Question

An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 950 m, horizontal distance 19.1 km, and 13.0° south of west. The second aircraft is at altitude 1050 m, horizontal distance 17.0 km, and 27.0° west of south.

(a) What the displacement vector FROM the first plane TO the second plane, letting i hat bold represent east, j hat bold north, and k hat bold up?

(b) How far apart are the two planes?

Answer 1

Answer on Question #38970, Physics, Mechanics | Kinamatics | Dynamics

An air-traffic controller observes two aircraft on his radar screen. The first is at altitude $950\mathrm{m}$ , horizontal distance $19.1\mathrm{km}$ , and $13.0{}^{\circ}$ south of west. The second aircraft is at altitude $1050\mathrm{m}$ , horizontal distance $17.0\mathrm{km}$ , and $27.0{}^{\circ}$ west of south.

(a) What the displacement vector FROM the first plane TO the second plane, letting i hat bold represent east, j hat bold north, and k hat bold up?

(b) How far apart are the two planes?

Solution:

A convenient way to specify the position of an object is with the help of a coordinate system. We choose a fixed point, called the origin and three directed lines, which pass through the origin and are perpendicular to each other. These lines are called the coordinate axes of a three-dimensional rectangular (Cartesian) coordinate system and are labeled the $x$ -, $y$ -, and $z$ -axis. Three numbers with units specify the position of a point $P$ . These numbers are the $x$ -, $y$ -, and $z$ -coordinates of the point $P$ . Here $\hat{\mathbf{i}}$ , $\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are unit vectors.

Find the xyz coordinates of each aircraft using:

$+x =$ east

$+y =$ north

$+z =$ altitude.

For first aircraft:

$$x _ {1} = - 1 9 1 0 0 \cdot \cos 1 3 {}^ {\circ} = - 1 9 1 0 0 \cdot 0. 9 7 4 3 7 \approx - 1 8 6 1 0. 5$$ $$y _ {1} = - 1 9 1 0 0 \cdot \sin 1 3 {}^ {\circ} = - 1 9 1 0 0 \cdot 0. 2 2 4 9 5 \approx - 4 2 9 6. 6$$ $$z _ {1} = 9 5 0$$

For second aircraft:

$$x _ {2} = - 1 7 0 0 0 \cdot \sin 2 7 {}^ {\circ} = - 1 7 0 0 0 \cdot 0. 4 5 3 9 9 \approx - 7 7 1 7. 8$$ $$y _ {2} = - 1 7 0 0 0 \cdot \cos 2 7 {}^ {\circ} = - 1 7 0 0 0 \cdot 0. 8 9 1 0 1 \approx - 1 5 1 4 7. 2$$ $$z _ {2} = 1 0 5 0$$

The displacement vector $\mathbf{d}$ from $\mathsf{P}_1$ to $\mathsf{P}_2$ may be written as

$$\vec {d} = \left(x _ {2} - x _ {1}\right) \hat {\mathbf {i}} + \left(y _ {2} - y _ {1}\right) \hat {\mathbf {j}} + \left(z _ {2} - z _ {1}\right) \hat {\mathbf {k}}$$ $$\vec {d} = (- 7 7 1 7. 8 \pm 1 8 6 1 0. 5) \hat {\mathbf {i}} + (- 1 5 1 4 7. 2 + 4 2 9 6. 6) \hat {\mathbf {j}} + (1 0 5 0 - 9 5 0) \hat {\mathbf {k}}$$ $$\vec {d} = (1 0 8 9 2. 7) \hat {\mathbf {i}} + (- 1 0 8 5 0. 6) \hat {\mathbf {j}} + (1 0 0) \hat {\mathbf {k}}$$

The magnitude of the position vector is its length $r$ . It depends on the choice of the origin of the coordinate system. It is the distance of $P$ from the origin.

The magnitude of the displacement is

$$d = \sqrt {(x _ {2} - x _ {1}) ^ {2} + (y _ {2} - y _ {1}) ^ {2} + (z _ {2} - z _ {1}) ^ {2}}$$ $$d = \sqrt {(10892.7) ^ {2} + (10850.6) ^ {2} + (100) ^ {2}} \approx 15375.2 \mathrm {m}$$

Answer. a) The displacement vector $\vec{\mathbf{d}} = (10892.7)\hat{\mathbf{i}} + (-10850.6)\hat{\mathbf{j}} + (100)\hat{\mathbf{k}}$ b) $d \approx 15375.2 \, \text{m}$