Unit tangent vector on points of a closed surface (e.g. ellipsoid)

Question

Assuming that we know the normal unit vectors on the points of a closed surface such as an ellipsoid, is it possible to find the corresponding tangent unit vectors on the same points?

Answer 1

Let's say you know a normal unit vector $[A, B, C]$ and want to find the coordinates of a tangent unit vector $[p_1, p_2, p_3]$. As these two vectors are orthogonal, their dot product must be zero:

$$ [A, B, C] · [p_1, p_2, p_3] = 0 $$

$$ Ap_1 + Bp_2 + Cp_3 = 0 $$

Basically, that's all. Here $A, B, C$ are the coeffitients and $p_1, p_2, p_3$ are the unknowns. You can express any unknown coordinate (if its coefficient is non-zero) through the other two, and then plug in arbitary values for them to get a tangent vector, for example:

$$ p_1 = \frac{-Bp_2 - Cp_3}{A} $$

let $p_2 = 2, p_3=5$, then $p_1 = \frac{-2B - 5C}{A}$

To make it a unit vector, simply divide each coordinate by the length of the vector, which is $\sqrt{p_1^2 + p_2^2 + p_3^2}$