The discussion focuses on writing a vector equation for a line that passes through the point P(–1, 0, 3) and is parallel to the y-axis. The correct vector equation is expressed as (x,y,z)=(-1,0,3)+t(0,1,0), where u ⃗=(0,1,0) represents the direction vector parallel to the y-axis. Participants confirm that the y-value can take any value, including 0, depending on the parameter t, which allows for reparametrization of the line. The conversation emphasizes the importance of using the support point P and a suitable direction vector to define the line accurately.
PREREQUISITESStudents in mathematics or physics, educators teaching vector calculus, and anyone interested in understanding the geometric representation of lines in three-dimensional space.
Do you mean the y-component of the tangent vector? In that case yes. It would only be a reparametrisation of the line.Physics345 said:basically in this case y can equal anything except for 0 for it to be parallel to the y-axis?
Orodruin said:Do you mean the y-component of the tangent vector? In that case yes. It would only be a reparametrisation of the line.
As for the y-value of the line, it can take any value - including 0 - depending on the value of the curve parameter.
It is exactly what he did.Tonyb24 said:if it passes through that point, then why not use that point as the support point?
Again, it is exactly what he did.Tonyb24 said:Then, what is a point on the Y axis that you can multiply to reach any y-axis point?
Which is exactly what he got:Tonyb24 said:So now, use your support point P and add to it a vector that can be multiplied by t where t E R, that will allow you to reach anypoint on the line (like a unit vector or something). L:(x,y,z)=P+t(x,y,z);t E R.
Physics345 said:(x,y,z)=(-1,0,3)+t(0,1,0)