I have read this question to find the parametric form for the line formed by the intersection of two planes.
But I have two equations that have bounds instead of 0 on the right hand side:
- $$B_{min} < a_1x + b_1y + c_1z + d_1 < B_{max}$$
- $$B_{min} < a_2x + b_2y + c_2z + d_2 < B_{max}$$
Instead of the normal way like as follows :
- $$a_1x + b_1y + c_1z + d_1 = 0$$
- $$a_2x + b_2y + c_2z + d_2 = 0$$
where $a_1, a_2, b_1, b_2, c_1, c_2, d_1, d_2, B_{min}\ and\ B_{max}$ are known constant coefficients.
Should I use the same procedure for finding the line for the intersection of the planes and then apply the bounds on the line segment? But I don't know how I can use scalars $B_{min} \ and\ B_{max}$ as bounds on a 3D line.
And I assume I would use the answer for this question as the parametric equation for a line segment in 3D with the two endpoints of the line segment being somehow related to $B_{min}$ and $B_{max}$. But I cannot see a way to formulate everything together.
The gradients of the line does not depend on D, $B_{max}$ and $B_{min}$ . we may write:
$\begin{cases}a_1x+b_1y+c_1z+ d_1-B_{max}+e_1=0\\a_2x+b_2y+c_2z+ d_2-B_{max}+e_2=0\end {cases}$
$l=\begin{vmatrix}b_1&c_1\\b_2&c_2\end{vmatrix}$
$l=\begin{vmatrix}c_1&a_1\\c_2&a_2\end{vmatrix}$
$l=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$
An equationof line is:
$\frac{x-x_0}l=\frac{y-y_0}m=\frac {z-z_0}n$
The bound defines the restrictions for $x$, $y$ and $z$.
Suppose we have plane $Ax+By+Cz-D=0$,
1- If $x=y=0$ then $z=\frac DC$; in our case, for first equation; $z=\frac {B_{max}-d_1=e_1}{c_1}$ , or $z<\frac {B_{max}-d_1}{c_1}$
If $x=y=0$ then $z=\frac DC$; in our case for second equation $z=\frac {B_{max}-d_2=e_2}{c_2}$ , or $z<\frac {B_{max}-d_2}{c_2}$
These are restrictions for $z$ in upper bound. Similarly you can find restrictions for upper and lower bound for $x$, $y$ and $z$.