Using a sketch, we can add vectors geometrically. On a vector-capable calculator, we can add them directly on the screen. A third way to add vectors is to combine their components axis by axis, which is the way we examine here.
To start, consider the statement
which says that the vector 
is the same as the vector (
+ 
). Thus, each component of must be the same as the corresponding component of ( + ):
Fig. 3-15 (a) The vector components of vector . (b) The vector components of vector .
In other words, two vectors must be equal if their corresponding components are equal. Equations 3-10 to 3-13 tell us that to add vectors and , we must (1) resolve the vectors into their scalar components; (2) combine these scalar components, axis by axis, to get the components of the sum ; and (3) combine the components of to get itself. We have a choice in step 3. We can express in unit-vector notation (as in Eq. 3-9) or in magnitude-angle notation (as in the answer to Sample Problem 3-3).
This procedure for adding vectors by components also applies to vector subtractions. Recall that a subtraction such as 
= − can be rewritten as an addition = + (−). To subtract, we add and − by components, to get
CHECKPOINT 3 (a) In the figure here, what are the signs of the x components of 
and 
? (b) What are the signs of the y components of and ? (c) What are the signs of the x and y components of + ?
Figure 3-16a shows the following three vectors:
What is their vector sum , which is also shown?
Solution: The Key Idea here is that we can add the three vectors by components, axis by axis. For the x axis, we add the x components of , , and 
to get the x component of the vector sum :
Similarly, for the y axis,
Another Key Idea is that we can combine these components of to write the vector in unit-vector notation:
where (2.6 m)
is the vector component of along the x axis and −(2.3 m)
is that along the y axis. Figure 3-16b shows one way to arrange these vector components to form . (Can you sketch the other way?)
A third Key Idea is that we can also answer the question by giving the magnitude and an angle for . From Eq. 3-6, the magnitude is
and the angle (measured from the positive direction of x) is
where the minus sign means that the angle is measured clockwise.
Fig. 3-16 Vector is the vector sum of the other three vectors.
According to experiments, the desert ant shown in the chapter opening photograph keeps track of its movements along a mental coordinate system. When it wants to return to its home nest, it effectively sums its displacements along the axes of the system to calculate a vector that points directly home. As an example of the calculation, let’s consider an ant making five runs of 6.0 cm each on an xy coordinate system, in the directions shown in Fig. 3-17a, starting from home. At the end of the fifth run, what are the magnitude and angle of the ant’s net displacement vector 
, and what are those of the home ward vector 
that extends from the ant’s final position back to home?
Fig. 3-17 (a) A search path of five runs. (b) The x and y components of . (c) Vector points the way to the home nest.
Solution: The Key Idea here consists of three parts. First, to find the net displacement , we need to sum the five individual displacement vectors:
Second, we evaluate this sum for the x components alone,
and for the y components alone,
Third, we construct from its x and y components.
To evaluate Eq. 3-14, we apply the x part of Eq. 3-5 to each run:
Equation 3-14 then gives us
Similarly, we evaluate the individual y components of the five runs using the y part of Eq. 3-5. The results are shown in Table 3-1. Substituting the results into Eq. 3-15 then gives us
Vector and its x and y components are shown in Fig. 3-17b. To find the magnitude and angle of from its components, we use Eq. 3-6. The magnitude is
To find the angle (measured from the positive direction of x), we take an inverse tangent:
Caution: Recall from Problem-Solving Tactic 3 that taking an inverse tangent on a calculator may not give the correct answer. The answer −24.86° indicates that the direction of is in the fourth quadrant of our xy coordinate system. However, when we construct the vector from its components (Fig. 3-17b), we see that the direction of is in the second quadrant. Thus, we must “fix” the calculator’s answer by adding 180°:
Thus, the ant’s displacement has magnitude and angle
Vector directed from the ant to its home has the same magnitude as but the opposite direction (Fig. 3-17c). We already have the angle (−24.86° ≈ −25°) for the direction opposite . Thus, has magnitude and angle
A desert ant traveling more than 500 m from its home will actually make thousands of individual runs. Yet, it somehow knows how to calculate (without studying this chapter).
Here is a problem involving vector addition that cannot be solved directly on a vector-capable calculator, using the vector notation of the calculator. Three vectors satisfy the relation
has a magnitude of 22.0 units and is directed at an angle of −47.0° (clockwise) from the positive direction of an x axis. 
has a magnitude of 17.0 units and is directed counterclockwise from the positive direction of the x axis by angle 
. 
is in the positive direction of the x axis. What is the magnitude of ?
Solution: We cannot answer the question by adding and directly on a vector-capable calculator, say, in the generic form of
because we do not know the value for the angle of . However, we can use this Key Idea: We can express Eq. 3-16 in terms of components for either the x axis or the y axis. Since is directed along the x axis, we choose that axis and write
We next express each x component in the form of the x part of Eq. 3-5 and substitute known data. We then have
However, this hardly seems to help, because we still cannot solve for B without knowing .
Let us now express Eq. 3-16 in terms of components along the y axis:
We then cast these y components in the form of the y part of Eq. 3-5 and substitute known data, to write
which yields
Solving for then gives us
Substituting this result into Eq. 3-17 leads us to
Note the technique of solution: When we worked with components on the x axis, we got stuck with two unknowns—the desired B and the undesired . We then worked with components on the y axis and were able to evaluate . We next moved back to the x axis, to evaluate B.
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