Subtracting Vectors Components of a Vector

3.2 Vector Addition and Subtraction Using Components 61
Choosing x- and y-Axes
A problem can be made easier to solve with a good choice of axes. We can choose any
direction we want for the x- and y-axes, as long as they are perpendicular to each other.
Three common choices are
• x-axis horizontal and y-axis vertical, when the vectors all lie in a vertical plane;
• x-axis east and y-axis north, when the vectors all lie in a horizontal plane; and
• x-axis parallel to an inclined surface and y-axis perpendicular to it.
Example 3.2
An Irish Adventure (2)
In the trip of Example 3.1, Charlotte and Shona drive at a
compass heading of 27° west of south for 18 km to
Kenmare, then directly south for 17 km to Glengariff, then
at a compass heading of 13° north of east for 48 km to
Cork. Use the component method to find the magnitude
and direction of the displacement vector for the entire trip.
Strategy As before, let’s call the three successive displacements
A, B, and C, respectively. To add the vectors
using components, we first choose directions for the x-
and y-axes. Then we find the x- and y-components of the
three displacements. Adding the x- or y-components of
the three displacements gives the x- or y-component of the
total displacement. Finally, from the components we find
the magnitude and direction of the total displacement.
Solution A good choice is the conventional one: x-
axis to the east and the y-axis to the north. The first displacement
( A) is directed 27° west of south. Both of its
components are negative since west is the –x-direction
and south is the –y-direction. Using the triangle in
Fig. 3.9, the side of the triangle opposite the 27° angle is
parallel to the x-axis. The sine function relates the opposite
side to the hypotenuse:
A x = –A sin 27° = –18 km × 0.454 = –8.17 km
where A is the magnitude of A. The cosine relates the
adjacent side to the hypotenuse:
A y = –A cos 27° = –18 km × 0.891 = –16.0 km
Displacement B has no x-component since its direction
is south. Therefore,
B x = 0 and B y = –17 km
The direction of C is 13° north of east. Both its components
are positive. From Fig. 3.9, the side of the triangle
opposite the 13° angle is parallel to the y-axis, so
y
A = 18 km 27° B = 17 km C = 48 km
A A
C
y
B
13°
C y
x
A x
C x
Figure 3.9
Resolving A, B, and C into x- and y-components.
C x = +C cos 13° = +48 km × 0.974 = +46.8 km
C y = +C sin 13° = +48 km × 0.225 = +10.8 km
Now we sum the x- and y-components separately to
find the x- and y-components of the total displacement:
∆x = C x + B x + A x
= 46.8 km + 0 + (–8.17 km) = +38.63 km
∆y = C y + B y + A y
= 10.8 km + (–17 km) + (–16.0 km) = –22.2 km
The magnitude and direction of ∆r can be found from
the triangle in Fig. 3.10. The magnitude is represented by
the hypotenuse:
∆r = (∆x) 2 + (∆y) 2 = (38.63 km) 2 + (–22.2km) 2
= 45 km
The angle q is
q = tan –1 o pposite
= tan –1 22.
2 km
= 30°
adjacent
3 8.
63
km
Since +x is east and –y is
south, the direction of the
displacement is 30° south
of east. The magnitude
and direction of the displacement
found using
components agree with
the displacement found
graphically in Fig. 3.3.
Discussion Note that the x-component of one displacement
was found using the sine function while another was
found using the cosine. The x-component (or the y-
component) of the vector can be related to either the sine or
the cosine, depending on which angle in the triangle is used.
Practice Problem 3.2
Coordinate Axes
38.63 km
θ
22.2 km
Changing the
Find the x- and y-components of the displacements for
the three legs of the trip if the x-axis points south and the
y-axis points east.
y
∆r
Figure 3.10
Finding the magnitude and
direction of ∆r.
x