Lesson Explainer: Graphical Operations on Vectors Mathematics

In this explainer, we will learn how to do operations on vectors graphically using triangle and parallelogram rules.

Vectors are objects that are entirely defined by their magnitude and direction. We recall that we can represent vectors by a directed line segment, in a suitable space, where the length of the line segment tells us the magnitude of the vector and the direction is indicated by the initial and terminal points of the directed line segment. In this space, we can think of vectors as representing displacement, from the initial point to the terminal point.

For example, the vector from 𝐴 to 𝐡 can be represented as the directed line segment from 𝐴 to 𝐡.

The length of this line segment is the magnitude of 𝐴𝐡, written ‖‖𝐴𝐡‖‖, and the direction is shown by the arrow. It is worth reiterating that a vector is defined entirely by its magnitude and direction, so any two vectors with the same magnitude and direction will be equal. In particular, we can draw the vector anywhere in the π‘₯𝑦-plane; this will not change its magnitude or direction, and hence it will still be the same vector.

We can also represent this vector by the horizontal change and vertical change. If 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦), then from the diagram (or the coordinates) we can find the horizontal and vertical changes when moving from 𝐴 to 𝐡.

The change in the horizontal coordinate is π‘₯βˆ’π‘₯ and the change in the vertical coordinate is π‘¦βˆ’π‘¦οŠ¨οŠ§. We write this as 𝐴𝐡=(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦). The first component tells us the horizontal displacement of the vector and the second component tells us the vertical displacement of the vector.

This then allows us to add vectors together. In terms of displacement, the sum of two vectors will be the total displacement of both vectors. In terms of the components, we recall that this means if we have ⃑𝑒=(π‘₯,𝑦) and ⃑𝑣=(π‘₯,𝑦), then ⃑𝑒+⃑𝑣=(π‘₯+π‘₯,𝑦+𝑦); we add the horizontal and vertical components separately. This then allows us to add vectors together graphically.

For example, if we have the vector 𝐴𝐡, then in this space, this is the displacement from point 𝐴 to point 𝐡. Similarly, οƒŸπ΅πΆ is the displacement from point 𝐡 to point 𝐢. Their sum, 𝐴𝐡+οƒŸπ΅πΆ, should be the total displacement of both vectors, that is, the displacement from 𝐴 to 𝐡 and then from 𝐡 to 𝐢. This is the vector 𝐴𝐢, and we can show this graphically.

We can see that the displacement from 𝐴 to 𝐢 is the same as the displacement from 𝐴 to 𝐡 and then from 𝐡 to 𝐢, since their initial and terminal points are the same. This is often referred to as the triangle rule for vectors. We can state this as follows.

Rule: Triangle Rule for Vectors

For any three points 𝐴, 𝐡, and 𝐢, we have 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢 as shown in the diagram.

As a result of the triangle rule, if we have two vectors ⃑𝑒 and ⃑𝑣 represented graphically, then we can add these vectors together by sketching the terminal point of ⃑𝑒 to be the initial point of ⃑𝑣, as shown.

Before we see how to apply this to problems involving vectors, there is one more rule we can show. Consider parallelogram 𝐴𝐡𝐢𝐷.

In a parallelogram, opposite sides are parallel and have the same length; therefore, 𝐴𝐷=οƒŸπ΅πΆοƒ π΄π΅=𝐷𝐢.and

We can apply the triangle rule for vectors to this diagram by considering the sides as vectors and adding vector 𝐴𝐢 to the diagram, as shown.

Applying the triangle rule for vectors to points 𝐴, 𝐡, and 𝐢, we have 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢.

We know that 𝐴𝐷=οƒŸπ΅πΆ since they have the same magnitude and direction. We can see this graphically as we can translate the vectors on top of each other. Substituting this into the equation gives 𝐴𝐡+𝐴𝐷=𝐴𝐢.

This is known as the parallelogram rule for vectors, and we can summarize this is as follows.

Rule: Parallelogram Rule for Vectors

For vectors 𝐴𝐡 and 𝐴𝐷 with the same initial point, 𝐴, 𝐴𝐡+𝐴𝐷=𝐴𝐢, where 𝐢 is the point that makes 𝐴𝐡𝐢𝐷 a parallelogram, as shown in the following diagram.

