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?
I ended up getting 2.54 km for i -3.482 km for j .3 km which is correct and 4.32 km for B but it was wrong. Please help me.
Change in k is just change in altitude. 300m.
what about j and i?
correction: i got i and j wrong
To find the displacement vector from the first plane to the second plane and calculate the distance between them, we can use vector addition and the Pythagorean theorem.
(a) To find the displacement vector from the first plane to the second plane, we need to calculate the difference between their position vectors. We'll start with the position vector of the second plane and subtract the position vector of the first plane.
Let's break down the given information for each plane into their x, y, and z-components:
First plane:
Altitude (z): 750 m
Horizontal distance (distance from the origin in the x-y plane): 20.0 km (or 20,000 m)
Direction (angle south of west): 25.0°
Second plane:
Altitude (z): 1050 m
Horizontal distance (distance from the origin in the x-y plane): 17.0 km (or 17,000 m)
Direction (angle west of south): 17.0°
Now, let's convert the given information into vectors:
First plane position vector:
r₁ = (20,000 m) * cos(25.0°) * (-i) + (20,000 m) * sin(25.0°) * (-j) + (750 m) * k
Second plane position vector:
r₂ = (17,000 m) * sin(17.0°) * (-i) + (17,000 m) * cos(17.0°) * (-j) + (1050 m) * k
Now, subtract the first plane position vector from the second plane position vector to get the displacement vector from the first plane to the second plane:
Δr = r₂ - r₁
= [(17,000 m) * sin(17.0°)] * (-i) + [(17,000 m) * cos(17.0°)] * (-j) + [(1050 m) - (750 m)] * k
Simplifying the expression:
Δr = (-2.905 km) * (-i) + (-16.922 km) * (-j) + (300 m) * k
= 2.905 km * i + 16.922 km * j + 0.300 km * k
Therefore, the displacement vector from the first plane to the second plane is 2.905 km * i + 16.922 km * j + 0.300 km * k.
(b) To find the distance between the two planes, we can calculate the magnitude of the displacement vector using the Pythagorean theorem:
Distance = √(Δr_x² + Δr_y² + Δr_z²)
= √((2.905 km)² + (16.922 km)² + (0.300 km)²)
≈ 17.472 km
Therefore, the two planes are approximately 17.472 km apart.