# physics

posted by amanda

You spot a plane that is 1.63 km north, 2.3 km east, and at an altitude 5.2 km above your position.

1) How far from you is the plane?

2) At what angle from due north (in the horizontal plane) are you looking?

3) Determine the plane's position vector (from your location) in terms of the unit vectors, letting 'i' be toward the east direction, 'j' be toward the north direction, and 'k' be in vertically upward.

6) At what elevation angle (above the horizontal plane of Earth) is the airplane?

answer : ° above the horizon

1. Steve

#1: distance in 2D is √(x^2+y^2)
in 3D it is √(x^2+y^2+z^2). In this case, that is

√(1.63^2 + 2.3^2 + 5.2^2) = 5.91

#2 The angle θ is such that
tanθ = 2.3/1.63

#3,4,5: Huh? They gave you the distances and directions. Just match them up

#6: the angle θ is such that
sinθ = z-distance/xyz-distance
= 5.2/√(1.63^2 + 2.3^2 + 5.2^2) = 0.879

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