# Graphical Vector Addition

*872*pages on

this wiki

**Graphical Vector Addition**Edit

#### What are "vectors," anyways?Edit

*Vectors are measurements that have both magnitude and direction. Vector quantities are used frequently in both math and physics; vector usage has both basic and complex applications. Vectors are represented by straight lines with a head on one end (denoted by an arrow,) and a tail (denoted by a “flat” end) on the other. As might have been assumed, the arrow represents the direction of the vector, and the length of the line represents the size or magnitude of the vector.*

*[The meaning of the word "vector" itself has a definition that lends itself to desciribing many other phyics related terms (each that complys with having both a direction and a magnitude) such as velocity as opposed to speed, which connotes only magnitude and not direction.]*

#### The history of Graphical VectorsEdit

Vectors come from the manipulation of the long standing theorem made by Pythagoras during the early years of mathematical foundation as well as the methods of Cartesian geometry found in Bolzano's work in the early 19th century. According to some, the Pythagorean Theorem is one of the greatest contributions to the mathematical and scientific community; it was also one of the earliest known theorems known to ancient civilizations. The actual date of Pythagoras’ discovery of the notion that the sum of the squares of a right triangle equals the hypotenuse is uncertain, yet many say that it came about around the time that he founded the Pythagorean School of Mathematics (500 BC). The basis of vector operations lie within the concept that (a2 + b2 = c2) assuming that “a” and “b” are the legs of a triangle where the angle between them is 90 degrees, and that “c” is the hypotenuse. Mathematicians soon realized that the components that make up Pythagora's special triangles could be rearranged in hopes of being added; this "sum" is called the resultant.

Bolzano's work delt specifically with the concept that points, lines and planes are undefined elements and can be operated on as distinct particles. This idea had a strong influence on the concept of linear space and visual abstraction. Bolzano was interested in creating concrete objects from spacial ones; he called vectors "abstract linear operators" that were limited to the coordinate plane only. With the ideas from both Bolzano and Pythagora, the mathematical and scientific world was forever changed; and the resulting technology was directly influenced. Although Bolzano inspired many works from genius' such as Poncelet and Chasles, the notion of vector addition has only been recently accepted, for many years the concept of vectors was murky and there was much confusion. However, early into the 20th century there was a rise of interest in vector analysis and its relation to architecture. [Many say that this may have been a result of the rise of technology and the need to understand two-dimensional figures before composing three-dimensonal ones. If you're interested, you can take a look at "The Evolution of the Idea of a Vectorial System" written by Michael J. Crowe.]

#### What's so fascinating about these vectors?Edit

What makes vector's such an interesting concept is their ability to have great impoartnace in ordinary life. Often, physics applications do not have much relevance on the common persons life; however, vectors are not as scary and complex as other physics terms may be. Vectors can be calculated with very little and simple mathematical procedures, yet have much significance in the "real world." Vectors can have such small importances as helping someone figure out which route is faster, and such big importances as solving for molecular distances within an object. The simple term "vector" connects numerous schools of thought together (mathematical, scientific, physical, and linguistical)!

#### Who else uses the concept of Graphical Vector Addition?Edit

Just as the Pythagorean Theorem is internationally famous, so too is graphical vector addition. Graphical vector addition has been around for centuries, thus it is not suprising that numerous countries utilize its functions and apply it to contemporary society (both in the real world and the simulated-school- world). In common society, most cultures use vector addition in very similar ways, and for very similar reasons. Either to solve for unknown distances, describe uniform motion, or create basic skeleton diagrams for 3-dimensional objects. For example, in modern society, students who are studying to become engineers use vector applications frequently as means of simplifying the visualiztion of 3-dimensional geometry, or calculating the area of a potential pyramid. However, a hundred years ago or so, vectors had various primary purposes depending on the culture it was used in. For example, in Ireland in the 1830's vectors were used as evidence to supporty the thoery of complex numbers, and their position on the plane. Hamilton, a devoted Irish mathematician, used vectors to represent complex numbers in a two-dimensional space, and attempted to add a third and fourth vector in hopes of finding a location for both the complex and imaginary numbers.

