Adding Vectors Analytically

OpenStax; Heath Hatch

30 Adding Vectors Analytically

OpenStax and Heath Hatch

Learning Objectives

This material is presented both as videos by Heath Hatch and with the relevant OpenStax textbook. Either method is fine. Regardless, by the end of this section, you should be able to:

Understand the rules of vector addition and subtraction using analytical methods.
Apply analytical methods to determine the magnitude and direction of a resultant vector.

Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 5, in which the vectors $\vec{A}$ and $\vec{B}$ are added to produce the resultant $\vec{R}.$

Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown.

If $\vec{A}$ and $\vec{B}$ represent two legs of a walk (two displacements), then $\vec{R}$ is the total displacement. The person taking the walk ends up at the tip of $\vec{R}.$ There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x– and y-components of the resultant, ${\vec{R}_x}$ and ${\vec{R}_y}.$ If we know ${\vec{R}_x}$ and ${\vec{R}_y},$ we can find $\vec{R}$ and ${\theta}$ using the equations ${\vec{A}=\sqrt{{\vec{A}_x}^2\:+\:{\vec{A}_y}^2}}$ and ${\theta=\tan ^{-1}(\vec{A}_y\:/\:\vec{A}_x)}.$ When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations ${\vec{A}_x=A \cos\:\theta}$ and ${\vec{A}_y=A \sin\:\theta}$ to find the components. In Figure 6, these components are ${\vec{A}_x},{\vec{A}_x},{\vec{B}_x},$ and ${\vec{B}_y}.$ The angles that vectors $\vec{A}$ and $\vec{B}$ make with the x-axis are ${\theta_{A}}$ and ${\theta_{B}},$ respectively.

Two vectors A and B are shown. The tail of the vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The horizontal and vertical components of the vectors A and B are shown with the help of dotted lines. The vectors labeled as A sub x and A sub y are the components of vector A, and B sub x and B sub y as the components of vector B..

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 7,

${\vec{R}_x=\vec{A}_x+\vec{B}_x}$

and

${\vec{R}_y=\vec{A}_y+\vec{B}_y.}$

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of $\vec{R}$ are known, its magnitude and direction can be found.

Step 3. To get the magnitude $\vec{R}$ of the resultant, use the Pythagorean theorem:

${\vec{R}=\sqrt{{\vec{R}_x}^2\:+\:\vec{R}_y^2}.}$

Step 4. To get the direction of the resultant:

${\theta=\tan ^{-1}(\vec{R}_y\:/\:\vec{R}_x).}$

The following example illustrates this technique for adding vectors using perpendicular components.

Example 1: Adding Vectors Using Analytical Methods

Add the vector $\vec{A}$ to the vector $\vec{B}$ shown in Figure 8, using perpendicular components along the x– and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively. Vector $\vec{A}$ represents the first leg of a walk in which a person walks ${53.0\, \mathrm{m}}$ in a direction ${20.0^o}$ north of east. Vector $\vec{B}$ represents the second leg, a displacement of ${34.0\vec{ min}}$ a direction ${63.0^o}$ north of east.

Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B.

Strategy

The components of $\vec{A}$ and $\vec{B}$ along the x– and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of $\vec{A}$ and $\vec{B}$ along the x– and y-axes. Note that ${\vec{A}=53.0\, \mathrm{m}},{\theta_A=20.0^o},{\vec{B}=34.0\, \mathrm{m}},$ and ${\theta_B=63.0^o}.$ We find the x-components by using ${\vec{A}_x=A \cos\:\theta},$ which gives

${\vec{A}_x=A \cos\:\theta_A=(53.0\, \mathrm{m})(\cos 20.0^o)}$

${=(53.0\, \mathrm{m})(0.940)=49.8\, \mathrm{m}}\:\:\:\:\:$

and

${\vec{B}_x=B \cos\:\theta_B=(34.0\, \mathrm{m})(\cos 63.0^o)}$

${=(34.0\, \mathrm{m})(0.454)=15.4\, \mathrm{m}.}\:\:\:\:\:$

Similarly, the y-components are found using ${\vec{A}_y=A \sin\:\theta}:$

${\vec{A}_y=A \sin\:\theta_A=(53.0\, \mathrm{m})(\sin 20.0^o)}$

${=(53.0\, \mathrm{m})(0.342)=18.1\, \mathrm{m}}\:\:\:\:\:$

and

${\vec{B}_y=B \sin\:\theta_B=(34.0\, \mathrm{m})(\sin 63.0^o)}$

${=(34.0\, \mathrm{m})(0.891)=30.3\, \mathrm{m}.}\:\:\:\:\:$

The x– and y-components of the resultant are thus

${\vec{R}_x=\vec{A}_x\:+\:\vec{B}_x=49.8\, \mathrm{m}\:+\:15.4\, \mathrm{m}=65.2\, \mathrm{m}}$

and

${\vec{R}_y=\vec{A}_y\:+\:\vec{B}_y=18.1\, \mathrm{m}\:+\:30.3\, \mathrm{m}=48.4\, \mathrm{m}.}$

