Section Learning Objectives
By the end of this section, you will be able to do the following:
 Define components of vectors
 Describe the analytical method of vector addition and subtraction
 Use the analytical method of vector addition and subtraction to solve problems
Teacher Support
Teacher Support
The learning objectives in this section will help your students master the following standards:
 (3) Scientific processes. The student uses critical thinking, scientific reasoning, and problem solving to make informed decisions within and outside the classroom. The student is expected to:
 (F) express and interpret relationships symbolically in accordance with accepted theories to make predictions and solve problems mathematically, including problems requiring proportional reasoning and graphical vector addition
 (4) Science concepts. The student knows and applies the laws governing motion in two dimensions for a variety of situations. The student is expected to:
 (E) develop and interpret freebody force diagrams;
 (F) identify and describe motion relative to different frames of reference.
In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:
 (3) Scientific processes. The student uses critical thinking, scientific reasoning, and problem solving to make informed decisions within and outside the classroom. The student is expected to:
 (F) express and interpret relationships symbolically in accordance with accepted theories to make predictions and solve problems mathematically, including problems requiring proportional reasoning and graphical vector addition.
Section Key Terms
analytical method  component (of a twodimensional vector) 
Components of Vectors
For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the headtotail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17.
Figure 5.17 For a right triangle, the sine, cosine, and tangent of θ are defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, x is the adjacent side, y is the opposite side, and h is the hypotenuse.
Teacher Support
Teacher Support
[BL][OL] Review trigonometric concepts of sine, cosine, tangent and the Pythagorean theorem.
Since, by definition, $\text{cos}\theta =x/h$, we can find the length x if we know h and $\theta $ by using $x=h\text{cos}\theta $ . Similarly, we can find the length of y by using $y=h\text{sin}\theta $ . These trigonometric relationships are useful for adding vectors.
When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a twodimensional vector, a component is a piece of a vector that points in either the x or ydirection. Every 2d vector can be expressed as a sum of its x and y components.
For example, given a vector like $A$ in Figure 5.18, we may want to find what two perpendicular vectors, ${A}_{x}$ and ${A}_{y}$, add to produce it. In this example, ${A}_{x}$ and ${A}_{y}$ form a right triangle, meaning that the angle between them is 90 degrees. This is a common situation in physics and happens to be the least complicated situation trigonometrically.
Figure 5.18 The vector $A$, with its tail at the origin of an x ycoordinate system, is shown together with its x and ycomponents, ${A}_{x}$ and ${A}_{y}.$ These vectors form a right triangle.
${A}_{x}$ and ${A}_{y}$ are defined to be the components of $A$ along the x and yaxes. The three vectors, $A$, ${A}_{x}$, and ${A}_{y}$, form a right triangle.
$${A}_{x}+{A}_{y}=A$$
If the vector $A$ is known, then its magnitude $A$ (its length) and its angle $\theta $ (its direction) are known. To find ${A}_{x}$ and ${A}_{y}$, its x and ycomponents, we use the following relationships for a right triangle:
$${A}_{x}=A\text{cos}\theta $$
and
$${A}_{y}=A\text{sin}\mathrm{\theta ,}$$
where ${A}_{x}$ is the magnitude of A in the xdirection, ${A}_{y}$ is the magnitude of A in the ydirection, and $\theta $ is the angle of the resultant with respect to the xaxis, as shown in Figure 5.19.
Figure 5.19 The magnitudes of the vector components ${A}_{x}$ and ${A}_{y}$ can be related to the resultant vector $A$ and the angle $\theta $ with trigonometric identities. Here we see that ${A}_{x}=A\text{cos}\theta $ and ${A}_{y}=A\text{sin}\theta .$
Teacher Support
Teacher Support
[BL][OL][AL] Derive the formula for getting the magnitude and direction of a vector.
Misconception Alert
Students might be confused between the relationship ${A}_{x}+{A}_{y}=A$, which shows the addition of vectors and $A=\sqrt{{A}_{x}{}^{2}+{A}_{y}{}^{2}}$ which shows the addition of magnitudes of vectors.
Suppose, for example, that $A$ is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20.
Figure 5.20 We can use the relationships ${A}_{x}=A\text{cos}\theta $ and ${A}_{y}=A\text{sin}\theta $ to determine the magnitude of the horizontal and vertical component vectors in this example.
Then A = 10.3 blocks and $\theta ={29.1}^{\circ}$, so that
$$\begin{array}{ccc}\hfill {A}_{x}& =& A\text{cos}\theta \hfill \\ & =& (10.3\phantom{\rule{0ex}{0ex}}\text{blocks)(cos29}{\text{.1}}^{\circ})\hfill \\ & =& (10.3\phantom{\rule{0ex}{0ex}}\text{blocks)(0}\text{.874)}\hfill \\ & =& \text{9}\text{.0blocks.}\hfill \end{array}$$
5.6
This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9block eastward displacement. Similarly,