**Working with Vectors**Edit

#### Vector Addition: part 1, the basicsEdit

The simplest type of a vector operation is the addition of vectors in order to determine the resultant. —The resultant can also be referred to as the "resultant displacement" of an object, which means the distance the object is from its original position after every tranformation it undergoes.— [Often times students are familiar with a little bit of previous knowledge concerning the idea that the net (total) force on an object is the vector sum of all the individual forces acting on that object.] Assuming that the student is completely unfamiliar with net forces, etc. lets start at the beginning. We first denote one direction as positive and the other as negative. Lets assume that right is positive, and left is negative. Next we draw a vector pointing in a specific direction and give it a defined magnitude. Lets assume that the vector points to the right (positive direction) with a magnitude of 5. This means that any vector also pointing towards the right can be added to the first vector, regardless of the magnitude. Lets propose that this second vector has a magnitude of 5 also and it too points toward the right. These two vectors will be added as positive integers: 5 + 5 = 10. The resultant vector is 15. (This concept is being shown at right)Now, when one vector points in one direction, and the other vector points in the opposite direction, we assume that one is a positive vector (as it points in the positive direction) and the other is a negative vector (as it points in the negative direction); thus adding these two vectors becomes more like a subtraction problem. Lets suppose that the vector pointing towards the right has a magnitude of 5, and the vector pointing towards the left has a magnitude of 10. In order to add these two vectors we must add positive 5 and negative 10, which looks like: 5 + (-10) = 5 — 10 = (-5). The resultant vector would have a magniture of 5 in the direction opposite from the direction of the first vector.

Assume two vectors are equal in magnitude, but opposite in direction. Lets say they each have a magnitude of 5, then in this case our mathematical statement looks like: 5 + (-5) = 0. Then the two vectors will cancel eachother out and the resultant vector will be zero.

Before we try to do some problems on our own let's review the steps for properly adding vectors.

1. Label one direction positive (+) and the other direction negative (-).

2. Draw proper vectors, preferably to scale.

3. Position vectors so that their heads point in their respecatble directions.

4. Add them together using the basic laws of arithmetic.

5. Convert pos/neg vectors back into proper directions

(for More Tips on how to go about approaching an uncomfortable problem, see the Universtiy of Tennessee's high school physics internet site! http://electron4.phys.utk.edu )

##### Regents Review Sample ProblemsEdit

These questions were taken from The Physics Classroom website (www.physicsclassroom.com)

question: What is the resultant of 5N pointing north and 18N pointing south? (Don't forget to draw a diagram and label each part!)

solution: Lets say that north is positive, and south is negative. To add the two vectors we have 5N + (-18N) = (-13N). Thus our answer is 13N pointing south.

question: A brid flies north 3 kilmoeters and then south 4 kilometers, what is the resultant displacement of the bird?

solution: If we say that north is negative, and south is positive then we have a negative vector of magnitude 3km and a positive vector of magnitude 4km. The resultant is what we get when we add the two vectors: (-3km) + 4km = 1km. The resultant displacement of the bird is 1km.

question: Suppose a child walking home from school walks 2blocks east, until he realizes that he left his textbook at school and decides to walk the 2 blocks back and then walk to his house 8 blocks from school. How great is his resultant displacement from the point at which he realizes that he left his book at school?

solution: Lets say that east is positive and west is negative. The kid walks 2 positive blocks, 2 negative blocks, and then 8 more positive blocks. Except, that the question specifically asks for the childs resultant distance from the point 2 blocks away from school. This means that in reality, we only have 2negative blocks and 8positive blocks to add together. Thus, our vector addition looks like this: (-2) + 8 = 6. The child has a total resultant displacement of 6 blocks.