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

${\vec{R}=\sqrt{\vec{R}_x^2+\vec{R}_y^2}=\sqrt{(65.2)^2+(48.4)^2\, \mathrm{m}}}$

so that

${\vec{R}=81.2\, \mathrm{m}.}$

Finally, we find the direction of the resultant:

${\theta=\tan ^{-1}(\vec{R}_y\:/\:\vec{R}_x)=+\tan ^{-1}(48.4/65.2).}$

Thus,

${\theta=\tan ^{-1}(0.742)=36.6^o.}$

The addition of two vectors A and B is shown. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. Vector B is of magnitude thirty four units and is inclined at angle sixty three degrees to the horizontal. The components of vector A are shown as dotted vectors A X is equal to forty nine point eight meter along x axis and A Y is equal to eighteen point one meter along Y axis. The components of vector B are also shown as dotted vectors B X is equal to fifteen point four meter and B Y is equal to thirty point three meter. The horizontal component of the resultant R X is equal to A X plus B X is equal to sixty five point two meter. The vertical component of the resultant R Y is equal to A Y plus B Y is equal to forty eight point four meter. The magnitude of the resultant of two vectors is eighty one point two meters. The direction of the resultant R is in thirty six point six degree from the vector A in anticlockwise direction.

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, ${\vec{A}-\vec{B}\equiv\vec{A}+(-\vec{B})}.$ Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of ${-\vec{B}}$ are the negatives of the components of $\vec{B}.$ The x– and y-components of the resultant ${\vec{A}-\vec{B}=\vec{R}}$ are thus

${\vec{R}_x=\vec{A}_x\:+\:(-\vec{B}_x)}$

and

${\vec{R}_y=\vec{A}_y\:+\:(-\vec{B}_y)}$

and the rest of the method outlined above is identical to that for addition. (See Figure 10.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Chapter 3.4 Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.

In this figure, the subtraction of two vectors A and B is shown. A red colored vector A is inclined at an angle theta A to the positive of x axis. From the head of vector A a blue vector negative B is drawn. Vector B is in west of south direction. The resultant of the vector A and vector negative B is shown as a black vector R from the tail of vector A to the head of vector negative B. The resultant R is inclined to x axis at an angle theta below the x axis. The components of the vectors are also shown along the coordinate axes as dotted lines of their respective colors. — **Figure 10.** The subtraction of the two vectors shown in Figure 5. The components of -B are the negatives of the components of B. The method of subtraction is the same as that for addition.

PHET EXPLORATIONS: VECTOR ADDITION

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

Summary

The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
The steps to add vectors $\vec{A}$ and $\vec{B}$ using the analytical method are as follows:
Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

${\vec{A}_x=A \cos\:\theta}$

${\vec{B}_x=B \cos\:\theta}$

and

${\vec{A}_y=A \sin\:\theta}$

${\vec{B}_y=B \sin\:\theta}.$

Step 2: Add the horizontal and vertical components of each vector to determine the components ${\vec{R}_x}$ and ${\vec{R}_y}$ of the resultant vector, $\vec{R}:$

${\vec{R}_x=\vec{A}_x+\vec{B}_x}$

and

${\vec{R}_y=\vec{A}_y+\vec{B}_y}.$

Step 3: Use the Pythagorean theorem to determine the magnitude, ${R},$ of the resultant vector $\vec{R}:$

${\vec{R}=\sqrt{{\vec{R}_x}^2\:+\:{\vec{R}_y}^2}.}$

Step 4: Use a trigonometric identity to determine the direction, ${\theta},$ of $\vec{R}:$

${\theta=\tan ^{-1}(\vec{R}_y\:/\:\vec{R}_x).}$

Conceptual Questions

1: Suppose you add two vectors $\vec{A}$ and $\vec{B}.$ What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?

2: Give an example of a nonzero vector that has a component of zero.

3: Explain why a vector cannot have a component greater than its own magnitude.

