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.
answer : km 'i'
4) answer : km 'j'
5) answer : km 'k'
6) At what elevation angle (above the horizontal plane of Earth) is the airplane?
answer : ° above the horizon
#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
To answer these questions, we can use the concept of vector addition and trigonometry. Let's break down each question and explain how to find the answer.
1) How far from you is the plane?
To find the distance between you and the plane, we can use the formula for the magnitude (length) of a 3D vector. Using the given coordinates for the plane (1.63 km north, 2.3 km east, 5.2 km above your position), we can calculate the magnitude of the vector using the Pythagorean theorem.
Distance = sqrt((1.63 km)^2 + (2.3 km)^2 + (5.2 km)^2)
2) At what angle from due north (in the horizontal plane) are you looking?
To find the angle, we can use trigonometry. Given the coordinates of the plane (1.63 km north, 2.3 km east), we can use the tangent function to calculate the angle. The angle from due north can be found as:
Angle = arctan((2.3 km) / (1.63 km))
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.
The position vector can be represented as a sum of component vectors. Taking i, j, and k as the unit vectors in the east, north, and vertical directions respectively, the position vector of the plane can be written as:
Position Vector = (2.3 km) * i + (1.63 km) * j + (5.2 km) * k
4) answer: km 'i'
5) answer: km 'j'
6) At what elevation angle (above the horizontal plane of Earth) is the airplane?
To find the elevation angle, we can again use trigonometry. Given the altitude of the plane (5.2 km) and the distance between you and the plane (found in question 1), we can calculate the angle using the tangent function. The elevation angle can be found as:
Elevation Angle = arctan((5.2 km) / (Distance))
Please note that for questions 1, 3, and 6, you need to substitute the values obtained through the calculations to get the final answers.