Question

An air traffic controller notices that one aircraft approaches the airport at an altitude 2500 m,horizontal distance 120 km and a bearing 20 degrees south of east. A second aircraft is at an altitude 3500 m,horizontal distance 110 km,and bearing 25 degrees south of east.Find the displacement vector from the first aircraft to the second in terms of altitude,horizontal distance and bearing.

Answer 1

On the picture: OE shows East direction, A is a projection of a position of the first airplane and B is a projection of a position of the second airplane. $\angle AOE = 20{}^{\circ}$ , $\angle BOE = 25{}^{\circ}$ , $AO = 120$ and $BO = 110$ .

Let we have 2 vectors $-v_{1}$ and $v_{2}$ . $v_{1}$ is the vector from the origin (airport) to the first airplane and $v_{2}$ is the vector from the origin (airport) to the second airplane. We will use them in Cartesian 3-dimensional coordinates $x, y$ and $z$ . Let OE is x-axis, SO is y-axis and z-axis is directed upwards. Then we have:

$$v _ {1} = \left(x _ {1}, y _ {1}, z _ {1}\right) = (1 2 0 \cos 2 0 {}^ {\circ}, 1 2 0 \sin 2 0 {}^ {\circ}, 2. 5) = (1 1 2. 7 6, 4 1, 2. 5)$$ $$v _ {2} = \left(x _ {2}, y _ {2}, z _ {2}\right) = (1 1 0 \cos 2 5 {}^ {\circ}, 1 1 0 \sin 2 5 {}^ {\circ}, 3. 5) = (9 9. 6 9, 4 6. 4 9, 3. 5)$$

The vector $u$ from the first airplane to the second is:

$$u = v _ {2} - v _ {1} = (- 1 3. 0 7, 5. 4 9, 1)$$

In terms of altitude, horizontal distance and bearing we have:

Altitude: $1000\mathrm{m}$

Horizontal distance: $d = \sqrt{(-13.07)^2 + 5.49^2} = 14.18\mathrm{km}$

Bearing: $\phi = \mathrm{atan}\frac{y}{x} = \mathrm{atan}\left(\frac{5.49}{-13.07}\right) = -22.78{}^{\circ} \approx -23{}^{\circ}$ which means $23{}^{\circ}$ to the south of west.