$$\begin{array}{ccc}\hfill {A}_{y}& =& A\text{sin}\theta \hfill \\ & =& (10.3\phantom{\rule{0ex}{0ex}}\text{blocks)(sin29}{\text{.1}}^{\circ})\hfill \\ & =& (10.3\phantom{\rule{0ex}{0ex}}\text{blocks)(0}\text{.846)}\hfill \\ & =& \text{5}\text{.0blocks,}\hfill \end{array}$$
5.7
indicating that the walker has traveled 5 blocks to the north—a 5block northward displacement.
Analytical Method of Vector Addition and Subtraction
Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components ${A}_{x}$ and ${A}_{y}$ of a vector $A$ are known, then we can find $A$ analytically. How do we do this? Since, by definition,
$$\text{tan}\theta =y/x\text{(orinthiscasetan}\theta ={A}_{y}/{A}_{x}\text{),}$$
we solve for $\theta $ to find the direction of the resultant.
$$\theta ={\mathrm{tan}}^{1}({A}_{y}/{A}_{x})$$
Note that ${\text{tan}}^{1}(\theta )$ gives an angle in the first quadrant if ${A}_{y}/{A}_{x}\phantom{\rule{0ex}{0ex}}>\phantom{\rule{0ex}{0ex}}0$ and in the fourth quadrant if ${A}_{y}/{A}_{x}\phantom{\rule{0ex}{0ex}}<\phantom{\rule{0ex}{0ex}}0$. If, in fact, both ${A}_{x}$ and ${A}_{y}$ are negative, or if ${A}_{x}$ is negative and ${A}_{y}$ positive, then $\theta $, measured from the positive $x$ direction, is ${\text{tan}}^{\u20131}(\theta )+180\text{\xb0}$.
Since this is a right triangle, the Pythagorean theorem (x^{2} + y^{2} = h^{2}) for finding the hypotenuse applies. In this case, it becomes
$${A}^{2}={A}_{x}^{2}+{A}_{y}^{2}\text{.}$$
Solving for A gives
$$A=\sqrt{{A}_{x}{}^{2}+{A}_{y}{}^{2}}\text{.}$$
In summary, to find the magnitude $A$ and direction $\theta $ of a vector from its perpendicular components ${A}_{x}$ and ${A}_{y}$, as illustrated in Figure 5.21, we use the following relationships:
$$\begin{array}{ccc}\hfill A& =& \sqrt{{A}_{x}{}^{2}+{A}_{y}{}^{2}}\hfill \\ & & \theta ={\mathrm{tan}}^{1}({A}_{y}/{A}_{x})\hfill \end{array}$$
Figure 5.21 The magnitude and direction of the resultant vector $A$ can be determined once the horizontal components ${A}_{x}$ and ${A}_{y}$ have been determined.
Teacher Support
Teacher Support
[BL][OL][AL] Demonstrate a problem involving displacement by physically walking along the specified direction. Show how this can be represented on a graph. Explain that even when solving problems analytically; representing it on a graph would make it easier to visualize the problem.
Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors $A$ and $B$ are added to produce the resultant $R\text{,}$ as illustrated in Figure 5.22.
Figure 5.22 Vectors $A$ and $B$ are two legs of a walk, and $R$ is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of $R$ .
If $A$ and $B$ represent two legs of a walk (two displacements), then $R$ is the total displacement. The person taking the walk ends up at the tip of $R$ . There are many ways to arrive at the same point. The person could have walked straight ahead first in the xdirection and then in the ydirection. Those paths are the x and ycomponents of the resultant, ${R}_{x}$ and ${R}_{y}.$ If we know ${R}_{x}$ and ${R}_{y}$, we can find $R$ and $\theta $ using the equations $R=\sqrt{{R}_{\text{x}}{}^{2}+{R}_{\text{y}}{}^{2}}$ and $\theta =ta{n}^{\u20131}({R}_{y}/{R}_{x})$ .
 Draw in the x and y components of each vector (including the resultant) with a dashed line. Use the equations ${A}_{x}=A\text{cos}\theta $ and ${A}_{y}=A\text{sin}\theta $ to find the components. In Figure 5.23, these components are ${A}_{x}$, ${A}_{y}$, ${B}_{x}$, and ${B}_{y}.$ Vector $A$ makes an angle of ${\theta}_{A}$ with the xaxis, and vector $B$ makes and angle of ${\theta}_{B}$ with its own xaxis (which is slightly above the xaxis used by vector A).
Figure 5.23 To add vectors $A$ and $B,$ first determine the horizontal and vertical components of each vector. These are the dotted vectors ${A}_{x},$ ${A}_{y}$ ${B}_{y}$ shown in the image.
 Find the x component of the resultant by adding the x component of the vectors
$${R}_{x}={A}_{x}+{B}_{x}$$
and find the y component of the resultant (as illustrated in Figure 5.24) by adding the y component of the vectors.
$${R}_{y}={A}_{y}+{B}_{y}\text{.}$$
Figure 5.24 The vectors ${A}_{x}$ and ${B}_{x}$ add to give the magnitude of the resultant vector in the horizontal direction, ${R}_{\text{x}}.$ Similarly, the vectors ${A}_{y}$ and ${B}_{y}$ add to give the magnitude of the resultant vector in the vertical direction, ${R}_{\text{y}}.$
Now that we know the components of $R,$ we can find its magnitude and direction.
 To get the magnitude of the resultant R, use the Pythagorean theorem.
$$R=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}$$
 To get the direction of the resultant
$$\theta ={\mathrm{tan}}^{1}({R}_{y}/{R}_{x})\text{.}$$
Watch Physics
Classifying Vectors and Quantities Example
This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions.
Access multimedia content
Three vectors, $\overrightarrow{\text{u}}$, $\overrightarrow{\text{v}}$, and $\overrightarrow{\text{w}}$, have the same magnitude of $5\phantom{\rule{0ex}{0ex}}\text{units}$. Vector $\overrightarrow{\text{v}}$ points to the northeast. Vector $\overrightarrow{\text{w}}$ points to the southwest exactly opposite to vector $\overrightarrow{\text{u}}$. Vector $\overrightarrow{\text{u}}$ points in the northwest. If the vectors $\overrightarrow{\text{u}}$, $\overrightarrow{\text{v}}$, and $\overrightarrow{\text{w}}$ were added together, what would be the magnitude of the resultant vector? Why?

