How do I define a 3D line that does not leave the horizontal plane? 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?
 A: Typically you would describe it parametrically:
$$(x,y,z) = (t, 2t, 0), \quad t \in \Bbb R$$
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):
$$(x,y,z) = (p_x, p_y, p_z) + (v_xt, v_yt, v_zt), \quad t \in \Bbb R$$
Since this notation starts to get cumbersome, we can simplify it by writing
$$\mathbf{x} = \mathbf{p} + \mathbf{v}t, \quad t \in \Bbb R$$
where $\mathbf{x}$ represents $(x,y,z)$ and similarly for $\mathbf{p}$ and $\mathbf{v}$.
A: The general rule-of-thumb is that in $n$-dimensional space $\Bbb R^n$, 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 $\Bbb R^3$, $y=x$ 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.