Since vectors have a number of uses, for example, as a tool to evaluate the resultant of forces or to solve geometric problems, the triangle and parallelogram rules can be thought of entirely in terms of the field we are working in. However, it is usually easier to convert the problem into vectors.

Let’s start by looking at some examples of adding vectors together graphically.

Example 1: Finding the Components of the Sum of End-to-End Vectors on a Diagram

Shown on the grid of unit squares are the vectors ⃑𝑒 and ⃑𝑣.

  1. What are the components of ⃑𝑒?
  2. What are the components of ⃑𝑣?
  3. What are the components of ⃑𝑒+⃑𝑣?

Answer

Part 1

We recall that the components of a vector represented graphically are the horizontal and vertical displacements of the vector in this space. In particular, for a vector from 𝐴(π‘₯,𝑦) to 𝐡(π‘₯,𝑦), its components will be the difference in the coordinates: 𝐴𝐡=(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦).

There are a few ways of using this to determine the components of vector ⃑𝑒. For example, we could introduce a coordinate system onto the grid of unit squares. However, this is not necessary. Instead, we only need to know the horizontal and vertical displacements when traveling from the initial point of ⃑𝑒 to the terminal point of ⃑𝑒. Let’s start with the horizontal displacement.

Moving from the tail to the tip of vector ⃑𝑒, we travel over three of the squares. Since these are unit squares and we travel to the right, we can write this as displacement of +3 units.

We can do the same vertically.

We move two units upward, so the displacement is +2 units. This gives us both components of our vector. Remember, we do not need to include the positive symbol; this gives us ⃑𝑒=(3,2).

Part 2

We can follow the same procedure for vector ⃑𝑣.

First, we see that we move two units to the right, so the first component of the vector will be 2. We need to be careful when checking the vertical displacement of the vector; since we move downward, the value will be negative.

Traveling from the tail to the tip of vector ⃑𝑣, we move three units downward. Therefore, the vertical component of ⃑𝑣 will be βˆ’3. Hence, ⃑𝑣=(2,βˆ’3).

Part 3

There are two methods we can use to add the vectors together.

First, we can recall that we add vectors together component-wise and then use our answers in the first two parts of the question; this gives ⃑𝑒+⃑𝑣=(3,2)+(2,βˆ’3)=(3+2,2+(βˆ’3))=(5,βˆ’1).

However, this result relies on us finding the components of both vectors correctly and then not making any mistakes in the calculation. This becomes more of a problem the more vectors we are asked to add together. Instead, we can also add the two vectors together graphically.

Since we add the components of each vector separately, we can add vectors together using the β€œtip-to-tail” method. We sketch the vectors so that the tip of one vector is aligned with the tail of the next one, so the sum of the vectors is then the vector from the initial point of the first vector to the terminal point of the final vector. Since the vectors are already sketched in this way, we have the following.

The vector ⃑𝑒+⃑𝑣 is represented by the third side in the triangle since the directions of ⃑𝑒 and ⃑𝑣 match. We can then find the components of ⃑𝑒+⃑𝑣 using the diagram.

Following the vector ⃑𝑒+⃑𝑣, we move 5 units right and 1 unit down, so the horizontal component is 5 and the vertical component is βˆ’1, giving us ⃑𝑒+⃑𝑣=(5,βˆ’1).

We can see that this agrees with the direct calculation we did above.

Example 2: Finding the Sum of Two Vectors Given Graphically

Which graph represents ⃑𝐴+⃑𝐡, where ⃑𝐴=(3,4) and ⃑𝐡=(4,1)?

Answer

Since we are given the components of ⃑𝐴 and ⃑𝐡, we can add the vectors component-wise to get ⃑𝐴+⃑𝐡=(3,4)+(4,1)=(3+4,4+1)=(7,5).

Looking at the options, we can then see that only option A has ⃑𝐴=(3,4), ⃑𝐡=(4,1), and ⃑𝐴+⃑𝐡=(7,5), so it is the correct answer. The problem with answering the question in this manner is that we will often be asked to add vectors together graphically without being given the options, so we will also construct the diagram ourselves.