$0\phantom{\rule{0ex}{0ex}}\text{units}$. All of them will cancel each other out.

$5\phantom{\rule{0ex}{0ex}}\text{units}$. Two of them will cancel each other out.

$10\phantom{\rule{0ex}{0ex}}\text{units}$. Two of them will add together to give the resultant.

$15$ units. All of them will add together to give the resultant.
Tips For Success
In the video, the vectors were represented with an arrow above them rather than in bold. This is a common notation in math classes.
Using the Analytical Method of Vector Addition and Subtraction to Solve Problems
Figure 5.25 uses the analytical method to add vectors.
Worked Example
An Accelerating Subway Train
Add the vector $A$ to the vector $B$ shown in Figure 5.25, using the steps above. The xaxis is along the east–west direction, and the yaxis is along the north–south directions. A person first walks $\text{53}\text{.0m}$ in a direction $\text{20}\text{.0\xb0}$ north of east, represented by vector $A.$ The person then walks $\text{34}\text{.0m}$ in a direction $\text{63}\text{.0}\xb0$ north of east, represented by vector $B.$
Figure 5.25 You can use analytical models to add vectors.
Strategy
The components of $A$ and $B$ along the x and yaxes represent walking due east and due north to get to the same ending point. We will solve for these components and then add them in the xdirection and ydirection to find the resultant.
Solution
First, we find the components of $A$ and $B$ along the x and yaxes. From the problem, we know that $A=53.0\phantom{\rule{0ex}{0ex}}\text{m,}$ ${\theta}_{\text{A}}={20.0}^{\circ ,}$ $B$ = $\text{34}\text{.0m}$, and ${\theta}_{\text{B}}={63.0}^{\circ}$ . We find the xcomponents by using ${A}_{x}=A\mathrm{cos}\theta $, which gives
$$\begin{array}{ccc}\hfill {A}_{x}& =& A\mathrm{cos}{\theta}_{A}=(53.0\phantom{\rule{0ex}{0ex}}\text{m})(\mathrm{cos}{20.0}^{\circ})\hfill \\ & =& (53.0\phantom{\rule{0ex}{0ex}}\text{m})(0.940)=49.8\phantom{\rule{0ex}{0ex}}\text{m}\hfill \end{array}$$
and
$$\begin{array}{ccc}\hfill {B}_{x}& =& B\mathrm{cos}{\theta}_{B}=(34.0\phantom{\rule{0ex}{0ex}}\text{m})(\mathrm{cos}{63.0}^{\circ})\hfill \\ & =& (34.0\phantom{\rule{0ex}{0ex}}\text{m})(0.454)=15.4\phantom{\rule{0ex}{0ex}}\text{m.}\hfill \end{array}$$
Similarly, the ycomponents are found using ${A}_{y}=A\mathrm{sin}{\theta}_{A}$
$$\begin{array}{ccc}\hfill {A}_{y}& =& A\mathrm{sin}{\theta}_{A}=(53.0\phantom{\rule{0ex}{0ex}}\text{m})(\mathrm{sin}{20.0}^{\circ})\hfill \\ & =& (53.0\phantom{\rule{0ex}{0ex}}\text{m})(0.342)=18.1\phantom{\rule{0ex}{0ex}}\text{m}\hfill \end{array}$$
and
$$\begin{array}{ccc}\hfill {B}_{y}& =& B\mathrm{sin}{\theta}_{B}=(34.0\phantom{\rule{0ex}{0ex}}\text{m})(\mathrm{sin}{63.0}^{\circ})\hfill \\ & =& (34.0\phantom{\rule{0ex}{0ex}}\text{m})(0.891)=30.3\phantom{\rule{0ex}{0ex}}\text{m}\text{.}\hfill \end{array}$$
The x and ycomponents of the resultant are
$${R}_{x}={A}_{x}+{B}_{x}=49.8\phantom{\rule{0ex}{0ex}}\text{m}+15.4\phantom{\rule{0ex}{0ex}}\text{m}=65.2\phantom{\rule{0ex}{0ex}}\text{m}$$
and
$${R}_{y}={A}_{y}+{B}_{y}=18.1\phantom{\rule{0ex}{0ex}}\text{m}+30.3\phantom{\rule{0ex}{0ex}}\text{m}=48.4\phantom{\rule{0ex}{0ex}}\text{m}.$$
Now we can find the magnitude of the resultant by using the Pythagorean theorem
$$R=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}=\sqrt{{(65.2)}^{2}+{(48.4)}^{2}}\text{m}$$
