IB Mathematics AHL 3.13 Definition and calculation of the scalar product-AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question uses vector methods to help an air traffic controller named Maria determine whether aircraft at her airport in 2025 comply with air traffic regulations.
Maria uses a coordinate system with the base of her air traffic control tower as the origin \((0, 0, 0)\). An aircraft’s position is given by \((x, y, z)\), where \(x\) and \(y\) are the aircraft’s displacements east and north of the tower, and \(z\) is the vertical displacement above the tower’s base, all in kilometres.
At 12:00, aircraft A and B are at points \(P(100, -82, 10.7)\) and \(Q(215, -197, 10.7)\), respectively.
(a) Calculate the distance between them. [2]
Aircraft A and B fly along the same straight-line flight path, with B behind A. Both have a constant velocity of \(\begin{pmatrix} -640 \\ 640 \\ 0 \end{pmatrix}\) km/h.
(b) Calculate their speed. [2]
Maria must ensure compliance with regulations: if two aircraft are on the same flight path, they must maintain at least a 10-minute gap. Aircraft too close are ‘in conflict’.
(c) Calculate the time for aircraft B to travel from \(Q\) to \(P\), and determine if A and B are in conflict. [3]
(d) Write the position vector of aircraft A, \(r_A\), \(t\) hours after 12:00. [1]
If two aircraft are on different flight paths, regulations state: when the vertical distance is less than 0.3 km, the horizontal distance must exceed 10 km; if at least 0.3 km, no restrictions apply.
Maria notices aircraft C on a different flight path near A, with position \(t\) hours after 12:00 given by: \(\begin{pmatrix} -400 \\ -41 \\ 9.1 \end{pmatrix} + t \begin{pmatrix} -140 \\ 604 \\ 2 \end{pmatrix}\).
(e)
(i) Find the times \(t\) when the distance between A and C is 10 km. [3]
(ii) Given that the distance between A and C is less than 10 km only between these times, determine if A and C violate regulations, justifying your answer. [2]
(i) Find the times \(t\) when the distance between A and C is 10 km. [3]
(ii) Given that the distance between A and C is less than 10 km only between these times, determine if A and C violate regulations, justifying your answer. [2]
Maria uses a new 2D coordinate system \((x, y)\) with origin \(R\) 2 km directly above the tower. For aircraft in the horizontal plane containing \(R\), \(x\) and \(y\) represent displacements east and north of \(R\), in km.
Aircraft D flies at a constant height of 2 km near the airport, awaiting landing permission, with position at time \(t\) given by: \(\begin{pmatrix} 6.4 \cos(3t) \\ 6.4 \sin(3t) \end{pmatrix}\).
(f) Describe the flight path of aircraft D. [1]
Simultaneously, a small aircraft E flies at the same height as D along the line with vector equation: \(\begin{pmatrix} 20 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
(g) Let \(\vec{b} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
(i) Calculate \(\vec{RE} \cdot \vec{b}\) in terms of \(\lambda\). [2]
(ii) Find the value of \(\lambda\) that minimizes the distance from \(R\) to the line of E’s path. [2]
(iii) Calculate this minimum distance. [2]
(iv) Show that aircraft D and E do not violate regulations for any \(t\). [2]
(i) Calculate \(\vec{RE} \cdot \vec{b}\) in terms of \(\lambda\). [2]
(ii) Find the value of \(\lambda\) that minimizes the distance from \(R\) to the line of E’s path. [2]
(iii) Calculate this minimum distance. [2]
(iv) Show that aircraft D and E do not violate regulations for any \(t\). [2]
▶️ Answer/Explanation
Markscheme
(a)
\(\vec{PQ} = \begin{pmatrix} 215 – 100 \\ -197 – (-82) \\ 10.7 – 10.7 \end{pmatrix} = \begin{pmatrix} -115 \\ 115 \\ 0 \end{pmatrix}\).
Distance = \(\sqrt{(-115)^2 + 115^2 + 0^2} = \sqrt{115^2 \times 2} = 115 \sqrt{2} \approx 162.634 \approx 163 \, \text{km}\).
Distance \(\approx 163 \, \text{km}\) M1A1
[2 marks]
\(\vec{PQ} = \begin{pmatrix} 215 – 100 \\ -197 – (-82) \\ 10.7 – 10.7 \end{pmatrix} = \begin{pmatrix} -115 \\ 115 \\ 0 \end{pmatrix}\).
Distance = \(\sqrt{(-115)^2 + 115^2 + 0^2} = \sqrt{115^2 \times 2} = 115 \sqrt{2} \approx 162.634 \approx 163 \, \text{km}\).
Distance \(\approx 163 \, \text{km}\) M1A1
[2 marks]
(b)
Speed = \(\sqrt{(-640)^2 + 640^2 + 0^2} = \sqrt{640^2 \times 2} = 640 \sqrt{2} \approx 905.096 \approx 905 \, \text{km/h}\).
