2.2 Coordinate systems and components of a vector  (Page 6/15)

Vectors in three dimensions

To specify the location of a point in space, we need three coordinates ( x , y , z ), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical positions above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x -axis $\widehat{i}$ and the unit vector of the y -axis $\widehat{j}$ . The third unit vector $\widehat{k}$ is the direction of the z -axis ( [link] ). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x - y - z , which is equivalent to the order $\widehat{i}$ - $\widehat{j}$ - $\widehat{k}$ , defines the standard right-handed coordinate system (positive orientation).

In three-dimensional space, vector $\overrightarrow{A}$ has three vector components: the x -component ${\overrightarrow{A}}_x=A_x\widehat{i}$ , which is the part of vector $\overrightarrow{A}$ along the x -axis; the y -component ${\overrightarrow{A}}_y=A_y\widehat{j}$ , which is the part of $\overrightarrow{A}$ along the y -axis; and the z -component ${\overrightarrow{A}}_z=A_z\widehat{k}$ , which is the part of the vector along the z -axis. A vector in three-dimensional space is the vector sum of its three vector components ( [link] ):

If we know the coordinates of its origin $b(x_b,y_b,z_b)$ and of its end $e(x_e,y_e,z_e)$ , its scalar components are obtained by taking their differences: $A_x$ and $A_y$ are given by [link] and the z -component is given by

Magnitude A is obtained by generalizing [link] to three dimensions:

This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in [link] , the diagonal in the xy -plane has length $\sqrt{A_x^2+A_y^2}$ and its square adds to the square $A_z^2$ to give $A^2$ . Note that when the z -component is zero, the vector lies entirely in the xy -plane and its description is reduced to two dimensions.

Takeoff of a drone

During a takeoff of IAI Heron ( [link] ), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?

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 $\widehat{i}$ to the east, the direction of the + y -axis is given by unit vector $\widehat{j}$ to the north, and the direction of the + z -axis is given by unit vector $\widehat{k}$ , 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 [link] and [link] to find the scalar components of the drone’s displacement vector:

$\{\begin{array}{l}{D}_x=x_e-x_b=1200.0\text{m}-300.0\text{m}=900.0\text{m},\\ D_y=y_e-y_b=2100.0\text{m}-200.0\text{m}=1900.0\text{m,}\\ D_z=z_e-z_b=250.0\text{m}-100.0\text{m}=150.0\text{m}.\end{array}$

We substitute these components into [link] to find the displacement vector:

$\overrightarrow{D}=D_x\widehat{i}+D_y\widehat{j}+D_z\widehat{k}=900.0\text{m}\widehat{i}+1900.0\text{m}\widehat{j}+150.0\text{m}\widehat{k}=(0.90\widehat{i}+1.90\widehat{j}+0.15\widehat{k})\text{km}.$

We substitute into [link] to find the magnitude of the displacement”

$D=\sqrt{D_x^2+D_y^2+D_z^2}=\sqrt{{(0.90\text{km})}^2+{(1.90\text{km})}^2+{(0.15\text{km})}^2}=4.44\text{km}.$

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