An air traffic controller has two aircraft on radar. The first is at an altitude

Question

An air traffic controller has two aircraft on radar. The first is at an altitude of 0.500 km, a horizontal distance of 3.00 km measured along the ground to the point directly underneath the plane, and a bearing of 115° west of north. The second aircraft is at an altitude of 1.00 km, a horizontal distance of 8.00 km, and a bearing of 35.0° east of north. Write the displacement vector from aircraft 1 to aircraft 2 in "cylindrical" coordinates. That is, write a 3-component vector (r, , z) whose first two components are the polar coordinates of the horizontal displacement between the planes, and whose third coordinate is the vertical displacement between the planes. Consider east to be the positive x axis and north to be the positive y axis. Degrees is NOT 115 or -150 or 70.51 or 140. First answer is 10.7 km and last answer is .5 km. The answer should be a little less than 50 degrees, the teacher said.

Explanation / Answer

In given coordinate system, position vectors are:

for 1st aircraft, A1: (3,205,0.5)

for 2nd aircraft, A2: (8,55,1)

converting to cartesian coordinates:

A1=3(cos205 i + sin205 j) + 0.5k

A2=8(cos55 i + sin55 j) + k

displacement from 1 to 2= A2-A1= 7.3075 i + 7.821 j +0.5k

In cylindrical coordinates, displacement=(10.7,46.944,0.5)

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