Determine the magnitude and direction of the following vector (starting at the original) express the direction in both polar form and as a bearing. OA=(6,3)

Question

Determine the magnitude and direction of the following vector (starting at the original) express the direction in both polar form and as a bearing. OA=(6,3)

Answer 1

Recall the magnitude of a vector is given by |(a,b)| = √(a² + b²) and the direction is simply the vector divided by its length. So |OA| = √(6² + 3²) = √45 = 3√5 and its direction is (2,1)/√5 .

Now, polar form is a = r e^iθ where r is the distance from the origin (which we already know) and θ is the angle from the horizontal. θ can be found using simple trig, if you plot the vector and drop a perpendicular line down to the x axis we have a right triangle with θ being the angle closest to the origin. With trig we find tanθ = opp/adj = 3/6 =1/2 so θ = tan^-1(1/2) = 26.6 deg. Then OA = 3√4 e^{i (26.6 deg)}.

To find the bearing we simply need to figure out the angle the vector makes with respect to North. To get this we can simply subtract the angle N makes with the x axis, 90 deg, from the angle our vector makes with the x axis, 26.6 deg. So OA has a bearing or 90 - 26.6 deg or 63.4 deg.

Answer 2

Draw and label a diagram.

arctan 0.5 =- 26.565

polar form (√45, 26.565°)

Bearing is 90° - 26.565°