Vector Addition and Subtraction: Analytical Methods

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

Section Key Terms

analytical method

component (of a two-dimensional 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 head-to-tail 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.

Since, by definition, cosθ=x/hcos𝜃=𝑥/ℎ, we can find the length x if we know h and θ𝜃 by using x=hcosθ𝑥=ℎcos𝜃 . Similarly, we can find the length of y by using y=hsinθ𝑦=ℎsin𝜃 . 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 two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components.

For example, given a vector like A𝐴 in Figure 5.18, we may want to find what two perpendicular vectors, Ax𝐴𝑥 and Ay𝐴𝑦, add to produce it. In this example, Ax𝐴𝑥 and Ay𝐴𝑦 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 A𝐴, with its tail at the origin of an x– y-coordinate system, is shown together with its x– and y-components, Ax𝐴𝑥 and Ay.𝐴𝑦. These vectors form a right triangle.

Ax𝐴𝑥 and Ay𝐴𝑦 are defined to be the components of A𝐴 along the x– and y-axes. The three vectors, A𝐴, Ax𝐴𝑥, and Ay𝐴𝑦, form a right triangle.

Ax + Ay = A𝐴𝑥 + 𝐴𝑦 = 𝐴

If the vector A𝐴 is known, then its magnitude A𝐴 (its length) and its angle θ𝜃 (its direction) are known. To find Ax𝐴𝑥 and Ay𝐴𝑦, its x– and y-components, we use the following relationships for a right triangle:

Ax=Acosθ𝐴𝑥=𝐴cos𝜃

and

Ay=Asinθ,𝐴𝑦=𝐴sinθ,

where Ax𝐴𝑥 is the magnitude of A in the x-direction, Ay𝐴𝑦 is the magnitude of A in the y-direction, and θ𝜃 is the angle of the resultant with respect to the x-axis, as shown in Figure 5.19.

Figure 5.19 The magnitudes of the vector components Ax𝐴𝑥 and Ay𝐴𝑦 can be related to the resultant vector A𝐴 and the angle θ𝜃 with trigonometric identities. Here we see that Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ.𝐴𝑦=𝐴sin𝜃.

Suppose, for example, that A𝐴 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 Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃 to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A = 10.3 blocks and θ=29.1∘𝜃=29.1∘, so that

Ax====Acosθ(10.3 blocks)(cos29.1∘)(10.3 blocks)(0.874)9.0 blocks.𝐴𝑥=𝐴cos𝜃=(10.3 blocks)(cos29.1∘)=(10.3 blocks)(0.874)=9.0 blocks.

5.6

This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly,

Ay====Asinθ(10.3 blocks)(sin29.1∘)(10.3 blocks)(0.846)5.0 blocks,𝐴𝑦=𝐴sin𝜃=(10.3 blocks)(sin29.1∘)=(10.3 blocks)(0.846)=5.0 blocks,

5.7

indicating that the walker has traveled 5 blocks to the north—a 5-block 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 Ax𝐴𝑥 and Ay𝐴𝑦 of a vector A𝐴 are known, then we can find A𝐴 analytically. How do we do this? Since, by definition,

tanθ=y/x (or in this case tanθ=Ay/Ax),tan𝜃=𝑦/𝑥 (or in this case tan𝜃=𝐴𝑦/𝐴𝑥),

we solve for θ𝜃 to find the direction of the resultant.

θ=tan−1(Ay/Ax)𝜃=tan−1(𝐴𝑦/𝐴𝑥)

Note that tan−1(θ)tan−1(𝜃) gives an angle in the first quadrant if Ay/Ax>0𝐴𝑦/𝐴𝑥>0 and in the fourth quadrant if Ay/Ax<0𝐴𝑦/𝐴𝑥<0. If, in fact, both Ax𝐴𝑥 and Ay𝐴𝑦 are negative, or if Ax𝐴𝑥 is negative and Ay𝐴𝑦 positive, then θ𝜃, measured from the positive x𝑥 direction, is tan–1(θ)+180°tan–1(𝜃)+180°.

Since this is a right triangle, the Pythagorean theorem (x² + y² = h²) for finding the hypotenuse applies. In this case, it becomes

A2=A2x+A2y.𝐴2=𝐴𝑥2+𝐴𝑦2.

Solving for A gives

A=Ax2+Ay2−−−−−−−−√.𝐴=𝐴𝑥2+𝐴𝑦2.

In summary, to find the magnitude A𝐴 and direction θ𝜃 of a vector from its perpendicular components Ax𝐴𝑥 and Ay𝐴𝑦, as illustrated in Figure 5.21, we use the following relationships:

A=Ax2+Ay2−−−−−−−−√θ=tan−1(Ay/Ax)𝐴=𝐴𝑥2+𝐴𝑦2𝜃=tan−1(𝐴𝑦/𝐴𝑥)

Figure 5.21 The magnitude and direction of the resultant vector A𝐴 can be determined once the horizontal components Ax𝐴𝑥 and Ay𝐴𝑦 have been determined.

Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors A𝐴 and B𝐵 are added to produce the resultant R,𝑅, as illustrated in Figure 5.22.

Figure 5.22 Vectors A𝐴 and B𝐵 are two legs of a walk, and R𝑅 is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R𝑅 .

If A𝐴 and B𝐵 represent two legs of a walk (two displacements), then R𝑅 is the total displacement. The person taking the walk ends up at the tip of R𝑅 . There are many ways to arrive at the same point. The person could have walked straight ahead first in the x-direction and then in the y-direction. Those paths are the x– and y-components of the resultant, Rx𝑅𝑥 and Ry.𝑅𝑦. If we know Rx𝑅𝑥 and Ry𝑅𝑦, we can find R𝑅 and θ𝜃 using the equations R=Rx2+Ry2−−−−−−−−√𝑅=𝑅x2+𝑅y2 and θ=tan–1(Ry/Rx)𝜃=𝑡𝑎𝑛–1(𝑅𝑦/𝑅𝑥) .

