#### Answer

a. 628km/h, 5.82$^o$ east of south.
b. 16.5km

#### Work Step by Step

Call east the positive x direction and north the positive y direction.
a. Let P denote the Plane, A the Air, and G the Ground. The pair “PG”, for example, represents the plane’s motion relative to the ground.
$$ \vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}} $$
$$= (0, -688) km/h + (90.0 cos 45 ^{\circ}, 90.0 sin 45 ^{\circ}) km/h$$
$$= (63.6, -624) km/h$$
$$ \vec{v_{PG}} = (63.6, -624) km/h $$
Find the magnitude using the Pythagorean Theorem and the direction using the definition of the tangent.
$$ v_{PG} = 628 km/h$$
The angle is at -84.2$^o$, or 5.82$^o$ east of south.
b. The distance the plane is “off” is just the wind speed multiplied by the time.
$$(90 \frac{km}{h})(\frac{11}{60} hour) = 16.5 km$$