Given the equation of a straight line, how would I find a direction vector? I have a straight line equation as given below
$$x + y = a$$
where $a$ is some constant. I want to find a direction vector for this ilne. How would I find it?
 A: If we have a line $ax+by=c$ then the vector $(a,b)$ is perpendicular to the line and the vectors $(-b,a)$ and $(b,-a)$ are direction vectors of the line.
Why? If $(x_0,y_0)$ is a point of the line then its equation can be written as $$(a,b)\cdot (x-x_0,y-y_0)=a(x-x_0)+b(y-y_0)=0.$$ A point $(x,y)$ of the line is a solution of the above equation. And thus $(x-x_0,y-y_0)$ is a direction vector of the line (assuming different from zero). Thus $(a,b)$ is perpendicular to the line. So you only need a vector perpendicular to $(a,b).$ And $(-b,a)$ and $(b,-a)$ are two options.
A: The direction vector for any (non-vertical) line in $2$-space is just $(1,t)$ where $t$ is the slope. You know this because for every $1$ unit of $x$ you move $t$ units of $y$ which is the definition of the slope. If the line is vertical, a direction vector is just $(0,1)$.
In particular in your case you can see $y=-x+a$, so a direction vector is just $(1,-1)$.

Addendum:  To see this let the parametric equation be

$$\begin{cases}
x(t) = x_0 +at \\
y(t) = y_0 +bt
\end{cases}$$

Then $(a,b)\sim (1,b/a)$ are direction vector--multiplying by a non-zero scalar changes the magnitude, but keeps it pointing along the same line. The point slope form of the line is then

$$(y-y_0) = {b\over a}(x-x_0)$$

So this verifies the claim.
A: You have $ y=-x+a$ , so the vector equation of the line is:
$$
(x,y)^T=(x,-x+a)^T=(1,-1)^Tx+(0,a)^T \quad \forall x\in \mathbb{R}
$$
and the direction vector is $(1,-1)^T$
A: The equation of a line in two space can be written using vectors in an analogous way to planes in three space, i.e. for some normal vector to the line $(b,c)$ we have
$$
(b,c)\cdot(x,y)=0
$$
In your case you have $(b,c)=(1,1)$ and since it is an affine line, you must translate the line by $a$ to get the y intercept a, or insure the line passes through $(0,a)$ by taking 
$$
(1,1)\cdot (x,y-a)=0
$$ 
This allows you to read off the parametrized equation, since the slope of the line will be 
$$
\vec{r}(t)=(0,a)+t(-1,1),\;t\in \mathbb{R}
$$
