$$\require{cancel}$$

Polar Coordinates

Strategy

We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given by unit vector $$\hat{i}$$ to the east, the direction of the +y-axis is given by unit vector $$\hat{j}$$ to the north, and the direction of the +z-axis is given by unit vector $$\hat{k}$$, which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.

Solution

We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use Equation 2.13 and Equation 2.20 to find the scalar components of the drone’s displacement vector:

$$\begin{cases} D_{x} = x_{e} - x_{b} = 1200.0\; m - 300.0\; m = 900.0\; m,\\ D_{y} =y_{e} - y_{b} = 2100.0\; m - 200.0\; m = 1900.0\; m, \\ D_{z} =z_{e} - z_{b} = 250.0\; m - 100.0\; m = 150\; m \ldotp \end{cases}$$

We substitute these components into Equation 2.19 to find the displacement vector:

$$\vec{D} = D_{x}\; \hat{i} + D_{y}\; \hat{j} + D_{z}\; \hat{k} = 900.0\; \hat{i} + 1900.0\; \hat{j} + 150.0\; \hat{k} = (0.90\; \hat{i} + 1.90\; \hat{j} + 0.15\; \hat{k})\; km \ldotp$$

We substitute into Equation 2.21 to find the magnitude of the displacement:

$$D = \sqrt{D_{x}^{2} + D_{y}^{2} + D_{z}^{2}} = \sqrt{(0.90\; km)^{2} + (1.90\; km)^{2} + (0.15\; km)^{2}} = 4.44\; km \ldotp$$

Exercise 2.7

If the average velocity vector of the drone in the displacement in Example 2.7 is $$\vec{u}$$ = (15.0 $$\hat{i}$$ + 31.7 $$\hat{j}$$ + 2.5 $$\hat{k}$$) m/s, what is the magnitude of the drone’s velocity vector?

Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).