# 5.13: Directed Line Segments

**At Grade**Created by: CK-12

^{%}

**Practice**Directed Line Segments

While babysitting your nephew one day you have a discussion regarding math. As a second grader, he is learning to add and subtract quantities. He asks you about what you're doing in math class, and you explain to him that you have just been introduced to vectors. Once you explain to him that a vector is a mathematical quantity that has both magnitude and direction, his question is both simple and yet brilliant: "Why do you need vectors? Can't everything just be described with numbers that don't have direction?"

At the end of this Concept, you'll know how to answer your nephew.

### Watch This

James Sousa: Introduction to Vectors

### Guidance

A vector is represented diagrammatically by a directed line segment or arrow. A **directed line segment** has both **magnitude** and **direction**. **Magnitude** refers to the length of the directed line segment and is usually based on a scale. The vector quantity represented, such as influence of the wind or water current may be completely invisible.

A 25 mph wind is blowing from the northwest. If \begin{align*}1\ cm = 5\ mph\end{align*}

An object affected by this wind would travel in a southeast direction at 25 mph.

A vector is said to be in **standard position** if its **initial point** is at the origin. The initial point is where the vector begins and the **terminal point** is where it ends. The axes are arbitrary. They just give a place to draw the vector.

If we know the coordinates of a vector’s initial point and terminal point, we can use these coordinates to find the magnitude and direction of the vector.

All vectors have **magnitude**. This measures the total distance moved, total velocity, force or acceleration. “Distance” here applies to the magnitude of the vector even though the vector is a measure of velocity, force, or acceleration. In order to find the magnitude of a vector, we use the distance formula. A vector can have a negative magnitude. A force acting on a block pushing it at 20 lbs north can be also written as vector acting on the block from the south with a magnitude of -20 lbs. Such negative magnitudes can be confusing; making a diagram helps. The -20 lbs south can be re-written as +20 lbs north without changing the vector. Magnitude is also called the **absolute value** of a vector.

#### Example A

If a vector starts at the origin and has a terminal point with coordinates (3,5), find the length of the vector.

If we know the coordinates of the initial point and the terminal point, we can find the magnitude by using the distance formula. Initial point (0,0) and terminal point (3,5).

**Solution:** \begin{align*}|\vec{v}| = \sqrt{(3 - 0)^2 + (5 - 0)^2} = \sqrt{9 + 25} = 5.8\end{align*}

If we don’t know the coordinates of the vector, we must use a ruler and the given scale to find the magnitude. Also notice the notation of a vector, which is usually a lower case letter (typically \begin{align*}u, v\end{align*}**direction** of that vector.

#### Example B

If a vector is in standard position and its terminal point has coordinates of (12, 9) what is the direction?

**Solution:** The horizontal distance is 12 while the vertical distance is 9. We can use the tangent function since we know the opposite and adjacent sides of our triangle.

\begin{align*}\tan \theta & = \frac{9}{12} \\ \tan^{-1} \frac{9}{12} & = 36.9^\circ\end{align*}

So, the direction of the vector is \begin{align*}36.9^\circ\end{align*}

If the vector isn’t in standard position and we don’t know the coordinates of the terminal point, we must a protractor to find the direction.

Two vectors are **equal** if they have the same magnitude and direction. Look at the figures below for a visual understanding of **equal vectors**.

#### Example C

Determine if the two vectors are equal.

\begin{align*}\vec{a}\end{align*}

\begin{align*}\vec{b}\end{align*}

**Solution:** You need to determine if both the magnitude and the direction are the same.

\begin{align*}\text{Magnitude}:\quad |\vec{a}| & = \sqrt{(0-(-4))^2 + (0 - 12)^2} = \sqrt{16+144} = \sqrt{160} = 4\sqrt{10} \\ |\vec{b}| & = \sqrt{(7-3)^2 + (-6-6)^2} = \sqrt{16+144} = \sqrt{160} = 4 \sqrt{10} \\ \text{Direction}:\quad \vec{a} & \rightarrow \tan \theta = \frac{12}{-4} \rightarrow \theta = 108.43^\circ \\ \vec{b} & \rightarrow \tan \theta = \frac{-6-6}{7-3} = \frac{-12}{4} \rightarrow \theta = 108.43^\circ\end{align*}

Because the magnitude and the direction are the same, we can conclude that the two vectors are equal.

