IB Mathematics AHL 3.13 Definition and calculation of the scalar product-AI HL Paper 2- Exam Style Questions- New Syllabus
Question
In this question, all distances are measured in kilometres, and \( t \) represents time in hours.
Let \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \) denote a displacement of 1 km eastward, \( \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \) a displacement of 1 km northward, and \( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) a vertical displacement of 1 km.
Highway 85 in Saudi Arabia is a long, straight, flat roadway.
Relative to the center of Arar town, point \( O \), the position vector of a car, \( C \), moving along this road is expressed as:
\[ \overrightarrow{OC} = \begin{bmatrix} 10 \\ -5 \\ 0 \end{bmatrix} + t \begin{bmatrix} 50 \\ -33 \\ 0 \end{bmatrix}. \]
(a) Determine the car’s speed. [2]
The police are evaluating a long-range drone, \( D \), to observe vehicles on this road. The drone is deployed at \( t = 0 \) from a point with position vector \( \begin{bmatrix} 200 \\ -100 \\ 0.02 \end{bmatrix} \) and moves in a straight line at a constant altitude of 0.02 km with a constant velocity of \( \begin{bmatrix} -15 \\ -20 \\ 0 \end{bmatrix} \).
(b) Calculate the angle between the car’s path and the drone’s path. [3]
(c) State the position vector \( \overrightarrow{OD} \) of the drone at time \( t \). [1]
(d) At time \( t_1 \), the drone reaches the point with position vector:
\[ \begin{bmatrix} 152 \\ p \\ 0.02 \end{bmatrix}. \]
Find the values of:
(i) \( t_1 \)
(ii) \( p \)[3]
(e)
(i) Derive an expression for \( \overrightarrow{CD} \), the position of the drone relative to the car.
(ii) Using this, determine the minimum distance between the car and the drone.[6]
▶️ Answer/Explanation
(a) The speed is the magnitude of the velocity vector \( \begin{bmatrix} 50 \\ -33 \\ 0 \end{bmatrix} \):
\( \text{Speed} = \sqrt{50^2 + (-33)^2} = \sqrt{2500 + 1089} = \sqrt{3589} \approx 59.9082 \approx 59.9 \, \text{km h}^{-1} \). (M1)A1
[2 marks]
(b) The angle between the car’s velocity \( \begin{bmatrix} 50 \\ -33 \\ 0 \end{bmatrix} \) and the drone’s velocity \( \begin{bmatrix} -15 \\ -20 \\ 0 \end{bmatrix} \) is found using the dot product:
\( \begin{bmatrix} 50 \\ -33 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} -15 \\ -20 \\ 0 \end{bmatrix} = (50 \times -15) + (-33 \times -20) + (0 \times 0) = -750 + 660 = -90 \). (A1)
Magnitudes: \( \left| \begin{bmatrix} 50 \\ -33 \\ 0 \end{bmatrix} \right| \approx 59.9082 \), \( \left| \begin{bmatrix} -15 \\ -20 \\ 0 \end{bmatrix} \right| = \sqrt{(-15)^2 + (-20)^2} = \sqrt{625} = 25 \).
\( \cos \theta = \frac{-90}{59.9082 \times 25} \approx \frac{-90}{1497.705} \approx -0.06007 \).
\( \theta = \cos^{-1}(-0.06007) \approx 93.4450^\circ \approx 93.4^\circ \). (M1)A1
[3 marks]
(c) The position vector of the drone:
\( \overrightarrow{OD} = \begin{bmatrix} 200 \\ -100 \\ 0.02 \end{bmatrix} + t \begin{bmatrix} -15 \\ -20 \\ 0 \end{bmatrix} \). A1
[1 mark]
(d) (i) Set the x-component of the drone’s position vector equal to 152:
\( 200 + t \times (-15) = 152 \implies -15t = 152 – 200 \implies -15t = -48 \implies t = \frac{48}{15} = \frac{16}{5} = 3.2 \). (M1)A1
(ii) For the y-component: \( -100 + (-20) \times 3.2 = -100 – 64 = -164 \).
Thus, \( p = -164 \). A1
[3 marks]
(e) (i) The relative position vector \( \overrightarrow{CD} = \overrightarrow{OD} – \overrightarrow{OC} \):
\( \overrightarrow{OD} = \begin{bmatrix} 200 – 15t \\ -100 – 20t \\ 0.02 \end{bmatrix}, \quad \overrightarrow{OC} = \begin{bmatrix} 10 + 50t \\ -5 – 33t \\ 0 \end{bmatrix} \).
\( \overrightarrow{CD} = \begin{bmatrix} (200 – 15t) – (10 + 50t) \\ (-100 – 20t) – (-5 – 33t) \\ 0.02 – 0 \end{bmatrix} = \begin{bmatrix} 190 – 65t \\ -95 + 13t \\ 0.02 \end{bmatrix} \). (M1)A1
(ii) The distance between the car and drone is the magnitude of \( \overrightarrow{CD} \):
\( d(t) = \sqrt{(190 – 65t)^2 + (13t – 95)^2 + 0.02^2} \).
To find the minimum, compute the derivative of the squared distance (ignoring the square root for simplicity):
\( f(t) = (190 – 65t)^2 + (13t – 95)^2 + 0.02^2 \).
Derivative: \( f'(t) = 2 \times (190 – 65t) \times (-65) + 2 \times (13t – 95) \times 13 \).
\( f'(t) = -130 \times (190 – 65t) + 26 \times (13t – 95) = -24700 + 8450t + 338t – 2470 = 8788t – 27170 \).
Set \( f'(t) = 0 \): \( 8788t – 27170 = 0 \implies t \approx 3.091 \).
Evaluate distance at \( t \approx 3.091 \): \( 190 – 65 \times 3.091 \approx -10.915 \), \( 13 \times 3.091 – 95 \approx -54.817 \).
\( d \approx \sqrt{(-10.915)^2 + (-54.817)^2 + 0.02^2} \approx \sqrt{119.137 + 3004.91 + 0.0004} \approx 55.8931 \approx 55.9 \, \text{km} \). (M1)(A1)(M1)A1
[6 marks]