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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:

  1. $$B_{min} < a_1x + b_1y + c_1z + d_1 < B_{max}$$
  2. $$B_{min} < a_2x + b_2y + c_2z + d_2 < B_{max}$$

Instead of the normal way like as follows :

  1. $$a_1x + b_1y + c_1z + d_1 = 0$$
  2. $$a_2x + b_2y + c_2z + d_2 = 0$$

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.

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  • $\begingroup$ Each of these inequalities represents a 'sandwich' of 3D space. Their intersection will be some kind of parallelepipid.. What is the question by the way? Find the corners of the region, find its volume, ...? $\endgroup$
    – Paul
    Commented Jun 3, 2023 at 15:11
  • $\begingroup$ My task right now is to find (x, y, z) that satisfy both these equations. Or a region of (x, y, z) that satisfies the two equations. $\endgroup$
    – RaZ0rr
    Commented Jun 3, 2023 at 15:26

1 Answer 1

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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 $z$.

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  • $\begingroup$ Please let me know if this is correct. Since $d_1\ -B_{max}+e_1=0$ and $d_1\ -B_{max}+e_1=0$ the two plane equations have turned into $a_1x+b_1y+c_1z=0$ and $a_2x+b_2y+c_2z=0$. We solve for these equations, that is find the parametric form for the line formed from the intersection. Then use the bounds to restrict values on the line. Like this answer math.stackexchange.com/a/475983/1187705 , if we take $z$ variable as a free parameter, say $t$, then the first bound you have written $z<\frac{B_{max}-d_1}{c_1}$ restricts this free parameter $t$ since $z=t$ and then use same for $x$, $y$. $\endgroup$
    – RaZ0rr
    Commented Jun 4, 2023 at 9:34
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    $\begingroup$ @RaZ0rr, this is similar reasoning in parametric form. so it is correct. $\endgroup$
    – sirous
    Commented Jun 5, 2023 at 4:50
  • $\begingroup$ Thanks for the quick replies. This may be a silly question, But just to confirm once again what I said earlier and your reply to it, I would have to solve $(r_0 + rt) < \frac{B_{max} - d1}{c1} $ and $(r_0 + rt) < \frac{B_{max} - d2}{c2} $ for z, where the line of intersection of the planes is given by $(x, y, z) = (p_0, q_0, r_0) + t(p, q, r) = (p_0+pt, q_0+qt, r_0-rt) $ $\endgroup$
    – RaZ0rr
    Commented Jun 5, 2023 at 13:37
  • $\begingroup$ I also think there is something wrong with your initial assumption. Taking $e_1 = B_{max} - d_1$ seems incorrect. I belive it should be $e_1 = B_{max} - d_1 - Ax - By -Cz$. Otherwise $Ax + By + Cz + d_1 + e_1 - B_{max} = 0$ is wrong. Example:- Take $Ax = By + Cz = 3, d_1 = 5\ \mathrm{and}\ B_{max} = 10$ then $3+5 < 10$ but from your initial assumption, $3+5+5-10=0$ where $e_1 = B_{max}-d_1=10-5=5$ $\endgroup$
    – RaZ0rr
    Commented Jun 5, 2023 at 13:51
  • $\begingroup$ @RaZ0rr, You are rught. I corrected it. $\endgroup$
    – sirous
    Commented Jun 6, 2023 at 4:46

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