4: If the vectors $\vec{A}$ and $\vec{B}$ are perpendicular, what is the component of $\vec{A}$ along the direction of $\vec{B}?$ What is the component of $\vec{B}$ along the direction of $\vec{A}?$

Problems & Exercises

1: Find the following for path C in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

A map of city is shown. The houses are in form of square blocks of side one hundred and twenty meter each. Four paths A B C and D are shown in different colors. The path c shown as blue extends to one block towards north, then five blocks towards east and then two blocks towards south then one block towards west and one block towards north and finally three blocks towards west. It is asked to find out the total distance traveled the magnitude and the direction of the displacement from start to finish for path C.

2: Find the following for path D in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

3: Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $\vec{A}$ and $\vec{B},$ as in Figure 14, then this problem asks you to find their sum ${\vec{R}=\vec{A}+\vec{B}}.)$

In the given figure displacement of a person is shown. First movement of the person is shown as vector A from origin along negative x axis. He then turns to his right. His movement is now shown as a vertical vector in north direction. The displacement vector R is also shown. In the question you are asked to find the displacement of the person from the start to finish.

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.

4: Repeat Exercise 4 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, ${\vec{B}+\vec{A} =\vec{A}+\vec{B}}.)$ Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

5: Do Exercise 4 again using analytical techniques and change the second leg of the walk to ${25.0\, \mathrm{m}}$ straight south. (This is equivalent to subtracting $\vec{B}$ from $\vec{A}$ —that is, finding ${\vec{R}^{\prime}=\vec{A} -\vec{B}})$ (b) Repeat again, but now you first walk ${25.0\, \mathrm{m}}$ north and then ${18.0\, \mathrm{m}}$ east. (This is equivalent to subtract $\vec{A}$ from $\vec{B}$ —that is, to find ${\vec{A}=\vec{B}+\vec{C}}.$ Is that consistent with your result?)

6: A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors $\vec{A}$ from $\vec{B}$ in Figure 15. She then correctly calculates the length and orientation of the third side $\vec{C}.$ What is her result?

In the given figure the sides of a triangular piece of land are shown in vector form. West corner is at origin. A vector starts from the origin towards south east direction and makes an angle twenty-one degrees with the horizontal. Then from the head of this vector another vector B making an angle eleven degrees with the vertical is drawn upwards. Then another vector C from the head of the vector B to the tail of the initial vector is drawn. The length and orientation of side C is indicated as unknown, represented by a question mark. — **Figure 15.**

7: A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as $\vec{A},\vec{B},$ and $\vec{C}$ in Figure 16, and then correctly calculates the length and orientation of the fourth side $\vec{D}.$
What is his result?

A quadrilateral with sides A, B, C, and D. A begins at the end of D and is 4 point seven zero kilometers at an angle of 7 point 5 degrees south of west. B begins at the end of A and is 2 point four eight kilometers in a direction sixteen degrees west of north. C begins at the end of B and is 3 point zero 2 kilometers in a direction nineteen degrees north of west. D begins at the end of C and runs distance and direction that must be calculated — **Figure 16.**

8: In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: ${2.50\vec{ km}\:45.0^o}$ north of west; then ${4.70\vec{ km}\:60.0^o}$ south of east; then ${1.30\vec{ km}\:25.0^o}$ south of west; then ${5.10\vec{ km}}$ straight east; then ${1.70\vec{ km}\:5.00^o}$ east of north; then ${7.20\vec{ km}\:55.0^o}$ south of west; and finally ${2.80\vec{ km}\:10.0^o}$ north of east. What is his final position relative to the island?

9: Suppose a pilot flies ${40.0\vec{ km}}$ in a direction ${60^o}$ north of east and then flies ${30.0\vec{ km}}$ in a direction ${15^o}$ north of east as shown in Figure 17. Find her total distance $\vec{R}$ from the starting point and the direction ${\theta}$ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

A triangle defined by vectors A, B, and R. A begins at the origin and run forty kilometers in a direction sixty degrees north of east. B begins at the end of A and runs thirty kilometers in a direction fifteen degrees north of east. R is the resultant vector and runs from the origin (the beginning of A) to the end of B for a distance and in a direction theta that need to be calculated. — **Figure 17.**

Glossary

analytical method: the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities

Solutions

Problems & Exercises

1:

(a) 1.56 km

(b) 120 m east

4:

30.8 m, 35.8 west of north

5:

(a) ${30.8\, \mathrm{m}},{54.2^o}$ south of west

(b) ${30.8\, \mathrm{m}},{54.2^o}$ north of east

8:

${7.34\vec{ km}},{63.5^o}$ south of east

Homework Problems

10. Adding vectors by components.

11. A bit more practice adding vectors by components.

License

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Physics 131: What Is Physics? by OpenStax and Heath Hatch is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.