Speed \(\approx 905 \, \text{km/h}\) M1A1
[2 marks]
Speed = \(\sqrt{(-640)^2 + 640^2 + 0^2} = \sqrt{640^2 \times 2} = 640 \sqrt{2} \approx 905.096 \approx 905 \, \text{km/h}\).
Speed \(\approx 905 \, \text{km/h}\) M1A1
[2 marks]
(c)
Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{162.634}{905.096} \approx 0.179687 \, \text{h} \times 60 \approx 10.7812 \, \text{min} \approx 10.8 \, \text{min}\).
Since 10.8 min > 10 min, not in conflict.
Time \(\approx 10.8 \, \text{min}\), not in conflict M1A1A1
[3 marks]
Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{162.634}{905.096} \approx 0.179687 \, \text{h} \times 60 \approx 10.7812 \, \text{min} \approx 10.8 \, \text{min}\).
Since 10.8 min > 10 min, not in conflict.
Time \(\approx 10.8 \, \text{min}\), not in conflict M1A1A1
[3 marks]
(d)
\(r_A = \begin{pmatrix} 100 \\ -82 \\ 10.7 \end{pmatrix} + t \begin{pmatrix} -640 \\ 640 \\ 0 \end{pmatrix}\).
\(r_A = \begin{pmatrix} 100 – 640t \\ -82 + 640t \\ 10.7 \end{pmatrix}\) A1
[1 mark]
\(r_A = \begin{pmatrix} 100 \\ -82 \\ 10.7 \end{pmatrix} + t \begin{pmatrix} -640 \\ 640 \\ 0 \end{pmatrix}\).
\(r_A = \begin{pmatrix} 100 – 640t \\ -82 + 640t \\ 10.7 \end{pmatrix}\) A1
[1 mark]
(e)(i)
\(r_A – r_C = \begin{pmatrix} (100 – 640t) – (-400 – 140t) \\ (-82 + 640t) – (-41 + 604t) \\ 10.7 – (9.1 + 2t) \end{pmatrix} = \begin{pmatrix} 500 – 500t \\ -41 + 36t \\ 1.6 – 2t \end{pmatrix}\).
Distance = \(\sqrt{(500 – 500t)^2 + (-41 + 36t)^2 + (1.6 – 2t)^2} = 10\).
Square both sides: \((500 – 500t)^2 + (-41 + 36t)^2 + (1.6 – 2t)^2 = 100\).
Compute: \(250000(1 – t)^2 + 1296(t – \frac{41}{36})^2 + 4(t – 0.8)^2 = 100\).
Simplify: \(250000(t^2 – 2t + 1) + 1296(t^2 – \frac{82}{36}t + \frac{1681}{1296}) + 4(t^2 – 1.6t + 0.64) = 100\).
Combine: \(251300 t^2 – 502592 t + 251296.97 \approx 100\).
Solve: \(251300 t^2 – 502592 t + 251196.97 = 0 \Rightarrow t^2 – 2.000368 t + 0.999602 = 0\).
\(t = \frac{2.000368 \pm \sqrt{(2.000368)^2 – 4 \times 0.999602}}{2} \approx 0.983441, 1.01799\).
Times \(t \approx 0.983, 1.02 \, \text{h}\) M1A1A1
[3 marks]
\(r_A – r_C = \begin{pmatrix} (100 – 640t) – (-400 – 140t) \\ (-82 + 640t) – (-41 + 604t) \\ 10.7 – (9.1 + 2t) \end{pmatrix} = \begin{pmatrix} 500 – 500t \\ -41 + 36t \\ 1.6 – 2t \end{pmatrix}\).
Distance = \(\sqrt{(500 – 500t)^2 + (-41 + 36t)^2 + (1.6 – 2t)^2} = 10\).
Square both sides: \((500 – 500t)^2 + (-41 + 36t)^2 + (1.6 – 2t)^2 = 100\).
Compute: \(250000(1 – t)^2 + 1296(t – \frac{41}{36})^2 + 4(t – 0.8)^2 = 100\).
Simplify: \(250000(t^2 – 2t + 1) + 1296(t^2 – \frac{82}{36}t + \frac{1681}{1296}) + 4(t^2 – 1.6t + 0.64) = 100\).
Combine: \(251300 t^2 – 502592 t + 251296.97 \approx 100\).
Solve: \(251300 t^2 – 502592 t + 251196.97 = 0 \Rightarrow t^2 – 2.000368 t + 0.999602 = 0\).
\(t = \frac{2.000368 \pm \sqrt{(2.000368)^2 – 4 \times 0.999602}}{2} \approx 0.983441, 1.01799\).
Times \(t \approx 0.983, 1.02 \, \text{h}\) M1A1A1
[3 marks]
(e)(ii)
Vertical distance: \(|1.6 – 2t|\).