We can start by sketching vectors ⃑𝐴=(3,4) and ⃑𝐡=(4,1) onto a diagram, where we recall that the first component tells us the horizontal displacement and the second component tells us the vertical displacement. It is also worth noting that we can sketch the vectors anywhere on the plane. However, for simplicity, we will start both vectors at the origin. First, ⃑𝐴=(3,4) will travel 3 units right and 4 units up to give us the following.

We can do the same with vector ⃑𝐡=(4,1), which will travel 4 units right and 1 unit upward, as shown.

It is worth noting that since we started our vectors from the origin, the coordinates of the terminal points will be equal to the corresponding components of the vector; these are often called position vectors.

To add these two vectors together graphically, we start by sketching them so that the terminal point of one vector is the initial point of the other. We will move ⃑𝐡 to have its initial point at (3,4); it will still move 4 units right and 1 unit up giving us the following.

Finally, the triangle rule for vectors tells us that the sum of these two vectors will be the vector having the initial point of ⃑𝐴 and the terminal point of ⃑𝐡 since these vectors are sketched tip to tail, giving us the following.

We can see in the diagram that ⃑𝐴+⃑𝐡 travels 7 units right and 5 units up, so ⃑𝐴+⃑𝐡=(7,5), and this is shown in option A.

In our previous examples, we used the graphical representations alongside their components to solve problems. In our next example, we will use only the graphical representation of the vectors to solve a geometric problem.

Example 3: Identifying the Correct Diagonal in the Parallelogram Law

Which vector is equivalent to ⃑𝑒+⃑𝑣?

Answer

Quadrilateral 𝐴𝐢𝐷𝐡 has opposite sides as equal vectors. Since equal vectors have the same magnitude and direction, we can conclude that 𝐴𝐢𝐷𝐡 is a parallelogram. We want to apply the parallelogram rule for vectors to add these vectors. To do this, we notice that ⃑𝑣 is the vector from 𝐴 to 𝐢, so ⃑𝑣=𝐴𝐢.

Similarly, ⃑𝑒=𝐴𝐡.

The parallelogram rule for adding vectors then tells us that if 𝐴𝐢 and 𝐴𝐡 have the same initial point, 𝐴, then 𝐴𝐢+𝐴𝐡=𝐴𝐷, where 𝐷 is the point that makes 𝐴𝐢𝐷𝐡 a parallelogram. Vector 𝐴𝐷 is the diagonal of the parallelogram as shown.

Therefore, ⃑𝑒+⃑𝑣=𝐴𝐷.

Using the above example, we can show a useful result from the parallelogram we constructed.

By applying the triangle rule for vectors on the points 𝐴, 𝐢, and 𝐷, we have 𝐴𝐢+𝐢𝐷=𝐴𝐷.

In terms of ⃑𝑒 and ⃑𝑣, this states that ⃑𝑣+⃑𝑒=⃑𝑒+⃑𝑣.

In other words, this is a geometric demonstration of the commutativity of vector addition.

Thus far, we have only considered adding to vectors together. We will now consider an example where we are asked to find the difference of two vectors.

Example 4: Identifying the Correct Graphical Representation of the Difference of Two Vectors

Which of the following parallelograms shows a valid way of obtaining βƒ‘π΄βˆ’βƒ‘π΅?

Answer

To subtract two vectors graphically, we will use two results. First, we can rewrite the expression as the sum of two vectors: βƒ‘π΄βˆ’βƒ‘π΅=⃑𝐴+ο€Ίβˆ’βƒ‘π΅ο†.

This tells us the difference between ⃑𝐴 and ⃑𝐡 is the same as the sum of ⃑𝐴 and βˆ’βƒ‘π΅, and we know how to add vectors together graphically by using both the triangle and parallelogram rules. Since the options have parallelograms included, we will use the parallelogram rule.

The parallelogram rule for vector addition tells us that if οƒŸπ‘ƒπ‘„ and οƒŸπ‘ƒπ‘† are vectors with the same initial point, 𝑃, then οƒŸπ‘ƒπ‘„+οƒŸπ‘ƒπ‘†=οƒŸπ‘ƒπ‘…, where 𝑅 is the point that makes 𝑃𝑄𝑅𝑆 a parallelogram, as shown in the following diagram.

