One general way of locating a particle (or particle-like object) is with a position vector 
, which is a vector that extends from a reference point (usually the origin of a coordinate system) to the particle. In the unit-vector notation of Section 3-5, can be written
where x
, y
, and z
are the vector components of and the coefficients x, y, and z are its scalar components.
The coefficients x, y, and z give the particle’s location along the coordinate axes and relative to the origin; that is, the particle has the rectangular coordinates (x, y, z). For instance, Fig. 4-1 shows a particle with position vector
and rectangular coordinates (−3 m, 2 m, 5 m). Along the x axis the particle is 3 m from the origin, in the − direction. Along the y axis it is 2 m from the origin, in the + direction. Along the z axis it is 5 m from the origin, in the + direction.
As a particle moves, its position vector changes in such a way that the vector always extends to the particle from the reference point (the origin). If the position vector changes—say, from 1 to 2 during a certain time interval—then the particle’s displacement Δ during that time interval is
Fig. 4-1 The position vector for a particle is the vector sum of its vector components.
Using the unit-vector notation of Eq. 4-1, we can rewrite this displacement as
where coordinates (x1, y1, z1) correspond to position vector 1 and coordinates (x2, y2, z2) correspond to position vector 2. We can also rewrite the displacement by substituting Δx for (x2 − x1), Δy for (y2 − y1), and Δz for (z2 − z1):
In Fig. 4-2, the position vector for a particle initially is
and then later is
What is the particle’s displacement Δ from 1 to 2?
Solution: The Key Idea is that the displacement Δ is obtained by subtracting the initial position vector 1 from the later position vector 2:
This displacement vector is parallel to the xz plane because it lacks a y component, a fact that is easier to see in the numerical result than in Fig. 4-2.
CHECKPOINT 1 (a) If a wily bat flies from xyz coordinates (−2 m, 4 m, −3 m) to coordinates (6 m, −2 m, −3 m), what is its displacement Δ
in unit-vector notation? (b) Is Δ parallel to one of the three coordinate planes? If so, which plane?
Fig. 4-2 The displacement Δ = 2 − 1 extends from the head of the initial position vector 1 to the head of the later position vector 2.
A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn. The coordinates of the rabbit’s position as functions of time t are given by
with t in seconds and x and y in meters.
(a) At t = 15 s, what is the rabbit’s position vector in unit-vector notation and in magnitude-angle notation?
Solution: The Key Idea here is that the x and y coordinates of the rabbit’s position, as given by Eqs. 4-5 and 4-6, are the scalar components of the rabbit’s position vector . Thus, we can write
(We write (t) rather than because the components are functions of t, and thus is also.)
At t = 15 s, the scalar components are
Thus, at t = 15 s,
which is drawn in Fig. 4-3a.
To get the magnitude and angle of , we can use a vector-capable calculator, or we can be guided by Eq. 3-6 to write
(Although θ = 139° has the same tangent as −41°, study of the signs of the components of indicates that the desired angle is in the fourth quadrant, given by 139° − 180° = −41°.)
(b) Graph the rabbit’s path for t = 0 to t = 25 s.
Fig. 4-3 (a) A rabbit’s position vector at time t = 15 s. The scalar components of are shown along the axes. (b) The rabbit’s path and its position at five values of t.
Solution: We can repeat part (a) for several values of t and then plot the results. Figure 4-3b shows the plots for five values of t and the path connecting them. We can also use a graphing calculator to make a parametric graph; that is, we would have the calculator plot y versus x, where these coordinates are given by Eqs. 4-5 and 4-6 as functions of time t.
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