Line intersecting spheroid I have two planes $(A): u_{1}x + v_{1}y + w_{1}z = d_{1}$ and $(B): u_{2}x + v_{2}y + w_{2}z = d_{2}$.
They intersect together, then they yield a line $(L)$ that has a direction vector $M (x_{M},y_{M},z_{M})$
$M$ is the cross product of the normal vectors of $A$ and $B$
$M = (u_{1},v_{1},w_{1})×(u_{2},v_{2},w_{2})$
The parametric equations of the line $L$ of a parameter $t$  are:
$x = x_{P} + x_{M}.t$
$y = y_{P} + y_{M}.t$
$z = t$
I intersect this line $L$ with an spheroid $E$ of semi-axis major $a$ in the equatorial plane $xoy-Plane$  along $x-axis$ and $y-axis$ and semi-axis minor $b$ along $z-axis$
My problem is how to find the coordinates of the two points of intersection?
 A: We have
$$
\Pi_1\to u_1 x+ v_1 y + w_1 z = d_1\\
\Pi_2\to u_2 x+v_2 y+ w_2 = d_2\\
L\to p = p_0 + \lambda \vec v 
$$
here
$$
p = (x,y,z)\\
\vec n_1 = (u_1, v_1, w_1)\\
\vec n_2 = (u_2, v_2, n_2)\\
\vec v = \vec n_1\times \vec n_2\\
p_0 = (x_P,y_P,z_P)
$$
then if $L \in \Pi_1 \cap \Pi_2$ follows
$$
u_1 x_P+ v_1 y_P + w_1 z_P = d_1\\
u_2 x_P+ v_2 y_P + w_2 z_P = d_2\\
$$
hence any $p_0$ obeying the two linear conditions above is a feasible $p_0$ as for instance
$$
\left\{
\begin{array}{rcl}
 x_P&=&\frac{d_2 v_1-d_1 v_2+(v_2 w_1 - w_2v_1)\lambda}{u_2 v_1-u_1 v_2} \\
 y_P&=&\frac{d_2 u_1-d_2 u_2+(u_2 w_1- w_2u_1)\lambda}{u_1 v_2-u_2v_1} \\
z_P & = & \lambda
\end{array}
\right.
$$
A: You need to restric your freedom, and things will get easier.
Introduce a 3rd plane, almost any will do as long as it its normal is independent from the other two normals.
Then this gives you a system that will intersect at a point, and that point will be on your line.
Probably the simplest $x = 0$
Then solve
$v_1y + w_1z = d_1\\v_2y + w_2z = d_2$
$(0, \frac {w_2d_1 - w_1d_2}{v_1w_2 - w_1v_2},\frac {v_1d_1 - v_2d_2}{v_1w_2 - w_1v_2})$ is on your line (unelss $v_1w_2 - w_1v_2= 0$, in which case you need a different plane).
A: $(x_p, y_p, z_p)$ is any point on the line of intersection. 
For example, suppose the two planes are x+ 2y+ z= 1 and 2x- y+ z= 3.  Subtracting the first equation from the second gives x- 3y= 2 so x= 3y+ 2.  Taking, arbitrarily, y= 0, x= 2.  Then both 2+ 0+ z= 1 and 4- 0+ z= 3 give z= -1. (2, 0, -1) is one of the infinitely many points on the line of intersection. 
