You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land. Your job is to make sure that airplanes are not closer to each other than a minimum safe separation distance of 2.00 km. You observe two small aircraft on your radar screen, out over the ocean surface. The first is at altitude 800 m above the surface, horizontal distance 18.0 km, and 25.0° south of west. The second aircraft is at altitude 1,100 m, horizontal distance 20.0 km, and 20.0° south of west.

1. Your supervisor is concerned that the two aircraft are too close together and asks for a separation distance (in km) for the two airplanes. (Place the x-axis west, the y-axis south, and the z-axis vertical.)

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

2 plane

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

d² = (x₂-x₁) ² + (y₂-y₁) ² + (z₂-z₁) ²2

Let's calculate

d² = (18126-16314) ² + (1100-800) ² + (8452-7607) ²

d² = 3,283 106 + 9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √ (408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km