If we label the vectors, which are the sides of the parallelogram, ⃑𝐴 and βˆ’βƒ‘π΅, we have the following.

Then, since βƒ‘π΄βˆ’βƒ‘π΅=⃑𝐴+ο€Ίβˆ’βƒ‘π΅ο†, we can replace the vector on the diagonal with βƒ‘π΄βˆ’βƒ‘π΅, which is the same as option C.

We can then ask the following question: given two vectors ⃑𝐴 and ⃑𝐡, graphically, how can we find βƒ‘π΄βˆ’βƒ‘π΅? To do this, we will start by sketching ⃑𝐴 and ⃑𝐡 to have the same initial point, say 𝑃; we will also label the terminal points of the vectors 𝑄 and 𝑅 as shown.

We want to determine βƒ‘π΄βˆ’βƒ‘π΅=οƒŸπ‘ƒπ‘…βˆ’οƒŸπ‘ƒπ‘„. Recall that βƒ‘π΄βˆ’βƒ‘π΅=⃑𝐴+ο€Ίβˆ’βƒ‘π΅ο† and that multiplying a vector by βˆ’1 switches its direction and does not change its magnitude. In our diagram, this means βˆ’βƒ‘π΅ is the vector from 𝑄 to 𝑃, as shown.

We can then see that the terminal point of βˆ’βƒ‘π΅ is the initial point of ⃑𝐴, so we can add these vectors by using the triangle rule for vector addition; we have the following.

We can then use commutativity of vector addition to write βˆ’βƒ‘π΅+⃑𝐴=βƒ‘π΄βˆ’βƒ‘π΅. This means οƒŸπ‘ƒπ‘…βˆ’οƒŸπ‘ƒπ‘„=οƒŸπ‘„π‘….

We can then represent this graphically as shown.

Let’s now see a few examples of applying this reasoning to subtract two vectors given graphically.

Example 5: Finding the Difference of Two Vectors Represented Graphically

Which of the following is equivalent to οƒ π΄π΅βˆ’οƒ π·π΅?

  1. 𝐷𝐴
  2. 𝐷𝐢
  3. 𝐴𝐢
  4. 𝐡𝐴
  5. οƒŸπ΅πΆ

Answer

We are asked to find the difference of two vectors. To do this, we start by noting that multiplying a vector by βˆ’1 switches its direction but leaves the magnitude unchanged, so βˆ’οƒ π·π΅=𝐡𝐷.

Hence, οƒ π΄π΅βˆ’οƒ π·π΅=𝐴𝐡+𝐡𝐷.

We can draw both of these vectors on the diagram.

Since the terminal point of 𝐴𝐡 is the initial point of 𝐡𝐷, we can add these vectors by using the triangle rule for vector addition, which tells us 𝐴𝐡+𝐡𝐷=𝐴𝐷, which we can show on the diagram.

However, this is not one of the given options. We need to also use the fact that 𝐴𝐡𝐢𝐷 is a parallelogram where opposite sides have equal size and are parallel. Therefore, 𝐴𝐷=οƒŸπ΅πΆ.

Hence, the answer is option E, οƒŸπ΅πΆ.

Example 6: Solving a Geometric Problem Involving Vectors as the Sides of a Parallelogram

Which of the following is equivalent to 12ο€ΊοƒŸπΆπ΅βˆ’οƒ π·πΆο†?

  1. 𝐢𝐴
  2. οƒŸπΆπΈ
  3. οƒŸπ΅πΈ
  4. 𝐴𝐸
  5. οƒŸπΈπΆ

Answer

We want to determine a vector equivalent to the given expression by using the diagram. To do this, we first note that for a vector ⃑𝑒, 12⃑𝑒 will have the same direction but half the magnitude. So, we will start by finding a vector equivalent to οƒŸπΆπ΅βˆ’οƒ π·πΆ. We can do this by first recalling that βˆ’οƒ π·πΆ=𝐢𝐷, so οƒŸπΆπ΅βˆ’οƒ π·πΆ=οƒŸπΆπ΅+𝐢𝐷.

We can add these vectors to the diagram.

To add vectors together graphically, we can use the parallelogram rule for vectors. Since the initial points of both vectors are the same, their sum will be the diagonal of the parallelogram.

