The discussion revolves around understanding the nature of the equations x + y = 1 and x + 2y + z = 3, specifically whether they represent lines or planes in three-dimensional space. Participants are tasked with writing a vector parametric equation for the line defined by these equations.
There is an ongoing exploration of the definitions and implications of the equations. Some participants have clarified their understanding of the problem, while others continue to question the initial assumptions about the nature of the equations.
Participants note that the problem statement specifies a line as the intersection of two planes, which has led to some misinterpretation of the equations as lines. There is acknowledgment of the need to reconsider the wording of the question.
Neither is a line. They are both planes in ##\mathbb R^3##.songoku said:Homework Statement: Let L be the line given by the equations x + y = 1 and x + 2y + z = 3. Write a vector parametric equation for L
Relevant Equations: Parametric equation:
##x=x_0 + ta##
##y=y_0+tb##
##z=z_0+tc##
I can imagine x + y = 1 to be line in xy - plane but how can x + 2y + z = 3 be a line, not a plane?
Thanks
Ok so it means the question is wrong to call those two equations as lines. Basically the question gives two equations of plane and asks for intersection of the two planes.Orodruin said:Neither is a line. They are both planes in ##\mathbb R^3##.
Two planes that are not parallel intersect each other in a line.
The question does not call those equations lines. The question specifies one line as the set of points satisfying both equations.songoku said:Ok so it means the question is wrong to call those two equations as lines.
Huh? I never claimed anything else. In fact, it is just a repetition of what I just said …FactChecker said:The points that satisfy both equations is a line. The problem statement is correct.
Sorry. Somehow I quoted the wrong post and didn't notice that I was responding to the wrong text. I must need more coffee. I'm going to fix that post.Orodruin said:Huh? I never claimed anything else. In fact, it is just a repetition of what I just said …
If you have two points on a line, then you can take one as the starting point, and the difference of two as the direction to travel along by the parameter. That gives you a natural parameterization.songoku said:Homework Statement: Let L be the line given by the equations x + y = 1 and x + 2y + z = 3. Write a vector parametric equation for L
Relevant Equations: Parametric equation:
##x=x_0 + ta##
##y=y_0+tb##
##z=z_0+tc##
I can imagine x + y = 1 to be line in xy - plane but how can x + 2y + z = 3 be a line, not a plane?
Thanks
It seemed I interpreted it too literally. "Let L be the line given by the equations x + y = 1 and x + 2y + z = 3" in my interpretation meant the given equation are lines, which is actually not what it means by the question.Orodruin said:The question does not call those equations lines. The question specifies one line as the set of points satisfying both equations.
Yes I canFactChecker said:The points that satisfy both equations is a line. The problem statement is correct.
Suppose you set ##x=t,\ \ t \in \mathbb R##.
Then from the first equation, you have ##y=1-x = 1-t,\ \ t \in \mathbb R##.
Can you use the first equation to convert the second equation into an equation for ##z##?
Which is actually not what it means at all. The question talks about the line given by two expressions. You have simply misread the question.songoku said:which is actually not what it means by the question.