Why is it that in the parametrization of a line, the point $P_0$ is indicated with brackets and not parenthesis? For example if I say I want the line perpendicular to the yz plane and through point $P = (4,9,8)$, then I have  $\langle4,9,8\rangle + t\langle1,0,0\rangle ={\langle4+t,9,8\rangle}$ 
Why when writing the equation, I use brackets around point P? I understand if P were a vector from the origin, but the line is just saying that it passes through point P. I also understand that the parmetrization equation is $r(t) = \vec{OP} + t\vec{v}$, but since I just want the line, then that means I shouldn't need $\vec{OP}$. It makes more sense to me if P were indicated using parenthesis (4,9,8). 
 A: The convention is that $(a, b, c)$ is a point, and $\langle a, b, c\rangle$ is a vector (from the origin).
You can consider $P$ to be either a point or a vector from the origin -- they will give you the same line. The difference is whether you imagine
1) tracing out the line by starting at $P$ and traveling left and right in units of $\langle 1, 0, 0\rangle$.
2) tracing out the line with the tip of a vector (from the origin) which is the sum of the vectors $P$ and $t\langle 1, 0, 0 \rangle$.
I prefer $1)$, personally. Note that it treats points as different objects from vectors, obeying the law "point + vector = point". But the "standard" way of teaching multivariable that I've seen is to treat everything as a vector (from the origin) and forget about points.
You can think about it either way, and probably no one will take offense.
(I'm a bit confused by your second paragraph -- you say "we shouldn't need $\vec{OP}$". But certainly the parameterization equation cannot be $r(t) = t\vec{v}$; that would give us an entirely different line, which passes through the origin instead of passing through $P$).