5.8
so that
$$R=\sqrt{6601\phantom{\rule{0ex}{0ex}}\text{m}}=81.2\phantom{\rule{0ex}{0ex}}\text{m}\text{.}$$
Finally, we find the direction of the resultant
$$\theta ={\mathrm{tan}}^{1}({R}_{y}/{R}_{x})=+{\mathrm{tan}}^{1}(48.4/65.2).$$
This is
$$\theta ={\mathrm{tan}}^{1}(0.742)={36.6}^{\circ}.$$
Discussion
This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, $\text{A}\text{B}\equiv \text{A}+(\text{B})$ . The components of – $\text{B}$ are the negatives of the components of $\text{B}$ . Therefore, the x and ycomponents of the resultant $\text{A}\text{B}=\text{R}$ are
$${R}_{x}={A}_{x}+{B}_{x}$$
and
$${R}_{y}={A}_{y}+{B}_{y}$$
and the rest of the method outlined above is identical to that for addition.
Practice Problems
5.
What is the magnitude of a vector whose xcomponent is 4 cm and whose ycomponent is 3 cm?
 1 cm
 5 cm
 7 cm
 25 cm
6.
What is the magnitude of a vector that makes an angle of 30° to the horizontal and whose xcomponent is 3 units?
 2.61 units
 3.00 units
 3.46 units
 6.00 units
Links To Physics
Atmospheric Science
Figure 5.26 This picture shows Bert Foord during a television Weather Forecast from the Meteorological Office in 1963. (BBC TV)
Atmospheric science is a physical science, meaning that it is a science based heavily on physics. Atmospheric science includes meteorology (the study of weather) and climatology (the study of climate). Climate is basically the average weather over a longer time scale. Weather changes quickly over time, whereas the climate changes more gradually.
The movement of air, water and heat is vitally important to climatology and meteorology. Since motion is such a major factor in weather and climate, this field uses vectors for much of its math.
Vectors are used to represent currents in the ocean, wind velocity and forces acting on a parcel of air. You have probably seen a weather map using vectors to show the strength (magnitude) and direction of the wind.
Vectors used in atmospheric science are often threedimensional. We won’t cover threedimensional motion in this text, but to go from twodimensions to threedimensions, you simply add a third vector component. Threedimensional motion is represented as a combination of x, y and z components, where z is the altitude.
Vector calculus combines vector math with calculus, and is often used to find the rates of change in temperature, pressure or wind speed over time or distance. This is useful information, since atmospheric motion is driven by changes in pressure or temperature. The greater the variation in pressure over a given distance, the stronger the wind to try to correct that imbalance. Cold air tends to be more dense and therefore has higher pressure than warm air. Higher pressure air rushes into a region of lower pressure and gets deflected by the spinning of the Earth, and friction slows the wind at Earth’s surface.
Finding how wind changes over distance and multiplying vectors lets meteorologists, like the one shown in Figure 5.26, figure out how much rotation (spin) there is in the atmosphere at any given time and location. This is an important tool for tornado prediction. Conditions with greater rotation are more likely to produce tornadoes.
Why are vectors used so frequently in atmospheric science?

