Intersecting Planes: Easiest Way to Make Up Equations

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What's the easiest way to make up two equations of a plane that intersect at a line?
 
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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.
 
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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.
 
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
 
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.
 
Indeed, so with two planes

[tex]\begin{array}{l}<br /> \alpha :ax + by + cz + d = 0 \\ <br /> \beta :a'x + b'y + c'z + d' = 0 \\ <br /> \end{array}[/tex]

the line of intersection is given by

[tex]\left\{ \begin{array}{l}<br /> ax + by + cz + d = 0 \\ <br /> a'x + b'y + c'z + d' = 0 \\ <br /> \end{array} \right[/tex]

under the condition that

[tex]\alpha \not\parallel \beta \Leftrightarrow \left( {a,b,c} \right) \ne k\left( {a',b',c'} \right)\forall k \in \mathbb{Z}[/tex]