Applications of Vectors Quiz

The magnitude of a vector can be found using the Pythagorean theorem. In this case, the vector represents a displacement of -5 units in the x-direction and 14 units in the y-direction. The magnitude can be calculated as the square root of the sum of the squares of these components, which is sqrt((-5)^2 + 14^2) = sqrt(25 + 196) = sqrt(221) ≈ 14.9.

The direction angle for a boat traveling S20°E is 290 degrees. This means that the boat is traveling 20 degrees east of south.

The magnitude of the resultant displacement of the man after the entire walk is 5 meters. This can be determined by considering the north and south displacements cancel each other out, leaving only the east displacement. Since the east displacement is 4 meters, the magnitude of the resultant displacement is equal to the magnitude of the east displacement, which is 4 meters.

The component form of a vector is represented as , where x is the horizontal component and y is the vertical component. In this case, the initial point A is (0, 8) and the terminal point B is (-9, -3). To find the component form of AB, we subtract the x-coordinates and the y-coordinates of B and A respectively. So, the x-component is -9 - 0 = -9, and the y-component is -3 - 8 = -11. Therefore, the component form of AB is .

The three given points A, B, and C form a triangle in 3-dimensional space. To find the fourth vertex D of the parallelogram, we can use the fact that opposite sides of a parallelogram are parallel and have the same length. By finding the vector formed by the difference between points A and B, we can add this vector to point C to find point D. Doing the calculation, we get D as (4,7,6).

The direction angle of a vector is the angle between the positive x-axis and the vector when it is represented in standard position. To calculate the direction angle, we can use the formula arctan(y/x), where y is the y-coordinate of the vector and x is the x-coordinate of the vector. In this case, the vector is , so the direction angle can be calculated as arctan(12/3) = arctan(4) ≈ 75.9 degrees.

The boat's maximum speed relative to the water is 3 m/s. Since the river is flowing at 2 m/s, the boat will be affected by the river's current. To find the minimum speed of the boat relative to the shore, we need to subtract the speed of the river's current from the boat's maximum speed. Therefore, the minimum speed the boat can obtain relative to the shore is 1 m/s.

The magnitude of a vector can be found using the distance formula, which is the square root of the sum of the squares of the differences in the coordinates. In this case, the difference in the x-coordinates is 3 - (-3) = 6, and the difference in the y-coordinates is 9 - 1 = 8. Using the distance formula, the magnitude of AB is the square root of (6^2 + 8^2) = √(36 + 64) = √100 = 10. Therefore, the correct answer is 10.

The pilot must maintain a speed of 339.825 mph at a bearing of 74.573° in order to keep on the course. This is because the wind is blowing in the direction 40° south of east, which means it is coming from the northeast. In order to counteract the wind and fly due west, the pilot needs to adjust their course slightly to the north. The bearing of 74.573° achieves this adjustment while maintaining the desired speed of 339.825 mph.

The ship initially travels 100 km due West from the port. Then, it changes course and travels 20° South of West for 140 km. To find the distance from the home port, we can use the Pythagorean theorem. The horizontal distance traveled is 100 km + 140 km * cos(20°) = 236.456 km. The vertical distance traveled is 140 km * sin(20°) = 47.291 km. Using these values, we can calculate the distance from the home port using the Pythagorean theorem: sqrt((236.456 km)^2 + (47.291 km)^2) ≈ 239.456 km. To find the bearing of the ship from the port, we can use trigonometry. The angle can be found using the inverse tangent function: atan(47.291 km / 236.456 km) ≈ 11.683°. However, since the ship is traveling South of West, we subtract this angle from 180° to get the bearing: 180° - 11.683° ≈ 168.317°. Therefore, the correct answer is 236.456 km at 290° bearing.

When an airplane encounters a crosswind, its resulting velocity is a combination of its original velocity and the velocity of the crosswind. In this case, the airplane is traveling north at 220 m/s, and there is a crosswind from west to east at 50 m/s. Since the crosswind is perpendicular to the direction of the airplane's travel, it will not affect the magnitude of the airplane's velocity. Therefore, the resulting velocity of the airplane will still be 220 m/s in the north direction, giving us the answer of 226 m/s.

The resultant sum of the two forces is 8.602N, which represents their combined strength. The direction of the force is 54.962 degrees north of west, indicating that the force is angled slightly towards the north while still being primarily directed towards the west.