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.
A vector parametric equation of a line expresses the line in terms of a vector equation that incorporates a parameter, typically denoted as \( t \). It is generally formulated as \( \mathbf{r}(t) = \mathbf{r_0} + t \mathbf{d} \), where \( \mathbf{r_0} \) is a position vector to a point on the line, \( \mathbf{d} \) is a direction vector indicating the line's direction, and \( t \) is a scalar parameter that varies along the line.
To derive the vector parametric equation of a line from two points \( \mathbf{P_1} \) and \( \mathbf{P_2} \), first find the direction vector \( \mathbf{d} = \mathbf{P_2} - \mathbf{P_1} \). Then, the vector parametric equation can be expressed as \( \mathbf{r}(t) = \mathbf{P_1} + t \mathbf{d} \), where \( t \) varies over all real numbers.
The components of a vector parametric equation of a line consist of a position vector \( \mathbf{r_0} \) and a direction vector \( \mathbf{d} \). The position vector \( \mathbf{r_0} \) represents a specific point on the line, while the direction vector \( \mathbf{d} \) indicates the line's direction. The parameter \( t \) is used to scale the direction vector, allowing the equation to represent all points along the line.
To convert a vector parametric equation of the form \( \mathbf{r}(t) = \mathbf{r_0} + t \mathbf{d} \) into a symmetric equation, you can express the individual components of the position vector in terms of the parameter \( t \). If \( \mathbf{d} = (d_x, d_y, d_z) \) and \( \mathbf{r_0} = (x_0, y_0, z_0) \), then the symmetric equations are given by \( \frac{x - x_0}{d_x} = \frac{y - y_0}{d_y} = \frac{z - z_0}{d_z} \), assuming \( d_x, d_y, d_z \) are non-zero.