Two vectors represent displacements of 60 km north and 80 km east. What is the resultant displacement's magnitude and bearing?

• A. 100 km bearing 123°
• B. 100 km bearing 090°
• C. 100 km bearing 053°
• D. 100 km bearing 270°

Answer: (C)

The main idea is combining perpendicular displacements using a right triangle. The north component is 60 km and the east component is 80 km. The resultant’s magnitude comes from the Pythagorean theorem: sqrt(60^2 + 80^2) = sqrt(3600 + 6400) = 100 km.

For the direction, think of the resultant as a vector that is 60 km north and 80 km east. The angle it makes with the north direction toward the east is arctan(opposite/adjacent) = arctan(80/60) = arctan(4/3) ≈ 53.13 degrees. Bearings are measured clockwise from north, so the bearing is 053°. This places the resultant in the northeast quadrant, which matches having both north and east components.

So the displacement is 100 km at a bearing of 053°.