Regulation: \(|1.6 – 2t| < 0.3 \Rightarrow -0.3 < 1.6 – 2t < 0.3 \Rightarrow 0.65 < t < 0.95\).
Distance < 10 km for \(0.983 < t < 1.02\), which is outside \(0.65 < t < 0.95\).
Thus, regulations are not violated.
No violation M1A1
[2 marks]
Vertical distance: \(|1.6 – 2t|\).
Regulation: \(|1.6 – 2t| < 0.3 \Rightarrow -0.3 < 1.6 – 2t < 0.3 \Rightarrow 0.65 < t < 0.95\).
Distance < 10 km for \(0.983 < t < 1.02\), which is outside \(0.65 < t < 0.95\).
Thus, regulations are not violated.
No violation M1A1
[2 marks]
(f)
Path: \(\begin{pmatrix} 6.4 \cos(3t) \\ 6.4 \sin(3t) \end{pmatrix}\).
Circle, radius 6.4 km, centered at \(R(0, 0)\).
Circular path, radius 6.4 km, centered at \(R\) A1
[1 mark]
Path: \(\begin{pmatrix} 6.4 \cos(3t) \\ 6.4 \sin(3t) \end{pmatrix}\).
Circle, radius 6.4 km, centered at \(R(0, 0)\).
Circular path, radius 6.4 km, centered at \(R\) A1
[1 mark]
(g)(i)
\(\vec{RE} = \begin{pmatrix} 20 – \lambda \\ 10 + \lambda \end{pmatrix}\), \(\vec{b} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
\(\vec{RE} \cdot \vec{b} = (20 – \lambda) \times (-1) + (10 + \lambda) \times 1 = -20 + \lambda + 10 + \lambda = 2\lambda – 10\).
\(\vec{RE} \cdot \vec{b} = 2\lambda – 10\) M1A1
[2 marks]
\(\vec{RE} = \begin{pmatrix} 20 – \lambda \\ 10 + \lambda \end{pmatrix}\), \(\vec{b} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
\(\vec{RE} \cdot \vec{b} = (20 – \lambda) \times (-1) + (10 + \lambda) \times 1 = -20 + \lambda + 10 + \lambda = 2\lambda – 10\).
\(\vec{RE} \cdot \vec{b} = 2\lambda – 10\) M1A1
[2 marks]
(g)(ii)
Minimum distance when \(\vec{RE} \cdot \vec{b} = 0\):
\(2\lambda – 10 = 0 \Rightarrow \lambda = 5\).
\(\lambda = 5\) M1A1
[2 marks]
Minimum distance when \(\vec{RE} \cdot \vec{b} = 0\):
\(2\lambda – 10 = 0 \Rightarrow \lambda = 5\).
\(\lambda = 5\) M1A1
[2 marks]
(g)(iii)
At \(\lambda = 5\), \(\vec{RE} = \begin{pmatrix} 20 – 5 \\ 10 + 5 \end{pmatrix} = \begin{pmatrix} 15 \\ 15 \end{pmatrix}\).
Distance = \(\sqrt{15^2 + 15^2} = \sqrt{450} = 15 \sqrt{2} \approx 21.2132 \approx 21.2 \, \text{km}\).
Minimum distance \(\approx 21.2 \, \text{km}\) M1A1
[2 marks]
At \(\lambda = 5\), \(\vec{RE} = \begin{pmatrix} 20 – 5 \\ 10 + 5 \end{pmatrix} = \begin{pmatrix} 15 \\ 15 \end{pmatrix}\).
Distance = \(\sqrt{15^2 + 15^2} = \sqrt{450} = 15 \sqrt{2} \approx 21.2132 \approx 21.2 \, \text{km}\).
Minimum distance \(\approx 21.2 \, \text{km}\) M1A1
[2 marks]
(g)(iv)
D’s maximum distance from \(R\): 6.4 km. E’s minimum distance from \(R\): 21.2 km.
Maximum distance D to E: \(21.2 + 6.4 = 27.6 \, \text{km} > 10 \, \text{km}\).
Minimum distance D to E: \(21.2 – 6.4 = 14.8 \, \text{km} > 10 \, \text{km}\).
Since both are at the same height (2 km), vertical distance = 0 < 0.3 km, but horizontal distance always exceeds 10 km.
No violation M1A1
[2 marks]
D’s maximum distance from \(R\): 6.4 km. E’s minimum distance from \(R\): 21.2 km.
Maximum distance D to E: \(21.2 + 6.4 = 27.6 \, \text{km} > 10 \, \text{km}\).
Minimum distance D to E: \(21.2 – 6.4 = 14.8 \, \text{km} > 10 \, \text{km}\).
Since both are at the same height (2 km), vertical distance = 0 < 0.3 km, but horizontal distance always exceeds 10 km.
No violation M1A1
[2 marks]
Total Marks: 16