For example you have the cartesian equation:
(x-2)/-2 = y/3 = (z-1)/3
One possible choice for a is the vector <2, 0, 1>. How can you find a different possible vector of a. I understand that another one for v is any scalar multiple of it as the pvf is a span right?
Note: Sorry for the maths not being typed well this is my first post and i am unsure how to type it.
Because the line doesn't go through the origin, you can't just scale up the vectors.
Instead, notice that you can freely set any one of the variables, e.g. $x$, and that will restrict the other values.
For example, set $x = p$, then you get $\frac{p - 2}{-2} = \frac{y}{3} \implies y = 3(1 - \frac{p}{2})$. Similarly, you'll be able to get an expression for $z$ in terms of $p$, so the possible vectors are $a = \langle p, 3(1 - \frac{p}{2}), ???\rangle$ where $???$ is whatever you get for $z$. To choose a particular value for $a$, just set $p$ to an actual value like 1 or -1.