Understanding Vector Displacement

Oct 13, 2025 by magicalceylontours.com 20 views

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Introduction

In this problem, we are given two points in a 2D coordinate system, $A (6,7)$ and $B (5,8)$ . We are asked to find the magnitude and direction of the vector $\overrightarrow{AB}$ , which represents the displacement from point $A$ to point $B$ .

Understanding Vector Displacement

Vector displacement is a fundamental concept in mathematics and physics, representing the change in position of an object from one point to another. In this case, the vector $\overrightarrow{AB}$ represents the displacement from point $A$ to point $B$ . To find the magnitude and direction of this vector, we need to use the concept of vectors and their properties.

Finding the Vector $\overrightarrow{AB}$

To find the vector $\overrightarrow{AB}$ , we need to subtract the coordinates of point $A$ from the coordinates of point $B$ . This can be done using the following formula:

\overrightarrow{AB} = \binom{x_B - x_A}{y_B - y_A}

where $(x_A, y_A)$ are the coordinates of point $A$ and $(x_B, y_B)$ are the coordinates of point $B$ .

Calculating the Vector $\overrightarrow{AB}$

Using the formula above, we can calculate the vector $\overrightarrow{AB}$ as follows:

\overrightarrow{AB} = \binom{5 - 6}{8 - 7} = \binom{-1}{1}

Understanding the Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. In this case, we need to find the magnitude of the vector $\overrightarrow{AB}$ . The magnitude of a vector $\overrightarrow{v} = \binom{x}{y}$ is given by the formula:

\| \overrightarrow{v} \| = \sqrt{x^2 + y^2}

Calculating the Magnitude of $\overrightarrow{AB}$

Using the formula above, we can calculate the magnitude of the vector $\overrightarrow{AB}$ as follows:

\| \overrightarrow{AB} \| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}

Understanding the Direction of a Vector

The direction of a vector is a measure of its orientation in space. In this case, we need to find the direction of the vector $\overrightarrow{AB}$ . The direction of a vector $\overrightarrow{v} = \binom{x}{y}$ can be found using the following formula:

\theta = \tan^{-1} \left( \frac{y}{x} \right)

Calculating the Direction of $\overrightarrow{AB}$

Using the formula above, we can calculate the direction of the vector $\overrightarrow{AB}$ as follows:

\theta = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1} (-1)

Conclusion

In this problem, we found the magnitude and direction of the vector $\overrightarrow{AB}$ , which represents the displacement from point $A$ to point $B$ . We used the concept of vectors and their properties to calculate the magnitude and direction of the vector. The magnitude of the vector $\overrightarrow{AB}$ is $\sqrt{2}$ , and the direction is $\tan^{-1} (-1)$ .

Final Answer

The final answer is $\boxed{\binom{-1}{1}}$ .

Step-by-Step Solution

Find the vector $\overrightarrow{AB}$ by subtracting the coordinates of point $A$ from the coordinates of point $B$ .
Calculate the magnitude of the vector $\overrightarrow{AB}$ using the formula $\| \overrightarrow{v} \| = \sqrt{x^2 + y^2}$ .
Calculate the direction of the vector $\overrightarrow{AB}$ using the formula $\theta = \tan^{-1} \left( \frac{y}{x} \right)$ .

Key Concepts

Vector displacement
Magnitude of a vector
Direction of a vector
Vectors and their properties

Q: What is vector displacement?

A: Vector displacement is a fundamental concept in mathematics and physics, representing the change in position of an object from one point to another. It is a vector quantity that has both magnitude and direction.

Q: How do I find the vector displacement between two points?

A: To find the vector displacement between two points, you need to subtract the coordinates of the initial point from the coordinates of the final point. This can be done using the following formula:

\overrightarrow{AB} = \binom{x_B - x_A}{y_B - y_A}

where $(x_A, y_A)$ are the coordinates of the initial point and $(x_B, y_B)$ are the coordinates of the final point.

Q: What is the magnitude of a vector?

A: The magnitude of a vector is a measure of its length or size. It is calculated using the formula:

\| \overrightarrow{v} \| = \sqrt{x^2 + y^2}

Q: How do I calculate the magnitude of a vector?

A: To calculate the magnitude of a vector, you need to plug in the values of the x and y components of the vector into the formula above. For example, if the vector is $\overrightarrow{v} = \binom{3}{4}$ , the magnitude would be:

\| \overrightarrow{v} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Q: What is the direction of a vector?

A: The direction of a vector is a measure of its orientation in space. It is calculated using the formula:

\theta = \tan^{-1} \left( \frac{y}{x} \right)

Q: How do I calculate the direction of a vector?

A: To calculate the direction of a vector, you need to plug in the values of the x and y components of the vector into the formula above. For example, if the vector is $\overrightarrow{v} = \binom{3}{4}$ , the direction would be:

\theta = \tan^{-1} \left( \frac{4}{3} \right)

Q: What is the difference between vector displacement and distance?

