Calculus II Magnitude and Direction of Vectors

Question

I have worked this problem at least 8 different times and keep getting the same answer every single time. Could somebody please explain how to work this problem?

Here is the problem:

The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 30 km/h. (This means that the direction from which the wind blows is 45° west of the northerly direction.) A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 200 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. Give your answers correct to one decimal place.

I keep getting 22.1 degrees for the direction and 209.8 for the magnitude.

 

Work:

 

v⃗ =[30cos(45°)−30sin(45°)]+[200cos(30°)200sin(30°)]=[194,4278,79]

||v⃗ ||||v→|| = 209,8 m/s under an angle of 22.05 degrees

Answer 1

Using i and j for E and N respectively,

 

Wind velocity:

 

w = 30cos45ºi - 30sin45ºj = 21.21i - 21.21j

 

Airplane in still air:

 

u = 200sin60ºi + 200cos60ºj = 173.21i + 100j

 

Resultant:

 

v = w + u = (21.21 + 173.21)i + (-21.21 + 100)j = 194.42i + 78.79j

 

v = √(194.42² + 78.79²) = 209.8 km/h

 

Direction (north of east):

 

tan^-1(v_y/v_x) = tan^-1(78.79/194.42) = 22.1°

 

I get the same answer as you.