Question

An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 850 m, horizontal distance 19.1 km, and 25.5° south of west. The second aircraft is at altitude 1300 m, horizontal distance 17.6 km, and 15.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.)

Answer 1

Answer:

s = 3.69 km

Explanation:

given,

altitudes of the aircraft

h₁ = 0.85 km , h₂ = 1.3 km

position vector of the first plane is

$s_1 = 19.1 cos 25.5^0 \hat{i} + 19.1 sin 25.5^0 \hat{j} + 0.85 \hat{k}$

$s_1 = 17.24 \hat{i} + 8.22 \hat{j} + 0.85 \hat{k}$

position vector of the second plane is

$s_2 = 17.6 cos 15^0 \hat{i} + 17.6 sin 15^0 \hat{j} + 1.3 \hat{k}$

$s_2= 17 \hat{i} + 4.56 \hat{j} + 1.3 \hat{k}$

net displacement is

$s = s_2 - s_1$

  =$17. \hat{i} + 4.56 \hat{j} + 1.3 \hat{k} - (17.24 \hat{i} + 8.22 \hat{j} + 0.85 \hat{k})$

 = $-0.24 \hat{i} - 3.66 \hat{j} +0.45 \hat{k}$

magnitude is

$s = \sqrt{-0.24^2+(-3.66)^2+0.45^2}$

s = 3.69 km