A direction vector can be determined through the difference of two vectors position.

Question

I am learning about vectorial equation, and i have a doubt about this statement:

A direction vector can be determined through the difference of two vectors position.

The direction vector that is determined is the original direction vector or the translated direction vector?

With 'traslated direction vector', i mean to: (in the two images the red segment is the direction vector)

And with 'original direction vector' i mean to:

And, what is the proof of this statement?

Any advice to better understand this topic is welcome, thanks in advance.

Answer 1

You're determining the original directional vector. For vector algebra only the direction and magnitude matter, not the position. This is because the components of a vector don't change if you translate it.

It helps to first look at this in one dimension.
On a real number line consider two points. Say $a=3$ and $b=5$.
We may think of these as vectors.

The quantity $b-a$ represents the distance between $a$ and $b$.
Also the vector $b-a$ points from $a$ towards $b$. If $b-a$ is positive then the vector points towards positive side, else it points towards negative side.

Similarly the vector $a-b$ points from $b$ towards $a$.

Next you can extrapolate this to $2D$.
For the vectors $a$ and $b$, the vector $b-a$ points from $a$ towards $b$ :

Adding vectors $a$ and $b-a$ gives the vector $b$.
So to get any vector between $a$ and $b$ we may scale $b-a$ and then add it to $a$ : $$a + t(b-a)$$ $0\le t\le 1$.

Answer 2

A line is parametrized by knowing some point on it (given by the vector $\vec x_0$ from the origin) and its direction vector $\vec v$. Then the vector from the origin to an arbitrary point on the line can be written as $$\vec x = \vec x_0 + t\vec v \quad\text{for some value of the scalar } t.$$

Note that if you know two points on the line, this corresponds to such expressions for two different values of $t$, and so you recover the direction vector (or a nonzero scalar multiple of it) by subtracting the two vectors: If $\vec x_1 = \vec x_0 + t\vec v$ and $\vec x_2 = \vec x_0 + s\vec v$ (where $s$ and $t$ are different scalars), then $$\vec x_2 - \vec x_1 = (\vec x_0 + s\vec v) - (\vec x_0 + t\vec v) = (s-t)\vec v,$$ as desired.