Respuesta :
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
[tex]s_1 = 19.1 cos 25.5^0 \hat{i} + 19.1 sin 25.5^0 \hat{j} + 0.85 \hat{k}[/tex]
[tex]s_1 = 17.24 \hat{i} + 8.22 \hat{j} + 0.85 \hat{k}[/tex]
position vector of the second plane is
[tex]s_2 = 17.6 cos 15^0 \hat{i} + 17.6 sin 15^0 \hat{j} + 1.3 \hat{k}[/tex]
[tex]s_2= 17 \hat{i} + 4.56 \hat{j} + 1.3 \hat{k}[/tex]
net displacement is
[tex]s = s_2 - s_1[/tex]
=[tex]17. \hat{i} + 4.56 \hat{j} + 1.3 \hat{k} - (17.24 \hat{i} + 8.22 \hat{j} + 0.85 \hat{k})[/tex]
= [tex]-0.24 \hat{i} - 3.66 \hat{j} +0.45 \hat{k}[/tex]
magnitude is
[tex]s = \sqrt{-0.24^2+(-3.66)^2+0.45^2}[/tex]
s = 3.69 km
