# 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?

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

$$\left\{ \begin{array}{rcl} x_P&=&\frac{d_2 v_1-d_1 v_2+(v_2 w_1 - w_2v_1)\lambda}{u_2 v_1-u_1 v_2} \\ y_P&=&\frac{d_2 u_1-d_2 u_2+(u_2 w_1- w_2u_1)\lambda}{u_1 v_2-u_2v_1} \\ z_P & = & \lambda \end{array} \right.$$

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).

• I tried to put the parametric equations of $L$ into the equation of $E: (x/a)^2 + (y/a)^2 + (z/b)^2 =1$ then solving for $t$, I get $t_{1}$ and $t_{2}$ the put them in the parametric equations of $L$, they should give the coordinates of the two points, but it fails when the line is parallel to $xoy$ plane Sep 21 '18 at 22:58
• @Khaled There’s no particular reason why your method should fail for that case. Please show your work so that someone has a chance of pointing out your error.
– amd
Sep 21 '18 at 23:59
• @Khaled You’ve run into problems with lines parallel to the $x$-$y$ plane in previous question. Whatever you’re doing wrong this time is probably similar to the past errors.
– amd
Sep 22 '18 at 0:15

$$(x_p, y_p, z_p)$$ 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.