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

Beginner2022-09-04Added 9 answers

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 $(x-2y{)}^{2}+{z}^{2}=0$, but I'm not sure this is an improvement.

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 $(x-2y{)}^{2}+{z}^{2}=0$, but I'm not sure this is an improvement.

KesseTher12

Beginner2022-09-05Added 6 answers

Typically you would describe it parametrically:

$(x,y,z)=(t,2t,0),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$

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):

$(x,y,z)=({p}_{x},{p}_{y},{p}_{z})+({v}_{x}t,{v}_{y}t,{v}_{z}t),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$

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.

$(x,y,z)=(t,2t,0),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$

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):

$(x,y,z)=({p}_{x},{p}_{y},{p}_{z})+({v}_{x}t,{v}_{y}t,{v}_{z}t),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}$

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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