# Calculus II Vectors Magnitude and Direction Problem

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 38.6 degrees for the direction and 194.4 for the magnitude.

The way that I worked it:

Resultant:

$$= (-30\sin45 + 200\sin60, 30\sin45 + 200\cos60)$$

$$= (151.9918773, 121.2132034)$$

Direction:

$$= tan^{-1}(\frac{121.2132034}{151.9918773})$$

Magnitude:

$$= \sqrt{121.2132034^2 + 151.9918773^2}$$

Any help would be great.

Thanks!

EDIT: I discovered that the solution to this problem is N67.9$^\circ$E and 209.8 km/h

• Just added my work! Jan 20, 2018 at 19:59
• Is this better? Jan 20, 2018 at 20:08
• when the wind blows from the north-west direction, its vector points towards the south-east direction. Jan 20, 2018 at 20:13

$$\vec{v} = \begin{bmatrix} 30\cos(45°) \\ -30\sin(45°)\end{bmatrix} + \begin{bmatrix} 200\cos(30°) \\200\sin(30°)\end{bmatrix} = \begin{bmatrix} 194,42 \\ 78,79 \end{bmatrix}$$
$||\vec{v}||$ = 209,8 m/s under an angle of 22.05 degrees
• Hey! I finally figured this out! You were correct about this, but they wanted the true course of the plane. That is basically 90$^\circ$ - 22.1$^\circ$ to get an answer of N67$^\circ$E Jan 20, 2018 at 21:09