The solution I came up with:
Let $v = (a, b)$ be orthogonal to $(−3, 1)$. Note $\langle(a, b),(−3, 1)\rangle = 0$ or $−3a+b> = 0$; i.e. $b = 3a$. All unit vectors orthogonal to $(−3, 1)$ are of the form $||v||^{-1}(a , 3a )$.
The solution manual does mostly the same thing, but doesn't multiply a with $||v||^{-1}$. Instead it does:
Choosing $a = 1$ and normalizing gives $(1\sqrt{10},3\sqrt{10})$. Hence there are $2$ unit vectors orthogonal to $(−3, > 1)$ ; namely, $(1\sqrt{10}, 3\sqrt{10})$ and $(−1\sqrt{10}, −3\sqrt{10})$.
I don't understand why are there only two unit vectors? Couldn't we have chosen $a = 2, 3$, etc and produced more?
All the vectors orthogonal to $(-3, 1)$ are of the form $t(1,3)$ i.e. $(t,3t)$. However you need unit vectors, so their length should be $1$, which means that $t^2 + (3t)^2 = 1$. This is where $\pm\frac{1}{\sqrt{10}}$ comes from.