Answer
The length of the displacement vector is 0.283 mi and the direction is $45^{\circ}$ north of west (which is directly northwest).
Work Step by Step
The west component of the displacement is 0.500 mi - 0.300 mi = 0.200 mi west.
The north component of the displacement is 0.200 mi north.
We can find the length of the displacement:
$\sqrt{(0.200~mi)^2+(0.200~mi)^2} = 0.283~mi$
We can find the direction north of west:
$tan~\theta = \frac{0.200~mi}{0.200~mi}$
$\theta = tan^{-1}(\frac{0.200~mi}{0.200~mi})$
$\theta = tan^{-1} (1)$
$\theta = 45^{\circ}$
