# Applications of Vectors Quiz

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Are you aware of the concept of vectors in geometry? Take the applications of vectors quiz to see how much you know about the use of vectors in real life. Here, we have got some questions based on practical examples. Give this quiz a try and check your knowledge. All the best to get a perfect score on this quiz. If you find this quiz informative and helpful, do not forget to share the quiz results with others.

• 1.

### The three vertices of a parallelogram ABCD are A (1, 2, 3), B (-1, -2, -1), and C (2, 3, 2). The coordinates of the fourth vertex, D are

• A.

(4,7,6)

• B.

(-2,-3,-4)

• C.

(-2,-3,0)

• D.

(2,3,4)

A. (4,7,6)
Explanation
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).

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• 2.

### Calculate the direction angle (in degrees) for the vector <3, 12>.

• A.

14.1

• B.

75.9

• C.

-14.1

• D.

-75.9

B. 75.9
Explanation
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.

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• 3.

### What is the direction angle (in degrees) for a boat traveling S20°E?

• A.

50

• B.

100

• C.

290

• D.

350

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

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• 4.

### Find the magnitude of the vector <-5, 14>.

• A.

10

• B.

14.9

• C.

15.2

• D.

20.7

B. 14.9
Explanation
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.

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• 5.

### Find the magnitude of AB if the initial point is (-3, 1) and the terminal point is (3, 9).

• A.

8

• B.

8.9

• C.

9.3

• D.

10

D. 10
Explanation
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.

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• 6.

### Find the component form of AB with initial point A(0, 8) and terminal point B(-9, -3)

• A.

<-9, -11>

• B.

<-9, 11>

• C.

<9, -5>

• D.

<-9, 5>

A. <-9, -11>
Explanation
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 < -9, -11 >.

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• 7.

### A man walks 5 meters north, then 4 meters east, and finally 2 meters south. What is the magnitude of the resultant displacement of the man after the entire walk?

• A.

2 m

• B.

5 m

• C.

7 m

• D.

10 m

B. 5 m
Explanation
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.

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• 8.

### An airplane traveling north at 220 meters per second encounters a 50 meters-per-second crosswind from west to east.

• A.

214 m/s

• B.

170 m/s

• C.

222 m/s

• D.

226 m/s

D. 226 m/s
Explanation
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.

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• 9.

### A boat with a maximum speed of 3 m/s relative to the water is in a river that is flowing at 2 m/s. What is the minimum speed the boat can obtain relative to the shore?

• A.

1 m/s

• B.

3 m/s

• C.

4 m/s

• D.

Data is insufficient

A. 1 m/s
Explanation
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.

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• 10.

### A ship travels 100 km due West from the port and then changes course going 20° South of West for 140 km. Find the distance the ship is from its home port and the bearing of the ship from the port.

• A.

Option 1236.456 km at 168.317o bearing

• B.

236.456 km at 171.683bearing

• C.

236.456 km at 290o bearing

• D.

Data is insufficient

C. 236.456 km at 290o bearing
Explanation
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.

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• 11.

### A force of 7 N North and a force of 5 N West are acting on a body. Find the resultant sum of the two forces and determine their strength. Find the direction of the force.

• A.

8.602N at 54.962 north of west

• B.

8.602N at 54.962 west of north

• C.

8.602N at 144.962 south

• D.

None of the above

A. 8.602N at 54.962 north of west
Explanation
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.

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• 12.

### A pilot must fly due west at a constant speed of 300 mph against a wind of 50mph blowing in the direction 40° south of east. What direction and speed must the pilot maintain to keep on the course?

• A.

339.825 mph at bearing of 34.573o

• B.

339.825 mph at bearing of 74.573o

• C.

339.825 mph at bearing of 40o

• D.

339.825 mph at bearing of 80.56o