Components of Vectors

Adding vectors geometrically can be tedious. A neater and easier technique involves algebra but requires that the vectors be placed on a rectangular coordinate system. The x and y axes are usually drawn in the plane of the page, as shown in Fig. 3-8a. The z axis comes directly out of the page at the origin; we ignore it for now and deal only with two-dimensional vectors.

A component of a vector is the projection of the vector on an axis. In Fig. 3-8a, for example, a is the component of vector  on (or along) the x axis and a_y is the component along the y axis. To find the projection of a vector along an axis, we draw perpendicular lines from the two ends of the vector to the axis, as shown. The projection of a vector on an x axis is its x component, and similarly the projection on the y axis is the y component. The process of finding the components of a vector is called resolving the vector.

A component of a vector has the same direction (along an axis) as the vector. In Fig. 3-8, a_x and a_y are both positive because  extends in the positive direction of both axes. (Note the small arrowheads on the components, to indicate their direction.) If we were to reverse vector , then both components would be negative and their arrowheads would point toward negative x and y. Resolving vector  in Fig. 3-9 yields a positive component b_x and a negative component b_y.

Fig. 3-8 (a) The components a_x and a_y of vector . (b) The components are unchanged if the vector is shifted, as long as the magnitude and orientation are maintained. (c) The components form the legs of a right triangle whose hypotenuse is the magnitude of the vector.

In general, a vector has three components, although for the case of Fig. 3-8a the component along the z axis is zero. As Figs. 3-8a and b show, if you shift a vector without changing its direction, its components do not change.

We can find the components of  in Fig. 3-8a geometrically from the right triangle there:

where θ is the angle that the vector  makes with the positive direction of the x axis, and a is the magnitude of . Figure 3-8c shows that  and its x and y components form a right triangle. It also shows how we can reconstruct a vector from its components: we arrange those components head to tail. Then we complete a right triangle with the vector forming the hypotenuse, from the tail of one component to the head of the other component.

Fig. 3-9 The component of  on the x axis is positive, and that on the y axis is negative.

Once a vector has been resolved into its components along a set of axes, the components themselves can be used in place of the vector. For example,  in Fig. 3-8a is given (completely determined) by a and θ. It can also be given by its components a_x and a_y. Both pairs of values contain the same information. If we know a vector in component notation (a_x and a_y) and want it in magnitude-angle notation (a and θ), we can use the equations

to transform it.

In the more general three-dimensional case, we need a magnitude and two angles (say, a, θ, and ) or three components (a_x, a_y, and a_z) to specify a vector.

CHECKPOINT 2 In the figure, which of the indicated methods for combining the x and y components of vector  are proper to determine that vector?

Sample Problem 3-2

A small airplane leaves an airport on an overcast day and is later sighted 215 km away, in a direction making an angle of 22° east of due north. How far east and north is the airplane from the airport when sighted?

Solution: The Key Idea here is that we are given the magnitude (215 km) and the angle (22° east of due north) of a vector and need to find the components of the vector. We draw an xy coordinate system with the positive direction of x due east and that of y due north (Fig. 3-10). For convenience, the origin is placed at the airport. The airplane’s displacement  points from the origin to where the airplane is sighted.

To find the components of , we use Eq. 3-5 with θ = 68° (= 90° − 22°):

Thus, the airplane is 81 km east and 2.0 × 10² km north of the airport.

Fig. 3-10 A plane takes off from an airport at the origin and is later sighted at P.

Sample Problem 3-3

For two decades, spelunking teams sought a connection between the Flint Ridge cave system and Mammoth Cave, which are in Kentucky. When the connection was finally discovered, the combined system was declared the world’s longest cave (more than 200 km long). The team that found the connection had to crawl, climb, and squirm through countless passages, traveling a net 2.6 km westward, 3.9 km southward, and 25 m upward. What was their displacement from start to finish?

