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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?

YuiTo Cheng
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Khaled
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3 Answers3

1

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

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

Cesareo
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0

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).

Doug M
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  • 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 – Khaled 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 00:15
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$(x_p, y_p, z_p)$ 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.

user247327
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