# Find Latitude of point given Longitude and Great Circle orientation

Given the orientation of a great circle as the cartesian components of the normal vector $(a,b,c)$ to its plane, i.e. all points on the circle described in Earth-Centered Earth-Fixed (ECEF) coordinates $(x_p, y_p, z_p)$ satisfy the condition

$$x_p\cdot a + y_p\cdot b + z_p\cdot c = 0 \;\;\;\;\;\;\;\;\; (1)$$

(I am working on the Earth surface as modeled in the WGS84 standard (oblate spheroid))

Does it exist a (probably not-so-simple) method to compute the latitude of one of such points given the longitude?

I know that if I use a description of a meridian in ECEF coordinates I can "simply" intersect it with the great circle plane by solving a system of equations, but I seem to fail at some point here.

I describe the meridian as an ellipse in its plane, with semi-axes the respective standard radii of the Earth

$$\frac{x'^2}{WGS84\_A^2} + \frac{z^2}{WGS84\_B^2} = 1 \;\;\;\;\;\;\;\;\; (2)$$

Where $x'$ and $z$ can be described in term of the Latitude (but I fear this is not fully correct): $$x' = WGS84\_A \cdot cos(Latitude) \;\;\;\;\;\;\;\;\;\;\;\;\;\; \\ z = WGS84\_B \cdot sin(Latitude) \;\;\;\;\;\;\;\;\; (3)$$ and I am looking for the intercept with a positive $x'$.

Obviously the $z$ of Eq.(1) is the same of Eq.(2) and, considering that no point of the meridian will lay outside its plane, thus $y' == 0$, we can say that (and here I fear I might have insterted a mistake)

$$x' = x \cdot cos(Long) + y \cdot sin(Long) \\ 0 = - x \cdot sin(Long) + y \cdot cos(Long)$$

from the rotation above, by substitution, we obtain $x' = \frac{y}{sin(Long)}$

At this point we can solve the above non-linear sistem for the Latitude:

$$Latitude = asin(\frac{a\cdot cos(Long) + b\cdot sin(Long)}{WGS84\_B \cdot c \cdot \sqrt(\frac{1}{WGS84\_A^2} + (\frac{a\cdot cos(Long) + b\cdot sin(Long)}{WGS84\_B\cdot cos(Long)} )^2)})$$

Unfortunately, when I try this out the results are blatantly wrong, I have obviously committed a mistake, but I am unable to spot it.

Is there anyone that would like to help me or that knows the correct solution/approach?

I found the solution by changing approach (thanks to this answer)

Using the notation used there, I have

$$P = (a,b,c)$$

and

$$Q = (cos(Long+\pi/2),sin(Long+\pi/2),0)$$

the (normalized!) product $D = Q\times P$ is pointing in the direction of the point I am interested in, to extract the Latitude all that remains to do is

$$Latitude = asin(D_z)$$