Defining a line I've read this: 
To describe a line, we needed a point $b$ and a vector $v$ along the line. We could also start with two points $b$ and $a$ and take $v=a−b$.
In the first sentence, why do we need a point $b$ and a vector $v$ along the line in order to define it? Isn't a vector $v$ along it just enough? I see we don't need point $b$ with it . Am I right?
 A: You're correct that each vector $v \in \mathbb{R}^3$ defines a line $$\{av : a \in \mathbb{R}\}.$$ But such lines always go through the origin. Thankfully, we can generalize. If we have a point $p$ and a vector $v$, there's a corresponding line $$\{p+av : a \in \mathbb{R}\},$$ and such lines needn't go through the origin. Indeed, every line can be described in the latter form.
A: A regular old vector simply just says "go right by x, up by y, and forwards by z". It's kind of like the gradient in y = mx + c, and is not to be confused with a position vector, which says "go from the origin to this point in a straight line".
We often combine a vector with a variable coefficient, such as $ t(1, 3, 5) $ in order to make it extend infinitely in both directions. However, it still has no set point. It's like a ruler , with a physically fixed angle of course, that you can place anywhere on a piece of paper to draw a line.
However, it has no definite equation until we introduce a point to create a vector equation of a line, where we say "it goes through this point, with this gradient". The same goes for y = mx + c when you have both a co-ordinate and the gradient.
One thing to note is that many fixed positions will create the same line, since the line extends infinitely, including negatively. If our equation is: $ r $1$ = (1, 2, 3) + t(2, 5, 10)$, this is equal to $ r $1$ = (3, 7, 13) + t(2, 5, 10)$. If our variable coefficient was to have a value, the specific fixed position would matter.
