Further it is asked to interpret the result geometrically. I attempted this using Lagrange multiplier method but it becomes very difficult to solve as there are so many variables, it simply seems unmanageable. I want to know if there is an alternative to such problems other than the Lagrange method.

My approach:
$F(x,y,z,λ,μ) = x^2+y^2+z^2 + λ(ax^2+by^2+cz^2)+μ(lx + my+nz)$. Then I solved the following equations to get the extreme points
$\partial F/\partial x = 0$
$\partial F/\partial y = 0$
$\partial F/\partial z = 0$
$\partial F/\partial λ = 0$
$\partial F/\partial μ = 0$

and then solving above five to get μ and λ and then calculate $(x,y,z)$ for the extreme points accordingly. It becomes too cumbersome to manage all these variables. I want to know if there is any other way to arrive at the solution a bit more comfortably. Thanks. P.S. I am not writing the entire solution here for obvious reasons. I can share the image of the same if required.

My understanding of geometric interpretation is that it is asked to find the nearest and farthest point from origin where the ellipsoid and the plane intersects?

  • $\begingroup$ I have linked this and this questions because their subject are similar $\endgroup$ Aug 10 '17 at 19:15
  • $\begingroup$ , whereas my answer with geometric interpretation and Will Jagy’s analytical answer are complementary. Maybe later I’ll find how we can simplify the analytical calculations. $\endgroup$ Aug 10 '17 at 19:15

Sure. Take a rotated basis $$ \vec{U} = \frac{1}{\sqrt {l^2 + m^2 + n^2}} (l,m,n), $$ then $$ u = \vec{U} \cdot (x,y,z). $$

Next $$ \vec{V} = \frac{1}{\sqrt { m^2 + n^2}} (0,n,-m), $$ then $$ v = \vec{V} \cdot (x,y,z). $$

Finally the cross product $\vec{W}.$

The outcome is that the plane is now $u=0,$ after which the ellipse of intersection becomes some $Av^2 + B vw + C w^2 = D$

  • $\begingroup$ Thanks for quick reply, but I am unable to fully understand could you please elaborate a bit. I am still a fresher. $\endgroup$ Jun 26 '17 at 1:05

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