Distance from Ellipsoid to Plane - Lagrange Multiplier

Question

Find the distance from the ellipsoid $x^2 + y^2 + 4z^2 = 4$ to the plane $x + y + z = 6$.

I'm trying to do it using Lagrange multipliers over the distance equation, but then it just gets overwhelming and I have no idea how to go on? Can someone walk me through the computation?

Answer 1

This is easy if you don't insist on using Lagrange multipliers.

The normal to the ellipse at the point $(x,y,z)$ is $\nabla(x^2+y^2+4z^2) = (2x, 2y, 8z)$. At minimum or maximum distance to the plane, this must be parallel to the normal to the plane, which is $(1,1,1).$ So $x = y = 4z$.

Plug this back into the equation for the ellipse to get $36z^2 = 4$, or $z = \pm \frac13$. So the nearest and farthest points are $\pm(\frac43,\frac43,\frac13)$.

Answer 2

I don't know how to use that maths language so please bear with me.

(I am editing his answer as this seems to be the most relevant one apart from latex. In some countries, the undergrads are unaware of this language but it shouldn't be the barrier.)

ellipsoid: $x^2 + y^2 + 4(z^2) = 4$
plane: $x + y + z = 6$ Now distance of a point $(p,q,r)$ from a plane $ax+by+cz+d=0$ is given by
$|\frac{(ap+bq+cr+d)}{\sqrt(a^2 + b^2 + c^2)}|.$
Now let the point on the ellipsoid at minimum distance from the plane be $(x,y,z)$
Therefore minimum distance $= |\frac{(x+y+z-6)}{\sqrt3}|$
Now define a function $f(x,y,z)=(x+y+z-6)/\sqrt3$
and $g(x,y,z)=x^2 + y^2 + 4(z^2) - 4 $ The minimum absolute value of f is our answer subject to the constraint $g(x,y,z)=0.$
Now, $\nabla(f) = (1/\sqrt3)i +(1/\sqrt3)j +(1/\sqrt3)k$
and $\nabla(g) = (2x)i + (2y)j + (8z)k $ Applying Lagrange's method
At minimum value of f subject to constraint g=0
$\nabla(f) = m(\nabla(g))$ ......I've used $m$ instead of $\lambda$
therefore,
$1/\sqrt3 = 2mx, 1/\sqrt3 = 2my$, $1/\sqrt3 = 8mz$
therefore, $x=y=4z$
and $x^2 + y^2 + 4(z^2) - 4 = 0$
Solving the above two equations we get two sets of points
$(4/3, 4/3, 1/3)$ and $(-4/3, -4/3, -1/3)$
Minimum and maximum distance occurs at these points on the ellipsoid
The minimum distance is given by $|f)|$
therefore, the point of minimum distance is $(4/3, 4/3, 1/3) $ and the distance is $|(f(4/3,4/3,1/3)) |=\sqrt3$
You can also solve this problem by using six variables if you don't want to use the plane point distance formula. It looks overwhelming using six variables but its easy to do but, this method is better. For the six variable method you have to solve for finding the points on the plane also.