An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 750 m, horizontal distance 20.0 km, and 25.0° south of west. The second aircraft is at altitude 1050 m, horizontal distance 17.0 km, and 17.0° west of south.
(a) Write the displacement vector FROM the first plane TO the second plane, letting i represent east, j north, and k up.
(b) How far apart are the two planes?
Please help me on this problem
a) find the points horizontal (E,N) or (x,y) of the two points based on horizontal distance and angle. Now add the third dimension for each point knowing altitude.
distance apart.
d=sqrt((x1-x2)^2 + (y1-y2)^2 + (zi-z2)^2 )
displacment vector (x1-x2)xbar+...
Sure, I'd be happy to help you with this problem!
(a) To find the displacement vector from the first plane to the second plane, we need to find the difference in their positions in the x, y, and z directions.
First, let's find the position vector of the first plane:
r1 = -20.0 km * cos(25.0°)i + (-20.0 km * sin(25.0°))j + 750 mi * k
Next, let's find the position vector of the second plane:
r2 = -17.0 km * sin(17.0°)i + (-17.0 km * cos(17.0°))j + 1050 m * k
Now, we can find the displacement vector by subtracting r1 from r2:
Displacement vector = r2 - r1
= (-17.0 km * sin(17.0°)i + (-17.0 km * cos(17.0°))j + 1050 m * k) - (-20.0 km * cos(25.0°)i + (-20.0 km * sin(25.0°))j + 750 m * k)
= (-17.0 km * sin(17.0°) + 20.0 km * cos(25.0°))i + (-17.0 km * cos(17.0°) - 20.0 km * sin(25.0°))j + (1050 m - 750 m)k
So the displacement vector FROM the first plane TO the second plane is approximately:
(-7.259 km)i + (-32.536 km)j + 300 m*k
(b) To find how far apart the two planes are, we can calculate the magnitude of the displacement vector.
Magnitude of displacement vector = sqrt((-7.259 km)^2 + (-32.536 km)^2 + (300 m)^2)
≈ 33.5 km
Therefore, the two planes are approximately 33.5 km apart.