### Guided Practice

1. Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.

initial ( 2, 4) terminal (8, 6)

2. Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.

initial (5, -2) terminal (3, 1)

3. Assume \begin{align*}\vec{a}\end{align*} is in standard position. For the terminal point (12, 18), find the magnitude and direction of thevector.

**Solutions:**

1. \begin{align*}|\vec{a}| = \sqrt{(2-8)^2 + (4-6)^2} = 6.3\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{4-6}{2-8} \right ) = 18.4^\circ\end{align*}

2. \begin{align*}|\vec{a}| = \sqrt{(5-3)^2 + (-2-1)^2} = 3.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{-2-1}{5-3} \right ) = 123.7^\circ\end{align*}. Note that when you use your calculator to solve for \begin{align*}\tan^{-1} (\frac{-2-1}{5-3})\end{align*}, you will get \begin{align*}-56.3^\circ\end{align*}. The calculator produces this answer because the range of the calculator’s \begin{align*}y = \tan^{-1} x\end{align*} function is limited to \begin{align*}-90^\circ < y < 90^\circ\end{align*}. You need to sketch a draft of the vector to see that its direction when placed in standard position is into the second quadrant (and not the fourth quadrant), and so the correct angle is calculated by moving the angle into the second quadrant through the equation \begin{align*}-56.3^\circ + 180^\circ=123.7^\circ.\end{align*}

3. \begin{align*}|\vec{a}| = \sqrt{12^2 + 18^2} = 21.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{18}{12} \right ) = 56.3^\circ \end{align*}

### Concept Problem Solution

Your nephew's thinking is quite good. Many things in the world can be described by numbers, without the use of direction. However, math needs to "line up" with reality. If something doesn't work with just numbers, there needs to be a new type of mathematical quantity to describe the behavior completely. For example, consider two cars that are both moving at 25 miles per hour. Will they collide?

You can see that there isn't enough information to answer the question. You don't know which way the cars are going. If the two cars are going in the same direction, then they won't collide. If, however, they are going directly at each other, then they will certainly collide. In order to describe the behavior of the cars completely, a quantity is needed that is not just the magnitude of the car's motion, but also the direction - which is why vectors are needed.

And that is how you should answer your nephew.

### Explore More

- What is the difference between the magnitude and direction of a vector?
- How can you determine the magnitude of a vector if you know its initial point and terminal point?
- How can you determine the direction of a vector if you know its initial point and terminal point?
- How can you determine whether or not two vectors are equal?
- If a vector starts at the origin and has a terminal point with coordinates (2, 7), find the magnitude of the vector.
- If a vector is in standard position and its terminal point has coordinates of (3, 9), what is the direction of the vector?
- If a vector has an initial point at (1, 6) and has a terminal point at (5, 9), find the magnitude of the vector.
- If a vector has an initial point at (1, 4) and has a terminal point at (8, 7), what is the direction of the vector?

Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.

- initial (4, -1); terminal (5, 3)
- initial (2, -3); terminal (4, 5)
- initial (3, 2); terminal (0, 3)
- initial (-2, 5); terminal (2, 1)

Determine if the two vectors are equal.

- \begin{align*}\vec{a}\end{align*} is in standard position with terminal point (1, 5) and \begin{align*}\vec{b}\end{align*} has an initial point (3, -2) and terminal point (4, 2).
- \begin{align*}\vec{c}\end{align*} has an initial point (-3, 1) and terminal point (1, 2) and \begin{align*}\vec{d}\end{align*} has an initial point (3, 5) and terminal point (7, 6).
- \begin{align*}\vec{e}\end{align*} is in standard position with terminal point (2, 3) and \begin{align*}\vec{f}\end{align*} has an initial point (1, -6) and terminal point (3, -9).

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.13.

Directed Line Segment

A directed line segment is a portion of a line that has both a magnitude and direction.Magnitude

The magnitude of a line segment or vector is the length of the line segment or vector.Vector

A vector is a mathematical quantity that has both a magnitude and a direction.### Image Attributions

## Description

## Learning Objectives

Here you'll learn to define a vector mathematically and draw it.

## Difficulty Level:

At Grade## Authors:

## Subjects:

## Search Keywords:

## Concept Nodes:

## Date Created:

Sep 26, 2012## Last Modified:

Feb 26, 2015## Vocabulary

**Save or share your relevant files like activites, homework and worksheet.**

To add resources, you must be the owner of the Modality. Click Customize to make your own copy.