question: Two forces of 5 newtons and 15 newtos acting concurrently could have a resultant with a magnitude of
1) 5N
2) 10N
3) 25N
4) 75N

solution: The only answer here that complys with the properties of vectors is choice (2) 10N. This is because two forces of 5N and 15N can either be added where both 5N and 15N are positive vectors, or with 5N as the positive vector, and 15N as the negative vector, or 5N as the negative vector and 15N as the positive vector. Here are what the two possibilites look like: 5N + (-15N) = (-10N), which means 10N in the opposite direction from that of the original vector; or (-5N) + 15N = 10N, once again where the resultant vector is in the opposite direction of that of the original vector; the last possible answer would be 5N + 15N = 20N pointing in the same direction as the original vector. The only choice that fits these requirements, is 10N. Since the answer choices don't specify direction, we say that it is negligible, and thus 10N is the answer.

question: If a woman runs 100 meters north and then 70 meters south, her total displacement will be 1) 30m [N] 2) 30m [S] 3) 170m [N] 4) 170m [S]

solution: Lets say that north is positive and that south is negative. Now we can write an equation that looks like this: 100m + (-70m) = 30m. Since our answer is positive, we know that the resultant vector must be pointing north. Thus, our answer is 30m [N].

#### Vector addition: part 2, Graphical AdditionEdit

Another method of adding vectors entails maipulating their graphical representations on paper. Graphical Vector Addition is used for solving more advanced problems, such as questions that involve the placement of vectors coming from the north/south and east/west. When vectors are added graphically, they are to be positioned such that the head of one vector touched the tail of another vector. This specific positioning is called the "head-to-tail" method (Glencoe Phycsics Textbook, 67) which also entiails using a scaled vector diagram, where each vector is proportionate to the others in the same diagram. This alignment of these two vectors draws two sides of a triangle.

The diagram below illustrates the proper way to align vectors such that their heads touch the tails of the vector they are to be added to:

##### Right Triangle Vector AdditionEdit

When the directions of two vectors form a right angle, (such that one points north and one points east) the resultant vector is the hypotenuse of the triangle (the side that connects one leg to the other leg). The magnitude of this resultant vector can be found by using the Pythagorean theorem, where the values of the two legs of the triangle are substituted for “a” and “b” and the resultant vector is the “c” value. Thus the resultant vector is the square-root of the sum of the squares of the two separate vectors.

Some students find it easier to plot the scaled vectors on a coordinate graph, and then use the distance formula between the two endpoints of the vectors to solve for the resultant. However, when the vectors become more complicated and do not produce right triangles, this method proves to be futile. Thus, it is important that we understand the proper and efficent method of calculating resultant vectors, even if they may appear to be obvious at first.

This type of graphical vector addition is the most simple; these are the "plug-and-chug" problems where you have a simple formula and you just substitute in the values according to their respective locations. We have adapted the pythagorean theorem for this to be:

**r ^{2} = a^{2} + b^{2}**

[ http://hyperphysics.phy-ast.gsu.edu ]

##### Obtuse & Acute-Triangle Vector AdditionEdit

When two vectors are added at an angle different from 90degrees, the resultant vector cannot be calculated using the altered pythagorean theorem formula, it uses its own formula— the Law of Cosines: **r ^{2} = a^{2} + b^{2} — 2abCosX** (where "a," and "b" are the vector magnitues, and x is the angle between the two vectors). This equation only works when the vectors are placed head-to-tail; if the vectors are not originally in this format, then they must be shifted around (up or down, left or right) in order to be connected properly.

Occasionally, questions that involve odd triangles ask for the resultant vector and the angle it is displaced at. When this is the case, you must use the law of sines after solving for the magnitude of the resultant.

**(a/sinA) = (b/sinB) = (c/sinC)**

Remember when the question asks for the direction and the magnitude to give the correct units; for example: 30N at 25 degrees N of E. Being specific in your answers is key when dealing with vector quantities, because if you are too vague your answers can be interpreted incorrectly.

