Motivation: A Line in $\Bbb R^2$
Any line can be uniquely determined by two points. In $\Bbb R^2$, a point is uniquely determined by two values (its $x$ and $y$ coordinates). Hence, to uniquely determine a line in $\Bbb R^2$ requires at most four values (in this case $x_0$, $y_0$, $x_1$, and $y_1$).
Those aren’t the only four values that can be used; we could also express the line parametrically, as the pair of equations:
$$x=x_0+at\\y=y_0+bt$$
Now the four values are $x_0$, $y_0$, $a$, and $b$. We can rearrange that pair of equations to get the symmetric-form equation of the line, which also determines the line by the same four values:
$$\frac{x-x_0}{a}=\frac{y-y_0}{b}$$
No doubt most of you see where I’m headed, though. We can rearrange this to get the good old slope–intercept form of the line’s equation:
$$y=mx+c$$
This uniquely determines the line with just two values ($m$ and $c$)!
Some lines (namely those parallel to the axes) can be determined with just one value, but two is the minimum information necessary to generally determine a line in $\Bbb R^2$—not four, after all.
The Problem: A Line in $\Bbb R^3$
A line in $\Bbb R^3$ is still able to be determined by two points, although each point now needs three values ($x$, $y$, and $z$). Hence, to uniquely determine a line in $\Bbb R^3$ requires at most six values.
These six values can be $x_0$, $y_0$, $z_0$, $x_1$, $y_1$, and $z_1$; or in parametric form…
$$x=x_0+at\\y=y_0+bt\\z=z_0+ct$$
…or symmetric form…
$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$
…they could be $x_0$, $y_0$, $z_0$, $a$, $b$, and $c$. But no matter where I’ve looked or what I’ve tried, I can’t do it in less than six. Can it be done?
Worth noting is that a plane in $\Bbb R^3$ can be determined in fewer than the naive nine values (three points): using one point and a normal vector (three values each), we can do it in six. It seems counterintuitive that a plane and a line should both be determined by six values… but then, the same could be said about a line and a point in $\Bbb R^2$.