IB Mathematics AHL 3.13 Definition and calculation of the scalar product-AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
When \(t=0\), \(r = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\).
Coordinates of A: (2, 1, 0).
(b)
(i) \(\vec{AB}\) is the displacement at \(t=5\).
\(\vec{AB} = 5 \begin{pmatrix} -4 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} -20 \\ 20 \\ 35 \end{pmatrix}\).
(ii) Magnitude \(|\vec{AB}| = \sqrt{(-20)^2 + 20^2 + 35^2} = \sqrt{400 + 400 + 1225} = \sqrt{2025}\).
\(|\vec{AB}| = \mathbf{45 \text{ m}}\).
(c)
Angle \(C\hat{A}B\) is the angle between the direction vectors of \(P_1\) and \(P_2\).
\(d_1 = \begin{pmatrix} -4 \\ 4 \\ 7 \end{pmatrix}\), \(d_2 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}\).
Dot product: \((-4)(-1) + (4)(2) + (7)(0) = 4 + 8 + 0 = 12\).
Magnitudes: \(|d_1| = \sqrt{16+16+49} = \sqrt{81} = 9\).
\(|d_2| = \sqrt{1+4} = \sqrt{5}\).
\(\cos \theta = \frac{12}{9\sqrt{5}}\).
\(\theta = \arccos\left(\frac{12}{9\sqrt{5}}\right) \approx \mathbf{53.4^\circ}\) (0.932 rad).
(d)
At \(t=5\):
\(P_1\) is at distance 45m from A (calculated in b).
\(P_2\) travels at \(12 \text{ ms}^{-1}\). Distance \(AP_2 = 12 \times 5 = 60 \text{ m}\).
Using Cosine Rule in \(\Delta AP_1P_2\) (with angle \(53.4^\circ\)):
\(x^2 = 45^2 + 60^2 – 2(45)(60)\cos(53.4^\circ)\).
Using exact cosine value \(\frac{12}{9\sqrt{5}}\):
\(x^2 = 2025 + 3600 – 5400(\frac{12}{9\sqrt{5}}) \approx 2405\).
\(x \approx \mathbf{49.0 \text{ m}}\).