An aircraft’s position is given by the coordinates $(x, y, z)$, where $x$ and $y$ are the aircraft’s displacement east and north of an airport, and $z$ is the height of the aircraft above the ground. All displacements are given in kilometres.
The velocity of the aircraft is given as $\begin{pmatrix} -150 \\ -50 \\ -20 \end{pmatrix} \text{ km h}^{-1}$.
At 13:00 it is detected at a position 30 km east and 10 km north of the airport, and at a height of 5 km. Let $t$ be the length of time in hours from 13:00.
If the aircraft continued to fly with the velocity given, when the aircraft is 4 km above the ground it continues to fly on the same bearing but adjusts the angle of its descent so that it will land at the point $(0, 0, 0)$.
Write down a vector equation for the displacement, $\mathbf{r}$, of the aircraft in terms of $t$.
- State the vector equation: $\mathbf{r} = \begin{pmatrix} 30 \\ 10 \\ 5 \end{pmatrix} + t \begin{pmatrix} -150 \\ -50 \\ -20 \end{pmatrix}$ A1 A1
2 marks total
Verify that it would pass directly over the airport.
- Substitute $x=0$: $30 - 150t = 0 \Rightarrow t = \frac{30}{150} = 0.2$ M1
- Substitute $y=0$: $10 - 50t = 0 \Rightarrow t = \frac{10}{50} = 0.2$ A1
Since the two values of $t$ are equal, the aircraft passes directly over the airport.
2 marks total
State the height of the aircraft at this point.
- Calculate height: $z = 5 - 20(0.2) = 1$ km A1
1 mark total
Find the time at which it would fly directly over the airport.
Find the time at which the aircraft is 4 km above the ground.
Find the direct distance of the aircraft from the airport at this point.
- Calculate displacement at $t = \frac{1}{20}$: $\mathbf{r} = \begin{pmatrix} 30 \\ 10 \\ 5 \end{pmatrix} + \frac{1}{20} \begin{pmatrix} -150 \\ -50 \\ -20 \end{pmatrix} = \begin{pmatrix} 22.5 \\ 7.5 \\ 4 \end{pmatrix}$ A1
- Calculate distance: $d = \sqrt{22.5^2 + 7.5^2 + 4^2}$ M1
- State result: $d = 24.1$ km A1
3 marks total
Given that the velocity of the aircraft, after the adjustment of the angle of descent, is $\begin{pmatrix} -150 \\ -50 \\ a \end{pmatrix} \text{ km h}^{-1}$, find the value of $a$.
METHOD 1
- Calculate time until landing: $12 - 3 = 9 \text{ minutes} = 0.15 \text{ hours}$ M1
- Identify height to descend: 4 km
- Use vertical velocity equation: $a = \frac{-4}{0.15}$ M1
- Calculate $a$: $a = -26.7$ A1
METHOD 2
- Relate velocity to position vector: $\begin{pmatrix} -150 \\ -50 \\ a \end{pmatrix} = s \begin{pmatrix} 22.5 \\ 7.5 \\ 4 \end{pmatrix}$ M1
- Solve for scale factor $s$: $-150 = 22.5s \Rightarrow s = -\frac{20}{3}$ M1
- Calculate $a$: $a = -\frac{20}{3} \times 4 = -26.7$ A1
3 marks total
