An airtraffic controller observes two aircraft on his radar

An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 750 m, horizontal distance 19.0 km, and 35.0° south of west. The second aircraft is at altitude 1100 m, horizontal distance 17.7 km, and 33.0° west of south.

b) How far apart are the two planes?

Solution

Lets assume air -traffic controller is at origin

Plane 1 : altitude 750 m, horizontal distance 19.0 km, and 35.0° south of west.

It means 35.0° south of west = 55.0 deg west of south.zSo, writing the

x direction component = -19 cos55 = - 10.89 km

y direction component = - 19 sin55 = - 15.56 km

height = 0.75km

So, Vector position for plane : - 10.89i -15.56j +0.75k------ V1

Plane2 : 100 m, horizontal distance 17.7 km, and 33.0° west of south.

x direction component = -17.7 cos33 = - 14.84 km

y direction component = -17.7 sin 33 = - 9.64 km

height = 1.1 km

So, Vector position for plane 2: -14.84i -9.64j + 1.1k -----> V2

a) displacement vector FROM the first plane TO the second plane = V2 - V1 = -3.95i + 5.92j +0.35k

b) distance between planes = |V2 - V1| = sqrt( 3.95^2 +5.92^2 +0.35^2) = 7.12 km