Converting a 3D Line Equation: Help Needed for Homework

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Homework Help Overview

The discussion revolves around a line in 3D space represented by the equation x+y+2z-3=2x+3y+4z-4=0. Participants are exploring the nature of this equation, questioning whether it describes a line or a plane, and seeking to convert it into a standard form.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the given equation, with some suggesting it represents the intersection of two planes, while others explore the concept of standard forms for lines in three dimensions. There are inquiries about parameterizing the line and expressing variables in terms of a parameter.

Discussion Status

Several participants have provided insights into the nature of the equations, indicating that they describe two planes and thus a line of intersection. There is ongoing exploration of how to express the variables in parametric form, with some participants questioning the representation of y in terms of the parameter.

Contextual Notes

Participants are working under the assumption that the equations provided are intended to describe a line, despite initial confusion regarding their form. There is a focus on clarifying definitions and interpretations of standard forms for lines in three-dimensional space.

utkarshakash
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Homework Statement


I have been given a line in 3D space whose equation is x+y+2z-3=2x+3y+4z-4=0.
But what kind of equation is this? This is not a standard equation and I have never seen such an equation of line. It seems to me a plane instead of a line. Can anyone help me convert this into standard form?
 
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You have two equations between the three coordinates, i.e., you have one independent parameter. You can take, e.g., x as this independent parameter and solve for y and z. Then you get a parameter representation of a curve
\vec{r}(x)=\begin{pmatrix} x \\ y(x) \\ z(x) \end{pmatrix}.
Because the system of equations is linear this curve is a straight line.
 
utkarshakash said:

Homework Statement


I have been given a line in 3D space whose equation is x+y+2z-3=2x+3y+4z-4=0.
But what kind of equation is this? This is not a standard equation and I have never seen such an equation of line. It seems to me a plane instead of a line. Can anyone help me convert this into standard form?

x+y+2z-3=0 and 2x+3y+4z-4=0 are equations of two planes. All common points of both planes make the intersection of those planes, which is a straight line.

ehild
 
The question is what do you consider the "standard form" for a line? There are two forms usually used for lines in three dimensions:
The "symmetric form" Ax+ By+ Cz+ D= Ex+ Fy+ Gz+ H= Ix+ Jy+ Kz+ L which is the form you already have and, as ehild says, can be interpreted as defining the line as the intersection of two planes.

The "parametric form" where x= f(t), y= g(t), z= h(t) for some parameter t. This can be interpreted as an object moving along a line with t giving the time.
 
utkarshakash said:

Homework Statement


I have been given a line in 3D space whose equation is x+y+2z-3=2x+3y+4z-4=0.
But what kind of equation is this? This is not a standard equation and I have never seen such an equation of line. It seems to me a plane instead of a line. Can anyone help me convert this into standard form?

It looks like you are given a badly-written pair of linear equations, namely, x+y+2z-3 = 0 and 2x+3y+4z-4=0. As others have said, this describes two (non-parallel) planes in 3D, so taken together they describe a line. You can get a parametric form by solving for two of the variables in terms of the third; for example, solve for x and y in terms of z. Again, as others have said, I don't know what you regard as a "standard form".
 
Thanks to all of you!
 
When I assume x=t z can be expressed in terms of t but what about y? By solving I get y=-2. How should I express it in terms of t?
 
x+y+2z-3=2x+3y+4z-4=0.

Taking x= t, we then have t+ y+ 2z- 3= 0 and 2t+ 3y+ 4z- 4= 0 or
y+ 2z= 3- t and 3y+ 4z= 4- 2t.
Multiplying the first equation by 2, 2y+ 4z= 6- 2t. Subtracting that from the second equation,
y= (4- 2t)- (6- 2t)= -2 for all t. That is what you have. Excellent! That says that y is a contstant not depending on t. If you like, you can write it as y= -2+ 0t but just "y= -2" is correct.
 
utkarshakash said:
When I assume x=t z can be expressed in terms of t but what about y? By solving I get y=-2. How should I express it in terms of t?

What is wrong with the simple parametric equation y = -2 + 0*t?
 

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