I have the following two sufaces:
- A not-rotated and not-translated ellipsoid, i.e. whose axes are parallel to the axes of the coordinate system and whose center is at (0,0,0): $$ S1:\quad\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2}=1 $$
- A plane perpendicular to the z-axis intersecting the ellipsoid: $$ S2:\quad z=k,\quad 0\leq k \leq c $$
How can I evaluate the angles between the two sufaces at the intersection?
I probably need to do the following steps:
- Find the equation of the tangent to the ellipsoid (S3).
- Find the angles between S2 and S3.
- Constrain the equation of the angles to only the intersecting points between the ellipsoid and the plane.
Yes those steps are correct. Please note the angle between two planes is same as the angle between their normal vectors.
The tangent plane to ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ at point $P(X, Y, k)$ is given by
$\frac{x X}{a^2} + \frac{y Y}{b^2} + \frac{z k}{c^2} = 1$, where $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = \frac{c^2 - k^2}{c^2}$
So the normal vector to the tangent plane is $ \ \vec n_1 = (\frac{X}{a^2}, \frac{Y}{b^2},\frac{k}{c^2})$
The normal vector to the plane $z = k$ is $\ \vec n_2 = (0, 0, 1)$.
Now to find angle between normal vectors, use
$\vec n_1 \cdot \vec n_2 = |\vec n_1| \ |\vec n_2| \cos\theta$