A plane is flying to Minnesota with a velocity of 373.52 km/h, N. The plane encounters a crosswind with a velocity of 64.90 km/h, W. What is the magnitude of the resultant velocity of the plane?

Answer:   379.12 km/h N 9.86° WExplanation:You want to know the resultant velocity when a plane's velocity of 373.52 km/h N is modified by a crosswind at 64.90 km/h W.Vector sumIn (N, E) coordinates, the sum of the velocities is ...   (373.52, 0) +(0, -64.90) = (373.52, -64.90)The magnitude of this vector is ...   |v| = √(373.52² +64.90²) ≈ 379.12 . . . . km/hThe direction will be west of north by the angle ...   θ = arctan(-64.90/373.52) ≈ -9.86° . . . . . . N = 0°, + = clockwiseThe resultant velocity is 379.12 km/h N 9.86° W.