IBDP Maths AHL 3.13 Scalar product of two vectors AA HL Paper 2- Exam Style Questions- New Syllabus
Consider the points \(A(5, 2, 1)\), \(B(6, 5, 3)\), and \(C(7, 6, a + 1)\), \(a \in \mathbb{R}\).
Let \(\theta\) be the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
a. Find
(i) \(\overrightarrow{AB}\);
(ii) \(\overrightarrow{AC}\). [3]
b. Find the value of \(a\) for which \(\theta = \frac{\pi}{2}\). [4]
c.
(i) Show that \(\cos \theta = \frac{2a + 14}{\sqrt{14(a^2 + 20)}}\).
(ii) Hence, find the value of \(a\) for which \(\theta = 1.2\). [8]
▶️ Answer/Explanation
a. [3 marks]
(i) Appropriate approach, e.g., \(\overrightarrow{AO} + \overrightarrow{OB}\) or \(\mathbf{B} – \mathbf{A}\) (M1).
\(\overrightarrow{AB} = \begin{pmatrix} 6 – 5 \\ 5 – 2 \\ 3 – 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}\) (A1).
(ii) \(\overrightarrow{AC} = \begin{pmatrix} 7 – 5 \\ 6 – 2 \\ (a + 1) – 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ a \end{pmatrix}\) (A1).
b. [4 marks]
Valid reasoning, e.g., scalar product is zero for \(\theta = \frac{\pi}{2}\), \(\cos \frac{\pi}{2} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}\) (R1).
Scalar product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\): \(1 \cdot 2 + 3 \cdot 4 + 2 \cdot a = 2 + 12 + 2a = 2a + 14\) (A1).
Set scalar product to zero: \(2a + 14 = 0 \Rightarrow 2a = -14\) (A1).
\(a = -7\) (A1).
c. [8 marks]
(i) Magnitudes: \(|\overrightarrow{AB}| = \sqrt{1^2 + 3^2 + 2^2} = \sqrt{14}\) (A1), \(|\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + a^2} = \sqrt{4 + 16 + a^2} = \sqrt{a^2 + 20}\) (A1).
Substitution into cosine formula: \(\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{1 \cdot 2 + 3 \cdot 4 + 2 \cdot a}{\sqrt{1^2 + 3^2 + 2^2} \sqrt{2^2 + 4^2 + a^2}} = \frac{2 + 12 + 2a}{\sqrt{14} \sqrt{a^2 + 20}}\) (M1).
Simplification: \(\cos \theta = \frac{14 + 2a}{\sqrt{14} \sqrt{a^2 + 20}} = \frac{2(a + 7)}{\sqrt{14} \sqrt{a^2 + 20}} = \frac{2(a + 7)}{\sqrt{14(a^2 + 20)}}\) (A1).
(ii) Set up equation: \(\cos 1.2 = \frac{2a + 14}{\sqrt{14(a^2 + 20)}}\) (A1).
Valid attempt to solve, e.g., square both sides or numerical method (M1).
Square both sides: \((\cos 1.2)^2 = \left( \frac{2a + 14}{\sqrt{14(a^2 + 20)}} \right)^2\).
\(\cos 1.2 \approx 0.362357\), so \((\cos 1.2)^2 \approx 0.131303\).
\((2a + 14)^2 = (0.131303) \cdot 14(a^2 + 20)\).
\(4(a + 7)^2 = 1.83824(a^2 + 20)\).
Expand: \(4(a^2 + 14a + 49) = 1.83824a^2 + 36.7648\).
\(4a^2 + 56a + 196 = 1.83824a^2 + 36.7648\).
\(2.16176a^2 + 56a + 159.2352 = 0\).
Solve quadratic: \(a = \frac{-56 \pm \sqrt{56^2 – 4 \cdot 2.16176 \cdot 159.2352}}{2 \cdot 2.16176}\).
Discriminant: \(3136 – 1377.28 = 1758.72\).
\(a = \frac{-56 \pm \sqrt{1758.72}}{4.32352}\).
\(a \approx -3.25\) or \(a \approx -22.66\).
Test \(a = -3.25\): \(\cos \theta = \frac{2(-3.25) + 14}{\sqrt{14((-3.25)^2 + 20)}} = \frac{7.5}{\sqrt{14 \cdot 30.5625}} \approx 0.362\), matches \(\cos 1.2\).
Test \(a = -22.66\): yields negative cosine, invalid for \(\theta = 1.2\).
\(a = -3.25\) (A2).
Markscheme Answers:
(a) (i) \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}\) (M1A1); (ii) \(\overrightarrow{AC} = \begin{pmatrix} 2 \\ 4 \\ a \end{pmatrix}\) (A1)
(b) \(a = -7\) (R1A1A1A1)
(c) (i) \(\cos \theta = \frac{2a + 14}{\sqrt{14(a^2 + 20)}}\) (A1A1M1A1); (ii) \(a = -3.25\) (A1M1A2)