Hence, οƒŸπΆπ΅βˆ’οƒ π·πΆ=οƒŸπΆπ΅+𝐢𝐷=𝐢𝐴.

We want to find 12 of this vector, so we need a vector in the same direction as 𝐢𝐴 but with half the magnitude. Since 𝐸 is the midpoint of 𝐢𝐴, there are two options for this; both οƒŸπΆπΈ and 𝐸𝐴 will have the same direction as 𝐢𝐴 but half the magnitude.

Hence, 12ο€ΊοƒŸπΆπ΅βˆ’οƒ π·πΆο†=12ο€ΊοƒŸπΆπ΅+𝐢𝐷=12𝐢𝐴=οƒŸπΆπΈ.

So, the answer is option B, οƒŸπΆπΈ.

It is worth noting that we can apply these rules for vector addition multiple times. For example, consider the following diagram.

If we wanted to find 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷, we can do this by applying the triangle rule for vector addition twice. First, we have 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢, as shown in the following diagram.

We can then add 𝐴𝐢 and 𝐢𝐷 with the triangle rule for vector addition. We have 𝐴𝐢+𝐢𝐷=𝐴𝐷, as shown.

Hence, 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷=𝐴𝐷.

Graphically, we can think about this as the three vectors 𝐴𝐡, οƒŸπ΅πΆ, and 𝐢𝐷 all being drawn tip to tail, so their sum is the vector that has the initial point 𝐴 and the terminal point 𝐷. We can apply this same reasoning to even more complicated shapes as we will see in our final example.

Example 7: Simplifying the Sum of Vectors That are the Sides of a Polygon

Complete: In the following figure, 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷+𝐷𝐸=.

  1. 𝐸𝐴
  2. 𝐴𝐸
  3. 𝐴𝐢
  4. 2𝐴𝐢
  5. 2οƒŸπ΅πΈ

Answer

We begin by adding all of the vectors in the sum onto the diagram.

There are two ways we can determine this sum: we can do this directly by recalling that the sum of any number of vectors drawn tip to tail will have the initial point of the first vector and the terminal point of the final vector. On our diagram, this will be vector 𝐴𝐸 as shown.

It might be easier to think of this as following the vectors from 𝐴 all the way to 𝐸. We have 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷+𝐷𝐸=𝐴𝐸, which is option B.

To see why this is true, we need to apply the triangle rule, which states that for any points 𝑃, 𝑄, and 𝑅, οƒŸπ‘ƒπ‘„+οƒŸπ‘„π‘…=οƒŸπ‘ƒπ‘….

We can apply this to the first two vectors in the sum to get 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢, so 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷+𝐷𝐸=𝐴𝐢+𝐢𝐷+𝐷𝐸.

We can then apply the triangle rule again and see that 𝐴𝐢+𝐢𝐷=𝐴𝐷, so 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷+𝐷𝐸=𝐴𝐢+𝐢𝐷+𝐷𝐸=𝐴𝐷+𝐷𝐸.

Finally, we apply the triangle rule one more time to get 𝐴𝐷+𝐷𝐸=𝐴𝐸, and hence 𝐴𝐡+οƒŸπ΅πΆ+𝐢𝐷+𝐷𝐸=𝐴𝐸.

Therefore, the answer is option B, 𝐴𝐸.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • The triangle rule for vectors tells us that, for any three points 𝐴, 𝐡, and 𝐢, 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢.
  • The triangle rule for vectors allows us to add two vectors graphically by drawing them such that the initial point of one vector is the terminal point of the other.
  • The parallelogram rule for vectors tells us that 𝐴𝐡+𝐴𝐷=𝐴𝐢,
    where 𝐢 is the point that makes 𝐴𝐡𝐢𝐷 a parallelogram.
  • The parallelogram rule for vectors allows us to add vectors together graphically by drawing them such that they have the same initial point. Then, their resultant will be given by the diagonal of the parallelogram spanned by the vectors.
  • We can subtract two vectors by using the fact that, for any three points 𝑃, 𝑄, and 𝑅, οƒŸπ‘ƒπ‘…βˆ’οƒŸπ‘ƒπ‘„=οƒŸπ‘„π‘….

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