Solution: The Key Idea here is that we have the components of a three-dimensional vector, and we need to find the vector’s magnitude and two angles to specify the vector’s direction. We first draw the components as in Fig. 3-11a. The horizontal components (2.6 km west and 3.9 km south) form the legs of a horizontal right triangle. The team’s horizontal displacement forms the hypotenuse of the triangle, and its magnitude d_h is given by the Pythagorean theorem:

Fig. 3-11 (a) The components of the spelunking team’s overall displacement and their horizontal displacement d_h. (b) A side view showing d_h and the team’s overall displacement vector .

Also from the horizontal triangle in Fig. 3-11a, we see that this horizontal displacement is directed south of due west by an angle θ_h given by

which is one of the two angles we need to specify the direction of the overall displacement.

To include the vertical component (25 m = 0.025 km), we now take a side view of Fig. 3-11a, looking northwest. We get Fig. 3-11b, where the vertical component and the horizontal displacement d_h form the legs of another right triangle. Now the team’s overall displacement forms the hypotenuse of that triangle, with a magnitude d given by

This displacement is directed upward from the horizontal displacement by the angle

Thus, the team’s displacement vector had a magnitude of 4.7 km and was at an angle of 56° south of west and at an angle of 0.3° upward. The net vertical motion was, of course, insignificant compared with the horizontal motion. However, that fact would have been of no comfort to the team, which had to climb up and down countless times to get through the cave. The route they actually covered was quite different from the displacement vector, which merely points in a straight line from start to finish.

PROBLEM-SOLVING TACTICS

TACTIC 1 : Angles—Degrees and Radians

Angles that are measured relative to the positive direction of the x axis are positive if they are measured in the counterclockwise direction and negative if measured clockwise. For example, 210° and −150° are the same angle.

Angles may be measured in degrees or radians (rad). To relate the two measures, recall that a full circle is 360° and 2π rad. To convert, say, 40° to radians, write

Fig. 3-12 A triangle used to define the trigonometric functions. See also Appendix E.

TACTIC 2 : Trig Functions

You need to know the definitions of the common trigonometric functions—sine, cosine, and tangent—because they are part of the language of science and engineering. They are given in Fig. 3-12 in a form that does not depend on how the triangle is labeled.

You should also be able to sketch how the trig functions vary with angle, as in Fig. 3-13, in order to be able to judge whether a calculator result is reasonable. Even knowing the signs of the functions in the various quadrants can be of help.

TACTIC 3 : Inverse Trig Functions

When the inverse trig functions sin⁻¹, cos⁻¹, and tan⁻¹ are taken on a calculator, you must consider the reasonableness of the answer you get, because there is usually another possible answer that the calculator does not give. The range of operation for a calculator in taking each inverse trig function is indicated in Fig. 3-13. As an example, sin⁻¹ 0.5 has associated angles of 30° (which is displayed by the calculator, since 30° falls within its range of operation) and 150°. To see both values, draw a horizontal line through 0.5 in Fig. 3-13a and note where it cuts the sine curve.

How do you distinguish a correct answer? It is the one that seems more reasonable for the given situation. As an example, reconsider the calculation of θ in Sample Problem 3-3, where tan θ_h = 3.9/2.6 = 1.5. Taking tan⁻¹ 1.5 on your calculator tells you that θ_h = 56°, but θ_h = 236° (= 180° + 56°) also has a tangent of 1.5. Which is correct? From the physical situation (Fig. 3-11a), 56° is reasonable and 236° is clearly not.

TACTIC 4 : Measuring Vector Angles

The equations for cos θ and sin θ in Eq. 3-5 and for tan θ in Eq. 3-6 are valid only if the angle is measured from the positive direction of the x axis. If it is measured relative to some other direction, then the trig functions in Eq. 3-5 may have to be interchanged and the ratio in Eq. 3-6 may have to be inverted. A safer method is to convert the angle to one measured from the positive direction of the x axis.

Fig. 3-13 Three useful curves to remember. A calculator’s range of operation for taking inverse trig functions is indicated by the darker portions of the colored curves.