[ http://csep10.phys.utk.edu ]

#### Sample ProblemsEdit

##### Regents Style PrepEdit

The following questions are taken from The Physics Classroom website, and the Regents Exam Prep Center (www.regentsprep.org)

question: Find the magnitude of the sum of a 15-km displacement and a 25-km displacement when the angle between them is 135 degrees.

solution: Since it is given that the angle between the two vectors is not 90degrees, we know that we have to use the Law of Cosines formula to solve for the resultant.

r^{2} = a^{2} + b^{2} — 2abCosX
r^{2} = 15km^{2} + 25^{2} — 2(15km)(25km)cos135
r^{2} = 1380km
r = 37km

question: A girl swims across a stream 15m wide and then walks down the beach 10m, what was the resultant displacement if her custom shot glasses path in the water was perpendicular with the beach?

solution: Here we know that the angle between the two vectors is 90degrees, because it specifys that the stream is perpendicular to the beach. Because we know this we can use the equation r^{2} = a^{2} + b^{2}; r^{2} = 225m + 100m; r^{2} = 325m; r = 18m.

question: A car drives 11km N and then makes a turn and travels 11km E. What is the resultant displacement of the car?

solution: Since the problem specifies that the car went north and then east, we know that the vectors we're dealing with will form a right triangle. Now we can use the r^{2} = a^{2} + b^{2} formula; in this case "a" and "b" are 11km each.
r^{2} = (11)2 + (11)2 = 121 + 121 = 242
r = 15.55 = 15.6

question: Two forces of 10-newtons act concurrently on a point at an angle of 180-degrees to eachother. The magnitude of the resultant of the two forces is
1) 0N 2) 10.0N 3) 18.0N 4) 20.0N

solution: The wording of the question gives us all the key information we need! It explicitly states that there are two forces of 10N that act concurrently to eachother, this means that the forces are pointing in opposite directions to on another. Since the question also states that they are at an angle of 180-degrees, we know that these forces are on a flat surface. Because the two vectors are equal in magnitude, and opposite in direction, we konw that their forces will cancel eachother out and the resultant will be 0N, or choice (1). We can also show this answer mathematically by saying that to add these two vectors we must write the statement: 10N + (-10N) = 0N.

question: The resultant of a 12-newton force and a 7-newton force is 5 newtons. The angle between the forces is
1) 0 2) 45 3) 90 4) 180

solution: Since the two forces are each greater than the resultant force, we know that the vectors cannot create a triangle in order to solve for the angle. This is because the resultant side does not fall under the conditions of a triangle inequality (where the resultant is greater than the difference of the other two sides, and less that the sum of the other two sides.) Now we know that when vectors do not form a triangle, the only other way they can be added is along a flat surface, otherwise known as a 180-degree surface. If we assumed that the two vector forces pointed in opposite directions along a flat plane, then our mathematical statement would be: 12N + (-7N) = 5N. This is the answer as stated in the problem, thus we know that the angle between the two vectors is 180-degrees, or choice (4).

question: A resultant force of 10N is made up of two component forces acting at right angles to each other. If the magnitude of one of the components is 6N, then the magnitude of the other component must be
1) 16N 2) 8.0N 3) 6.0N 4) 4N

solution: Since the question tells us that the vectors meet at right angles, we know that we can use the equation r^{2} = a^{2} + b^{2}, in this case the value for "r" and "a" are known and "b" is what we are solving for. If we plug values into our equation we see that 100N = 36N + b2; 64N = r^{2}; 8N = b, or choice (2).

question: What is the resultant vector of two vector forces one of 30N and one of 17N with an angle of 135-degrees between them.

solution: Since the question states that the angle between the two vector forces is different from 90-degrees, we know that to sole this problem we must use the law of cosines equation. Plugging our values into the equation we have:
r^{2} = a^{2} + b^{2} — 2abCosX; r^{2} = 900N + 289N — (2)(30N)(17N)(cos135); r^{2} = 1910N; r = 43.7N; r = 44N. The displacement of the two vector forces is 44N.

