Answer
(b)
Work Step by Step
The only unit vector is the one given in (b). This is because in (b), we have the magnitude (length) of the vector:
$$\|F\|=\sqrt{\frac{y^2}{x^2+y^2}+\frac{x^2}{x^2+y^2}}=1.$$
But in (a) and (c), $\|F\|\neq 1$. We know that a unit vector must have a length of $1$, so only (b) is a unit vector.
