Intersecting Planes: Easiest Way to Make Up Equations

  • Thread starter Thread starter cscott
  • Start date Start date
  • Tags Tags
    Planes
AI Thread Summary
To create two equations of planes that intersect at a line, ensure their normal vectors are not scalar multiples of each other. This means that for planes represented as P1 = Ax + By + Cz + D and P2 = Wx + Xy + Yz + Z, the conditions [A, B, C] cannot equal k[W, X, Y] for any scalar k. The intersection line is defined by the simultaneous equations of the two planes. The discussion emphasizes understanding the representation of planes in three dimensions and ensuring they are not parallel to achieve the desired intersection. This approach is essential for solving problems related to intersecting planes in geometry.
cscott
Messages
778
Reaction score
1
What's the easiest way to make up two equations of a plane that intersect at a line?
 
Physics news on Phys.org
What are you looking for, the equation of the line of intersection? Because "two equations of a plane" doesn't really make sense to me...
 
TD said:
What are you looking for, the equation of the line of intersection? Because "two equations of a plane" doesn't really make sense to me...

I need two equations (Ax + By + Cz + D = 0) that describe two planes who's intersection point is a line.
 
Last edited:
Lines which are not parallel intersect with each other. so don't make them parallel. lines are parallel if their slopes are same. so think what have you to change or not to? A, B, C or D
 
vaishakh said:
Lines which are not parallel intersect with each other. so don't make them parallel. lines are parallel if their slopes are same. so think what have you to change or not to? A, B, C or D

I understand what you're saying but I'm not sure how the slope is represented in that form for 3-dimensions.
 
How do you generally write the equation of a plane?
 
cscott said:
I need two equations (Ax + By + Cz + D = 0) that describe two planes who's intersection point is a line.
I'm still not 100% sure what you mean but I *think* you mean that you're looking for the equations of two planes which, together as a system, form the equation of their intersection line (assuming the planes weren't parallel). Is this correct?
 
TD said:
I'm still not 100% sure what you mean but I *think* you mean that you're looking for the equations of two planes which, together as a system, form the equation of their intersection line (assuming the planes weren't parallel). Is this correct?

That is correct! How do I make sure they aren't parallel?
 
Scott: This is pretty easy!

The only thing u do is make sure both bormal vectors are not scalar multiples of each other... in other words

if P1 = Ax+By+Cz+D
and P2 = Wx + Xy +Yz +Z

then to intesect in a line... [A,B,C] canot equal k[W,X,Y] k is any num

Anton
 
  • #10
fomenkoa said:
Scott: This is pretty easy!
The only thing u do is make sure both bormal vectors are not scalar multiples of each other... in other words
if P1 = Ax+By+Cz+D
and P2 = Wx + Xy +Yz +Z
then to intesect in a line... [A,B,C] canot equal k[W,X,Y] k is any num
Anton

Alright, thanks.
 
  • #11
Indeed, so with two planes

\begin{array}{l}<br /> \alpha :ax + by + cz + d = 0 \\ <br /> \beta :a&#039;x + b&#039;y + c&#039;z + d&#039; = 0 \\ <br /> \end{array}

the line of intersection is given by

\left\{ \begin{array}{l}<br /> ax + by + cz + d = 0 \\ <br /> a&#039;x + b&#039;y + c&#039;z + d&#039; = 0 \\ <br /> \end{array} \right

under the condition that

\alpha \not\parallel \beta \Leftrightarrow \left( {a,b,c} \right) \ne k\left( {a&#039;,b&#039;,c&#039;} \right)\forall k \in \mathbb{Z}
 
Back
Top