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Intersection of planes in R3

  • Thread starter maccha
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  • #1
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thought I understood equations of planes in R3 and their intersections, but apparently not. I'm very confused by what seems to be a basic problem:

find a vector equation for the line of intersection of x + y + z= 0 and x + z = 0.

Is x + z= 0 still a plane even though it doesn't have the form Ax + By + Cz = D?

I notice that if you set the two equations equal to each other you find that y = 1. Does this mean that the planes intersect on a line where y =1 and all x coordinates are equal to negative z coordinates? Thanks!
 

Answers and Replies

  • #2
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The equation [tex]x + z = 0[/tex] does in fact have the form [tex]Ax + By + Cz = D[/tex]; what is [tex]B[/tex] here?

Also, how did you combine [tex]x + y + z = 0[/tex] with [tex]x + z = 0[/tex] to conclude that [tex]y = 1[/tex]? If you suppose that [tex]x + z = 0[/tex] and [tex]y = 1[/tex], what is [tex]x + y + z[/tex]?
 
  • #3
557
1
thought I understood equations of planes in R3 and their intersections, but apparently not. I'm very confused by what seems to be a basic problem:

find a vector equation for the line of intersection of x + y + z= 0 and x + z = 0.

Is x + z= 0 still a plane even though it doesn't have the form Ax + By + Cz = D?

I notice that if you set the two equations equal to each other you find that y = 1. Does this mean that the planes intersect on a line where y =1 and all x coordinates are equal to negative z coordinates? Thanks!
As a hint you have to use vectors and vector products.
 
  • #4
HallsofIvy
Science Advisor
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No, you don't have to use vectors and vector products. (Except that the problem specifically asked for a vector as solution. If it had not you could write the line as parametric equations.) But I see no reason for "vector products".
 

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