#### Answer

(a) $d = (-62\hat{i}-65\hat{j})~mi$
(b) The magnitude of the displacement vector is 90 miles and it is directed at an angle of $46.4^{\circ}$ south of west.

#### Work Step by Step

We can find a displacement vector from the origin to Ward's home as;
$W = (100\hat{i}+50\hat{j})~mi$
We can find Ruth's displacement vector from the origin as;
$R = (38\hat{i}-15\hat{j})~mi$
(a) We can find the displacement vector $d$ that Ward needs to fly.
$W+d = R$
$d = R-W$
$d = (38\hat{i}-15\hat{j})~mi-(100\hat{i}+50\hat{j})~mi$
$d = (-62\hat{i}-65\hat{j})~mi$
(b) We then find the magnitude of $d$:
$d = \sqrt{(-62~mi)^2+(-65~mi)^2}$
$d = 90~mi$
We can find the angle $\theta$ south of west;
$tan(\theta) = \frac{65}{62}$
$\theta = arctan(\frac{65}{62})$
$\theta = 46.4^{\circ}$
The magnitude of the displacement vector is 90 miles and it is directed at an angle of $46.4^{\circ}$ south of west.