}{11}=1.2792$

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    $\begingroup$ ya,you get furthest points. since the plane "cut" the ellipsoid, which means the shortest d=0. there is a curve like ellipse and there are not some countable points. Lagrange multipliers can find local min or max but for globe max and min, you need to check boundary. $\endgroup$ – chenbai Nov 4 '15 at 3:33
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    $\begingroup$ Sorry, i find a problem: when you divide $\mu$, you suppose it is none zero. but when it is zero,you have to consider this case which give you the shortest case. $\endgroup$ – chenbai Nov 4 '15 at 3:43
  • $\begingroup$ Thank you @Chenbai! Could you please expand upon this? I understand that we're considering the value of $\mu$ when it is 0 i'm then guessing that we assume that $\mu x = 2\mu y = 3\mu z =\frac{\lambda}{2} = 0$ when considering $\mu = 0$ $\endgroup$ – kingportable Nov 4 '15 at 3:52
  • $\begingroup$ @chenbai hey what do you think? $\endgroup$ – kingportable Nov 4 '15 at 5:23
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    $\begingroup$ ya,you are right. $\endgroup$ – chenbai Nov 4 '15 at 12:34

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