dripcima24

2022-09-03

If I want to define the line y=2x when discussing 3D lines this is actually a plane. We should be able to turn it into a line with cartesian equation x=2y=z/0 but that would give us something undefined. How do you resolve that problem?

Xavier Jennings

The general rule-of-thumb is that in n-dimensional space ${\mathbb{R}}^{n}$, every equation you add cuts the dimension by 1
In the plane, y=x gives a line. A system of 2 equations like x=y and x=3 determines a point.
In ${\mathbb{R}}^{3}$, y=x gives a plane. The system x=y and y=z defines a line, and something like x=y, y=z, and z=3 gives just a point.
In 7-dimensional space, y=x gives a 6-dimensional hyperplane, you need 6 equations to describe a line, etc.
All you need to do here is write $x=2y$ and z=0. I suppose with some elbow grease you could force this into one equation like $\left(x-2y{\right)}^{2}+{z}^{2}=0$, but I'm not sure this is an improvement.

KesseTher12

Typically you would describe it parametrically:
$\left(x,y,z\right)=\left(t,2t,0\right),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$
To elaborate on this, two pieces of information that describe a line are a starting point (which you can choose anywhere on the line) as well as a direction vector (which you can scale by any non-zero number):
$\left(x,y,z\right)=\left({p}_{x},{p}_{y},{p}_{z}\right)+\left({v}_{x}t,{v}_{y}t,{v}_{z}t\right),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$
Since this notation starts to get cumbersome, we can simplify it by writing
$\mathbf{x}=\mathbf{p}+\mathbf{v}t,\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$
where x represents (x,y,z) and similarly for p and v.

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