An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 750 m, horizontal distance 19.8 km, and 23.0° south of west. The second aircraft is at altitude 1300 m, horizontal distance 17.8 km, and 18.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.)

letting the origin be the center of the radar, then positions are

First: Change the ranges to distances along the axis given. Then, having two positions in the form of (xxxx,yyyy,zzzz)

you use the distance equation

distance= sqrt [(xxxx1-xxxx2)^2+ (yyyy1-yyyy2)^2 + (zzzz1-zzzz2))

remember to keep all numbers in the same units, either km, or meters.

To find the distance between the two aircraft, we can use the concept of vector addition. We can break down the given information into two vectors (one for each aircraft) and then find the resultant vector by adding these two vectors.

Let's represent the position of the first aircraft as vector R1 and the position of the second aircraft as vector R2. We can then find the resultant vector R by summing R1 and R2.

R1 = (-19.8 km) i + (-750 m) k
R2 = (-17.8 km) i + (-1300 m) k

Here, i, j, and k represent the unit vectors in the x, y, and z directions, respectively.

Now, let's calculate the resultant vector R by adding R1 and R2:

R = R1 + R2 = (-19.8 km) i + (-750 m) k + (-17.8 km) i + (-1300 m) k
= (-37.6 km) i + (-2050 m) k

To find the distance between the two aircraft, we need to find the magnitude (length) of the resultant vector R. The magnitude of a vector can be calculated using the Pythagorean theorem:

|R| = sqrt((Rx)^2 + (Rz)^2)

Here, Rx and Rz represent the components of the vector R in the x and z directions, respectively.

|R| = sqrt((-37.6 km)^2 + (-2050 m)^2)
= sqrt(1409.76 km^2 + 4,202,500 m^2)
= sqrt(1409.76 km^2 + 4.2025 km^2) (Converting m^2 to km^2)
= sqrt(1414.9625 km^2)
≈ 37.65 km

Therefore, the distance between the two aircraft is approximately 37.65 km.