**SAT II Graphical Vector Addition Problems**Edit

*Vector problems on national exams (such as the SAT Subject Test) tend to be rather straight-forawrd, as they do not deman a very sophisticared understanding of vectors. Occasionally, the Collegeboard will stick in a vector addition problem that requires the use of the Law of Cosines equation in order to properly solve for the resultant vector; however, most national exam questions are simple questions that are purely Pythagorean Theorem questions in "hiding." In fact NY State regents exam questions regarding vectors use more complicated fuctions! *

question: A hiker walks 4.5km in one direction, then makes a 45-degree turn to the right and walks another 6.4km. What is the magnitude of her displacement?

solution: The information in the problem describes a situation in which two vectors enclose an angle different from 90-degrees; this means that we must use the Law of Cosines equation to solve for the resultant.
r^{2} = a^{2} + b^{2} — 2(a)(b)CosX
r^{2} = (4.5Kkm)2 + (6.4km)2 — 2(4.5km)(6.4km)Cos45
r^{2} = 61.21km — 40.73km
r^{2} = 20.48km
r = 4.52km

question: Vector *A* has a magnitude of 9 and points due north, vector *B* has a magnitude of 3 and points due north, and vector *C* has a magnitude of 5 and points due west. What is the magnitude of the resultant vector *A + B + C*?

solution: First we must add the two parallel vectors that both point north. Since vectors *A* and *B* have magnitudes 9 and 3, respectively, the single vector that points north has a magnitude of 12. Now we need to find the resultant vector between vector *AB,* point due north and *C,* (which has a magnitude of 5) pointing due west. These directions tell us that the two vectors form a 90-degree angle; thus we must use the pythagorean theorem.
r^{2} = a^{2} + b^{2}
r^{2} = (12)2 + (5)2
r^{2} = 169
r = 13

question: A powerboad heads due northwest at 13m/s with respect to the water across a river that flows due north at 5.0 m/s. What is the velocity (both magnitude and direction) of the motorbaot with respect to the shore?

solution: In this question we know that we are dealing with the configuration of a right triangle, and of the information needed to solve for the unknown we have the *hypotenuse* value, (aka the resultant displacement) and one leg, our variable is the other leg. Plugging into the equation, our statement looks like this:
r^{2} = a^{2} + b^{2}
13m/s^{2} = 5m/s^{2} + b^{2}
144m/s = b^{2}
12m/s = b

question: An bird flies due north at 35km/h with respect to the air. There is a wind blowing at 65km/h to the northeast with respect to the ground. What are the bird's speed and direction with respect to the ground?

solution: Once again, we know that this problem sets up the need for a manipulation of the Pythagorean theorem. In our equation we can substitute in the information we are given, and solve for the unknown (one of the legs of the triangle).
r^{2} = a^{2} + b^{2}

65km/h2 = (35km/h)2 + b2

4225km/h = 1225 + b^{2}
3000km/h = b^{2}
54.77km/h = b
55km/h = b
Since the wind was blowing north*east* the bird 's speed must be directed east at 55km/h.

## RefrencesEdit

- More sample problems
- More Tips
- Graphical Vector Addition
- Right Triangle and vector addition
- Vectors that enclose angles other than 90 degrees
- Help if your completely confused
- {| border="0" cellpadding="0" cellspacing="0" width="146"

|- height="20" | class="xl65" height="20" style="height:15.0pt;width:110pt" width="146"|bingo online |}

## ResourcesEdit

- [1] An excellent website that reviews physics concepts on television and online. A great way to freshen-up on topics you might be a little confused with!
- [2] Terrific visuals, both 2-dimensional and 3-dimensional displays of graphical vector addition
- [3] A concise power-point presentation of the simple head-to-tail method of graphically adding vectors. This site is great for beginners, or anyone who needs to learn the material slowly.
- Barron's Regents Review Book Plenty of practice problems ranging from easy to hard. Perfect for preparing for the regents exam.
- Glencoe Physics Textbook Conscisely compiles all information regarding vector addition and relates it to the real world. An excellent source for the student who is interested in the "real world application."
- Kaplan's SAT II Physics Review Book A very comprehensive resource, covers all topics addressed in the physics SAT II —has more than just vector information in it! This study aid also comes with 2 full length tests and plenty of practice problems with tips!
- 10 Real SAT Subject Tests The closest you can get to taking the actual Physics SAT Subject test; the best source for a student who is preparing to take the physics SAT II.