Given 3 orthogonal vectors how to calculate the ellipsoid induced by them? Given 3 orthogonal vectors, how can I define their inscribing ellipsoid?
$$ax^2 + by^2 + cz^2= 1$$
or 
$$(\mathbf{x-v})^\mathrm{T}\! A\, (\mathbf{x-v}) = 1$$
Meaning, given 3 principal orthogonal directions how to craft the ellipsoid that inscribes them in the two form above.

Let's say $v_1,v_2,v_3$ are your principal directions.
If you're using the equation $(x-v)^t A (x-v) = 1$ then $v$ is the center of the ellipsoid, and $A$ is a SPD-matrix that has $v_i$ as eigenvectors. The eigenvalues of $A$ are the inverse squared lengths of the principal directions. So if we want an ellipsoid with $0$ as center the equation simplifies to $x^t A x = 1$.
Let $w_i = \frac{1}{||v_i||} v_i$ be the normalized principal directions, and let $W = [w_1 | w_2 | w_3]$. Then $W^t W  = I$ i.e. $W$ is orthogonal.
Surely now $AW = WD$ for some diagonal matrix $D = diag(d_1,d_2,d_3)$ since the columns of $W$ are eigenvectors of $A$, i.e. $A = WDW^t$. Now we just need to find the $d_i$.
We obviously want $1= v_i^t A v_i$ since $v_i$ are on the ellipsoid.
Let $e_i$ be the $i$-th unit vector i.e. $e_1 = (1,0,0)^t, e_2 = (0,1,0)^t, e_3 = (0,0,1)^t$.
So $1 = v_i^t Av_i = v_i^t WDW^t v_i = ||v_i ||e_i^t D e_i ||v_i|| = d_i \cdot ||v_i||^2$.
This holds since $v_i^t W = v_i^t [w_1|w_2|w_3] = v_i^t \left[\frac{1}{||v_1||} v_1 |\frac{1}{||v_2||} v_2 |\frac{1}{||v_3||} v_3 \right] = e_i \frac{v_i^t v_i}{||v_i||} = e_i \frac{||v_i||^2}{||v_i||} = e_i ||v_i||$.
So we have to choose $d_i = \frac{1}{||v_i||^2}$. With that choice of $W$ and $D$ the matrix $A$ satisfies $1=v^t A v$ for all points $v$ on the ellipsoid.
Only of $v_i$ are multiples of $e_i$ we can write this equation as $ax^2+by^2+cz^2=1$, for general $v_i$ and $v$ this becomes $ax^2+by^2+cz^2+dxy+exz+fyz+gx+hy+iz+d=1$.
This is easy to see: Let $v_i$ be multiples of $e_i$ then $W = I$ so $A = D$. Then let $v = (x,y,z)^t$ and we get $1= v^t A v = v^t D v = d_1 x^2 + d_2 y^2 + d_3 z^2$.
