IB Mathematics AHL 3.10 Concept of a vector AI HL Paper 2- Exam Style Questions- New Syllabus
Question
A protected nature reserve features three peaks located at points A, B, and C.
Based on a coordinate grid, the location of A is \( A = (0, 0, 2.8) \) and the location of B is \( B = (7.2, 5.1, 2.4). \) All measurements are in kilometres.
(a) (i) Determine the vector \(\overrightarrow{AB}\).
(ii) Use this to calculate \(AB\), the distance separating A and B.
The vector \(\overrightarrow{AC}\) aligns with the vector \( \begin{pmatrix} 1.1 \\[6pt] 8.4 \\[6pt] 0.2 \end{pmatrix}. \)
(b) Compute the angle formed between \( \begin{pmatrix} 1.1 \\[6pt] 8.4 \\[6pt] 0.2 \end{pmatrix} \quad \text{and} \quad \overrightarrow{AB}. \)
The angle between \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) measures \(55.2^\circ\).
(c) Apply the sine rule to determine the length of \(AC\).
▶️ Answer/Explanation
(a) (i) \( \overrightarrow{AB}=B-A= \begin{pmatrix}7.2\\[2pt]5.1\\[2pt]2.4\end{pmatrix}- \begin{pmatrix}0\\[2pt]0\\[2pt]2.8\end{pmatrix} = \begin{pmatrix}7.2\\[2pt]5.1\\[2pt]-0.4\end{pmatrix}. \)
(a) (ii) \( |AB|=\left\lVert\overrightarrow{AB}\right\rVert =\sqrt{7.2^{2}+5.1^{2}+(-0.4)^{2}} =\sqrt{51.84+26.01+0.16} =\sqrt{77.99}=8.83322\ldots \text{ km} \approx \boxed{8.83\text{ km}}. \)
(b) \( \mathbf{u}= \begin{pmatrix}1.1\\[2pt]8.4\\[2pt]0.2\end{pmatrix}, \qquad \mathbf{v}=\overrightarrow{AB}= \begin{pmatrix}7.2\\[2pt]5.1\\[2pt]-0.4\end{pmatrix}. \) Magnitude of \(\mathbf{u}\): \( \|\mathbf{u}\|=\sqrt{1.1^{2}+8.4^{2}+0.2^{2}} =\sqrt{71.81}=8.47407\ldots \) Dot product: \( \mathbf{u}\cdot\mathbf{v}=1.1(7.2)+8.4(5.1)+0.2(-0.4)=50.68. \) Hence \( \cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|} =\frac{50.68}{(8.47407\ldots)(8.83322\ldots)}=0.67706\ldots \) \( \therefore\quad \boxed{\theta=47.38^\circ\ (\text{to }2\,\mathrm{d.p.})}. \)
(c) \( \angle ACB=180^\circ-55.2^\circ-47.3805^\circ =77.4194^\circ\ (\text{approx}). \) By the Sine Rule, \( \frac{AC}{\sin(55.2^\circ)}=\frac{AB}{\sin(77.4194^\circ)} \quad\Rightarrow\quad AC=\frac{\sin(55.2^\circ)}{\sin(77.4194^\circ)}\times 8.83322\ldots =7.43107\ldots \text{ km}. \) \( \boxed{AC\approx 7.43\text{ km}}. \)