1. Draw in the x and y components of each vector (including the resultant) with a dashed line. Use the equations Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃 to find the components. In Figure 5.23, these components are Ax𝐴𝑥, Ay𝐴𝑦, Bx𝐵𝑥, and By.𝐵𝑦. Vector A𝐴 makes an angle of θA𝜃𝐴 with the x-axis, and vector B𝐵 makes and angle of θB𝜃𝐵 with its own x-axis (which is slightly above the x-axis used by vector A).Figure 5.23 To add vectors A𝐴 and B,𝐵, first determine the horizontal and vertical components of each vector. These are the dotted vectors Ax,𝐴𝑥, Ay𝐴𝑦 By𝐵𝑦 shown in the image.
2. Find the x component of the resultant by adding the x component of the vectorsRx=Ax+Bx𝑅𝑥=𝐴𝑥+𝐵𝑥and find the y component of the resultant (as illustrated in Figure 5.24) by adding the y component of the vectors.Ry=Ay+By.𝑅𝑦=𝐴𝑦+𝐵𝑦.Figure 5.24 The vectors Ax𝐴𝑥 and Bx𝐵𝑥 add to give the magnitude of the resultant vector in the horizontal direction, Rx.𝑅x. Similarly, the vectors Ay𝐴𝑦 and By𝐵𝑦 add to give the magnitude of the resultant vector in the vertical direction, Ry.𝑅y.Now that we know the components of R,𝑅, we can find its magnitude and direction.
3. To get the magnitude of the resultant R, use the Pythagorean theorem.R=R2x+R2y−−−−−−−√𝑅=𝑅𝑥2+𝑅𝑦2
4. To get the direction of the resultantθ=tan−1(Ry/Rx).𝜃=tan−1(𝑅𝑦/𝑅𝑥).

Classifying Vectors and Quantities Example

This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions.

Three vectors, u→u→, v→v→, and w→w→, have the same magnitude of 5units5units. Vector v→v→ points to the northeast. Vector w→w→ points to the southwest exactly opposite to vector u→u→. Vector u→u→ points in the northwest. If the vectors u→u→, v→v→, and w→w→ were added together, what would be the magnitude of the resultant vector? Why?

1. 0units0units. All of them will cancel each other out.
2. 5units5units. Two of them will cancel each other out.
3. 10units10units. Two of them will add together to give the resultant.
4. 1515 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 A𝐴 to the vector B𝐵 shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and the y-axis is along the north–south directions. A person first walks 53.0 m53.0 m in a direction 20.0°20.0° north of east, represented by vector A.𝐴. The person then walks 34.0 m34.0 m in a direction 63.0°63.0° north of east, represented by vector B.𝐵.

Figure 5.25 You can use analytical models to add vectors.

STRATEGY

The components of A𝐴 and B𝐵 along the x– and y-axes 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 x-direction and y-direction to find the resultant

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, A−B≡A+(−B)A−B≡A+(−B) . The components of – BB are the negatives of the components of BB . Therefore, the x– and y-components of the resultant A−B=RA−B=R are

Rx=Ax+−Bx𝑅𝑥=𝐴𝑥+-𝐵𝑥

and

Ry=Ay+−By𝑅𝑦=𝐴𝑦+-𝐵𝑦

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 x-component is 4 cm and whose y-component is 3 cm?

1. 1 cm
2. 5 cm
3. 7 cm
4. 25 cm

6.

What is the magnitude of a vector that makes an angle of 30° to the horizontal and whose x-component is 3 units?

1. 2.61 units
2. 3.00 units
3. 3.46 units
4. 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 three-dimensional. We won’t cover three-dimensional motion in this text, but to go from two-dimensions to three-dimensions, you simply add a third vector component. Three-dimensional 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?

1. Vectors have magnitude as well as direction and can be quickly solved through scalar algebraic operations.
2. Vectors have magnitude but no direction, so it becomes easy to express physical quantities involved in the atmospheric science.
3. Vectors can be solved very accurately through geometry, which helps to make better predictions in atmospheric science.
4. 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?

1. The analytical method is less accurate than the graphical method, because the former involves geometry and trigonometry.
2. The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations.
3. The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale.
4. 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?

1. A component is a piece of a vector that points in either the x or y direction.
2. A component is a piece of a vector that has half of the magnitude of the original vector.
3. A component is a piece of a vector that points in the direction opposite to the original vector.
4. 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 θ𝜃 (measured counter-clockwise from positive x𝑥) if we know Ax𝐴𝑥 and Ay𝐴𝑦?

1. θ=cos−1AyAx𝜃=cos−1⁡𝐴𝑦𝐴𝑥
2. θ=cot−1AyAx𝜃=cot−1⁡𝐴𝑦𝐴𝑥
3. θ=sin−1AyAx𝜃=sin−1⁡𝐴𝑦𝐴𝑥
4. θ=tan−1AyAx𝜃=tan−1⁡𝐴𝑦𝐴𝑥

10.

How can we determine the magnitude of a vector if we know the magnitudes of its components?

1. ∣∣∣A→∣∣∣=Ax+Ay|A→|=𝐴𝑥+𝐴𝑦
2. ∣∣∣A→∣∣∣=Ax2+Ay2|A→|=𝐴𝑥2+𝐴𝑦2
3. ∣∣∣A→∣∣∣=Ax2+Ay2−−−−−−−−√|A→|=𝐴𝑥2+𝐴𝑦2
4. ∣∣∣A→∣∣∣=(Ax2+Ay2)2