Consider the vectors W, X, Y and Z in R3, Where w = (3,2,6) and x = (4, 6, 8). We have to find WX(WX vector, not bold text is used instead of the arrow head), So, using the triangular law we take OX-OW to get: WX = (1, 4, 2). Taking this digrammatically:
So, How do we get a vector, say L, with it's head point at (1, 4, 2), which is exactly parallel to WX?
(Note that I have presented the diagrams in a sort of planar form, even though the vectors are in spaces).
On
When we are subtracting the coordinates of P from Q, we are taking the units along each axis, therefore, this would obviously be equal to the vector taken from the origin.
Another way to interpret is that, subtraction actually gives the slope, and thus the slope being equal makes the vector parallel.
I think you should consider all your vectors in this case as "difference" vectors ${\bf OW}=(3,2,6)-(0,0,0)$, ${\bf OX}=(4,6,8)-(0,0,0)$, so then ${\bf WX}=(4,6,8)-(3,2,6)$. The length and direction of ${\bf WX}$ is the same as for ${\bf OL}$, but ${\bf OL}$ is the "difference" from origin ${\bf OL}=(1,4,2)-(0,0,0)$. The only distinction between ${\bf WX}$ and ${\bf OL}$ is the origin, so in many cases the two will be interchangeable.