Find the parametric eq of the line through points $(1,3, -4)$ and $(3,2,1)$ Find the parametric eq of the line through points $(1,3, -4)$ and $(3,2,1)$.
Constructing a vector, we get, $[3-1, 2-3, 1+4] = [2, -1, 5]$ (point on line)
Let $r$ represent a point on the line $l$.
Then, 
$r = [2, -1, 5] + t[1,3,-4]$
or 
$r = [2, -1, 5] + t[3,2,1]$
Which can I choose as my parallel vector in this case?
 A: Actually, what you call "a point on the line" is your direction vector. So we rather have something of the form
$l=p+\begin{bmatrix} 2\\-1\\5 \end{bmatrix}t$, where $p$ is one of your points. To easily see this, just make a sketch!
A: You’ve gotten the direction vector and points on the line mixed up. The vector you computed, $[2,-1,5]$, isn’t a point on the line, it’s the direction vector for the line—what gets multiplied by $t$. So, your equation should be $r=P+t[2,-1,5]$, where $P$ is either of the two given points.  
However, you could just as well have chosen to subtract the two points in the opposite order, getting $[-2,1,-5]$ for the direction vector, so the parameterization would instead be $r=P+t[-2,1,-5]$. Then again, you might’ve normalized the direction vector and chosen to start from the midpoint of the two given points, or written the equation as $r=(1-t)[1,3,-4]+t[3,2,1]$, or ... My point is that there’s no such thing as the parametric equation of a line. There is in fact an infinite number of them for any single line.
A: *

*Strategy:
Given two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$, you can find a vector that specifies the line's direction, namely $\mathbf{d} = \left<x_1 - x_0, y_1 - y_0, z_1 - z_0 \right> = \left<a, b, c\right>$. Then, choose a point on the line (for which $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ are intuitive choices) -- call this point $P = (x', y', z')$ and construct the parametric equations as:
$\begin{cases}x(t) = x' + at \\ y(t) = y' + bt \\ z(t) = z' + ct  \end{cases}$
For some parameter $t$. 


*

*Direction Vector:
$\mathbf{d} = \left<2, -1, 5 \right>$


*

*Point on Line:
Since this line goes through $(1, 3, -4)$, choose that as your point.


*

*Construction:
Your equations are:
$\begin{cases}x(t) = 1 + 2t \\ y(t) = 3 - t \\ z(t) = 5t - 4  \end{cases}$