Vectors have magnitude as well as direction and can be quickly solved through scalar algebraic operations.

Vectors have magnitude but no direction, so it becomes easy to express physical quantities involved in the atmospheric science.

Vectors can be solved very accurately through geometry, which helps to make better predictions in atmospheric science.

Vectors have magnitude as well as direction and are used in equations that describe the three dimensional motion of the atmosphere.
Check Your Understanding
7.
Between the analytical and graphical methods of vector additions, which is more accurate? Why?
 The analytical method is less accurate than the graphical method, because the former involves geometry and trigonometry.
 The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations.
 The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale.
 The analytical method is more accurate than the graphical method, because the latter is limited by the precision of the drawing.
8.
What is a component of a two dimensional vector?
 A component is a piece of a vector that points in either the x or y direction.
 A component is a piece of a vector that has half of the magnitude of the original vector.
 A component is a piece of a vector that points in the direction opposite to the original vector.
 A component is a piece of a vector that points in the same direction as original vector but with double of its magnitude.
9.
How can we determine the global angle $\theta $ (measured counterclockwise from positive $x$) if we know ${A}_{x}$ and ${A}_{y}$?

$\theta ={\mathrm{cos}}^{1}\frac{{A}_{y}}{{A}_{x}}$

$\theta ={\mathrm{cot}}^{1}\frac{{A}_{y}}{{A}_{x}}$

$\theta ={\mathrm{sin}}^{1}\frac{{A}_{y}}{{A}_{x}}$

$\theta ={\mathrm{tan}}^{1}\frac{{A}_{y}}{{A}_{x}}$
10.
How can we determine the magnitude of a vector if we know the magnitudes of its components?
 $\left\overrightarrow{\text{A}}\right={A}_{x}+{A}_{y}$
 $\left\overrightarrow{\text{A}}\right={{A}_{x}}^{2}+{{A}_{y}}^{2}$
 $\left\overrightarrow{\text{A}}\right=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}$
 $\left\overrightarrow{\text{A}}\right=({{A}_{x}}^{2}+{{A}_{y}}^{2}{)}^{2}$
Teacher Support
Teacher Support
Use the Check Your Understanding questions to assess whether students achieve the learning objectives for this section. If students are struggling with a specific objective, the Check Your Understanding will help identify which objective is causing the problem and direct students to the relevant content.