Vector \[$\mathbf{u}\$\] Has An Initial Point At \[$(4, 4)\$\] And A Terminal Point At \[$(-12, 8)\$\].Which Are The Magnitude And Direction Of \[$\mathbf{u}\$\]?A. \[$\|\mathbf{u}\| = 14.422\$\]; \[$\theta =

Introduction

In mathematics, vectors are used to represent quantities with both magnitude and direction. The magnitude of a vector is its length or size, while the direction is the angle it makes with a reference line. In this article, we will discuss how to calculate the magnitude and direction of a vector given its initial and terminal points.

Calculating the Magnitude of a Vector

The magnitude of a vector can be calculated using the distance formula. The distance formula is given by:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Example

Let's consider the vector $\mathbf{u}$ with an initial point at $(4, 4)$ and a terminal point at $(-12, 8)$. To calculate the magnitude of this vector, we can use the distance formula:

$$\|\mathbf{u}\| = \sqrt{(-12 - 4)^2 + (8 - 4)^2}$$

$$\|\mathbf{u}\| = \sqrt{(-16)^2 + (4)^2}$$

$$\|\mathbf{u}\| = \sqrt{256 + 16}$$

$$\|\mathbf{u}\| = \sqrt{272}$$

$$\|\mathbf{u}\| = 16.5$$

Calculating the Direction of a Vector

The direction of a vector can be calculated using the tangent function. The tangent function is given by:

$$\tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Example

Let's consider the vector $\mathbf{u}$ with an initial point at $(4, 4)$ and a terminal point at $(-12, 8)$. To calculate the direction of this vector, we can use the tangent function:

$$\tan \theta = \frac{8 - 4}{-12 - 4}$$

$$\tan \theta = \frac{4}{-16}$$

$$\tan \theta = -0.25$$

To find the angle $\theta$, we can use the inverse tangent function:

$$\theta = \tan^{-1}(-0.25)$$

$$\theta = -14.036$$

Converting the Angle to a Standard Angle

Since the angle $\theta$ is negative, we can convert it to a standard angle by adding $180^\circ$:

$$\theta = -14.036 + 180$$

$$\theta = 165.964$$

Conclusion

In this article, we discussed how to calculate the magnitude and direction of a vector given its initial and terminal points. We used the distance formula to calculate the magnitude of the vector and the tangent function to calculate the direction. We also converted the angle to a standard angle by adding $180^\circ$. The magnitude of the vector $\mathbf{u}$ is $16.5$ and the direction is $165.964^\circ$.

Calculating the Magnitude and Direction of a Vector: A Step-by-Step Guide

Step 1: Calculate the Magnitude of the Vector

To calculate the magnitude of the vector, we can use the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Step 2: Calculate the Direction of the Vector

To calculate the direction of the vector, we can use the tangent function:

$$\tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Step 3: Convert the Angle to a Standard Angle

If the angle $\theta$ is negative, we can convert it to a standard angle by adding $180^\circ$:

$$\theta = -\theta + 180$$

Step 4: Calculate the Magnitude and Direction of the Vector

Using the distance formula and the tangent function, we can calculate the magnitude and direction of the vector.

Example

Let's consider the vector $\mathbf{u}$ with an initial point at $(4, 4)$ and a terminal point at $(-12, 8)$. To calculate the magnitude and direction of this vector, we can follow the steps above:

Step 1: Calculate the Magnitude of the Vector

$$\|\mathbf{u}\| = \sqrt{(-12 - 4)^2 + (8 - 4)^2}$$

$$\|\mathbf{u}\| = \sqrt{(-16)^2 + (4)^2}$$

$$\|\mathbf{u}\| = \sqrt{256 + 16}$$

$$\|\mathbf{u}\| = \sqrt{272}$$

$$\|\mathbf{u}\| = 16.5$$

Step 2: Calculate the Direction of the Vector

$$\tan \theta = \frac{8 - 4}{-12 - 4}$$

$$\tan \theta = \frac{4}{-16}$$

$$\tan \theta = -0.25$$

Step 3: Convert the Angle to a Standard Angle

$$\theta = -14.036 + 180$$

$$\theta = 165.964$$

Step 4: Calculate the Magnitude and Direction of the Vector

The magnitude of the vector $\mathbf{u}$ is $16.5$ and the direction is $165.964^\circ$.

Conclusion

$$$$$$$$

Q: What is the magnitude of a vector?

A: The magnitude of a vector is its length or size. It is a measure of how long the vector is.

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

A: To calculate the magnitude of a vector, you can use the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Q: What is the direction of a vector?

A: The direction of a vector is the angle it makes with a reference line. It is a measure of how the vector is oriented in space.

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

A: To calculate the direction of a vector, you can use the tangent function:

$$\tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$$

where $(x_1, y_1)$ is the initial point and $(x_2, y_2)$ is the terminal point of the vector.

Q: What is the difference between magnitude and direction?

A: The magnitude of a vector is its length or size, while the direction is the angle it makes with a reference line. The magnitude and direction of a vector are independent of each other.

Q: Can I have a vector with zero magnitude?

A: Yes, a vector can have zero magnitude if it has no length or size. This can happen if the initial and terminal points of the vector are the same.

Q: Can I have a vector with zero direction?

A: No, a vector cannot have zero direction. The direction of a vector is always defined as the angle it makes with a reference line.

Q: How do I convert the direction of a vector to a standard angle?

A: To convert the direction of a vector to a standard angle, you can add $180^\circ$ to the angle if it is negative.

Q: Can I use the magnitude and direction of a vector to find its components?

A: Yes, you can use the magnitude and direction of a vector to find its components. The components of a vector are the x and y coordinates of its terminal point.

Q: How do I find the components of a vector?

A: To find the components of a vector, you can use the following formulas:

$$x = \|\mathbf{u}\| \cos \theta$$

$$y = \|\mathbf{u}\| \sin \theta$$

where $\|\mathbf{u}\|$ is the magnitude of the vector and $\theta$ is its direction.

Q: Can I use the components of a vector to find its magnitude and direction?

A: Yes, you can use the components of a vector to find its magnitude and direction. The magnitude of a vector is the square root of the sum of the squares of its components, and the direction is the inverse tangent of the ratio of its y-component to its x-component.

Q: How do I find the magnitude and direction of a vector from its components?

A: To find the magnitude and direction of a vector from its components, you can use the following formulas:

$$\|\mathbf{u}\| = \sqrt{x^2 + y^2}$$

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

where $x$ and $y$ are the components of the vector.

Conclusion

In this article, we answered some frequently asked questions about the magnitude and direction of vectors. We discussed how to calculate the magnitude and direction of a vector, how to convert the direction to a standard angle, and how to find the components of a vector. We also discussed how to use the components of a vector to find its magnitude and direction.