Suppose you have the following problem: Find the line segment from $(1,0,3)$ to $(0,2,1)$. Let $A = (1,0,3)$ and $B = (0,2,1)$.
The correct way to this problem is to solve this problem is to use $\overrightarrow{r}(t) = \overrightarrow{r_0} + t\overrightarrow{v}$ with the initial position vector $\overrightarrow{r_0} = <0,2,1>$ and $ \overrightarrow{v} =\overrightarrow{AB} = <-1,2,-2>$, since it is the direction vector, making $$\overrightarrow{r}(t) = <-1,2,-2> + \space t<-1,2,-2>$$
Wouldn't it also be equally accurate to define the direction vector $\overrightarrow{v}$ as the unit vector of $\overrightarrow{AB}$, since we are only concerned with the direction of $\overrightarrow{v}$, not so much its magnitude, making
$$\overrightarrow{r}(t) = <-1,2,-2> + \space t<\frac{-1}{3},\frac{2}{3},\frac{-2}{3}>$$ Any insight into this would help. Thanks.
OK, let's make it an answer. Yes, you can use any positive multiple of $\langle−1,2,−2\rangle$ but in order to get the segment from $A$ to $B$, you also need to specify the domain of $t$, and with $\langle-1,2,-2\rangle$ the domain is $[0,1]$ whereas with the unit vector you've given it is $[0,3]$ (nothing wrong with the latter answer but you get $[0,1]$ automatically by using the vector from $A$ to $B$).