An air-traffic controller observes two aircraft on his radar screen. The first is at altitude $800\mathrm{m},$ horizontal distance $19.2\mathrm{km},$ and $25.0^{\circ }$ south of west. The second aircraft is at altitude $1100\mathrm{m}$, horizontal distance $17.6\mathrm{km},$ and $20.0^{\circ }$ 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.)

Key Concepts: Distance Formula, Trigonometry, 3D Coordinate Geometry Explanation: To solve this question, we can use the distance formula from 3D coordinate geometry. We can first start by converting the given information into Cartesian coordinates. Step by Step Solution: Step 1. The first aircraft is at altitude $800\mathrm{m}$, horizontal distance $19.2\mathrm{km},$ and $25.0^{\circ }$ south of west. This means that the angle it makes with the negative $x$-axis is $25.0^{\circ }+90^{\circ }=115.0^{\circ }$ (measured counterclockwise from the negative $x$-axis since the $y$-axis is assumed to be south). Hence, the corresponding direction vector is $\left(\begin{array}{c}\cos 115.0^{\circ }\\ \sin 115.0^{\circ }\\ 0\end{array}\right)=\left(\begin{array}{c}-0.4226\\ 0.9063\\ 0\end{array}\right)$. Using the given horizontal distance, we can obtain the position vector of the first aircraft as $\left(\begin{array}{c}0\\ -19200\\ 800\end{array}\right)\text{km}\cdot \left(\begin{array}{c}-0.4226\\ 0.9063\\ 0\end{array}\right)=\left(\begin{array}{c}8101.28\\ -17381.64\\ 0\end{array}\right)\text{km}$ Step 2. Similarly, the second aircraft is at altitude $1100\mathrm{m}$, horizontal distance $17.6\mathrm{km},$ and $20.0^{\circ }$ south of west. This means that the angle it makes with the negative x-axis is $20.0^{\circ }+90^{\circ }=110.0^{\circ }.$ Hence, the corresponding direction vector is $\left(\begin{array}{c}\cos 110.0^{\circ }\\ \sin 110.0^{\circ }\\ 0\end{array}\right)=\left(\begin{array}{c}-0.3420\\ 0.9397\\ 0\end{array}\right)$. Using the given horizontal distance, we can obtain the position vector of the second aircraft as $\left(\begin{array}{c}0\\ -17600\\ 1100\end{array}\right)\text{km}\cdot \left(\begin{array}{c}-0.3420\\ 0.9397\\ 0\end{array}\right)=\left(\begin{array}{c}6035.20\\ -16554.24\\ 0\end{array}\right)\text{km}$ Step 3. The distance between the two aircraft is the magnitude of the difference between their position vectors, i.e., $\left|\left(\begin{array}{c}8101.28\\ -17381.64\\ 0\end{array}\right)-\left(\begin{array}{c}6035.20\\ -16554.24\\ 0\end{array}\right)\right|=3421.63\text{ km}$ Final Answer: The distance between the two aircraft is $3421.63\mathrm{km}$.