A: Vector displacement and distance are two related but distinct concepts. Distance is a scalar quantity that represents the length of a path between two points, while vector displacement is a vector quantity that represents the change in position of an object from one point to another.

Q: Can you give an example of vector displacement in real-life?

A: Yes, a common example of vector displacement is the movement of an object from one location to another. For example, if you are driving from your home to the store, the vector displacement would be the change in position of your car from your home to the store.

Q: How do I use vectors to solve problems in physics?

A: Vectors are a powerful tool for solving problems in physics. They can be used to represent forces, velocities, and accelerations, and to solve problems involving motion, energy, and momentum.

Q: What are some common applications of vector displacement?

A: Vector displacement has many common applications in fields such as physics, engineering, and computer science. Some examples include:

Calculating the trajectory of a projectile
Determining the force required to move an object
Calculating the velocity and acceleration of an object
Modeling the motion of objects in computer simulations

Q: Can you give an example of a problem that involves vector displacement?

A: Yes, a common example of a problem that involves vector displacement is the following:

Problem: A car is traveling from point A to point B. The car's initial position is at (0,0) and its final position is at (3,4). What is the vector displacement of the car?

Solution: To find the vector displacement, we need to subtract the initial position from the final position. This gives us:

\overrightarrow{AB} = \binom{3 - 0}{4 - 0} = \binom{3}{4}

The magnitude of the vector displacement is:

\| \overrightarrow{AB} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The direction of the vector displacement is:

\theta = \tan^{-1} \left( \frac{4}{3} \right)

Q: What are some common mistakes to avoid when working with vector displacement?

A: Some common mistakes to avoid when working with vector displacement include:

Confusing vector displacement with distance
Failing to account for the direction of the vector
Not using the correct formula for calculating the magnitude and direction of the vector
Not checking the units of the vector displacement

Q: Can you give an example of a problem that involves vector displacement and distance?

A: Yes, a common example of a problem that involves vector displacement and distance is the following:

Problem: A person is walking from point A to point B. The person's initial position is at (0,0) and their final position is at (3,4). The distance between the two points is 5 units. What is the vector displacement of the person?

Solution: To find the vector displacement, we need to subtract the initial position from the final position. This gives us:

\overrightarrow{AB} = \binom{3 - 0}{4 - 0} = \binom{3}{4}

The magnitude of the vector displacement is:

\| \overrightarrow{AB} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The direction of the vector displacement is:

\theta = \tan^{-1} \left( \frac{4}{3} \right)

Note that the vector displacement is the same as the distance between the two points, but the direction of the vector displacement is different.

Q: What are some common applications of vector displacement in real-life?

A: Vector displacement has many common applications in real-life, including:

Navigation: Vector displacement is used in navigation systems to determine the position and direction of a vehicle or person.
Physics: Vector displacement is used in physics to calculate the motion of objects and the forces acting on them.
Engineering: Vector displacement is used in engineering to design and optimize systems, such as bridges and buildings.
Computer Science: Vector displacement is used in computer science to model the motion of objects in computer simulations.

Q: Can you give an example of a problem that involves vector displacement in real-life?

A: Yes, a common example of a problem that involves vector displacement in real-life is the following:

Problem: A pilot is flying a plane from point A to point B. The pilot's initial position is at (0,0) and their final position is at (3,4). The distance between the two points is 5 units. What is the vector displacement of the plane?

Solution: To find the vector displacement, we need to subtract the initial position from the final position. This gives us:

\overrightarrow{AB} = \binom{3 - 0}{4 - 0} = \binom{3}{4}

The magnitude of the vector displacement is:

\| \overrightarrow{AB} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The direction of the vector displacement is:

\theta = \tan^{-1} \left( \frac{4}{3} \right)

Note that the vector displacement is the same as the distance between the two points, but the direction of the vector displacement is different.

Q: What are some common challenges when working with vector displacement?

A: Some common challenges when working with vector displacement include:

Confusing vector displacement with distance
Failing to account for the direction of the vector
Not using the correct formula for calculating the magnitude and direction of the vector
Not checking the units of the vector displacement

Q: Can you give an example of a problem that involves vector displacement and a challenge?

A: Yes, a common example of a problem that involves vector displacement and a challenge is the following:

Problem: A person is walking from point A to point B. The person's initial position is at (0,0) and their final position is at (3,4). The distance between the two points is 5 units. However, the person takes a detour and walks in a different direction. What is the vector displacement of the person?

Solution: To find the vector displacement, we need to subtract the initial position from the final position. However, we need to account for the detour and the different direction of the vector. This gives us:

\overrightarrow{AB} = \binom{3 - 0}{4 - 0} = \binom{3}{4}

However, we need to add the vector displacement of the detour to the vector displacement of the original path. This gives us:

\overrightarrow{AB} = \binom{3}{4} + \