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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.30

Vectors: exam review Exam review Answers

Section A: short questions

    1. (a) \(\overrightarrow{PQ} = \begin{pmatrix} -3 \\ 4 \\ -6 \end{pmatrix}\)
    2. (b) \(\sqrt{61}\) metres
    3. (c) \(\begin{pmatrix} -3.84 \\ 5.12 \\ -7.68 \end{pmatrix}\)
    1. (a) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
    2. (b) Shown.
    3. (c) \(\overrightarrow{MQ} = 3\overrightarrow{MP}\), so M, P, Q are collinear.
    1. (a) \(p = 15\), \(q = -2\)
    2. (b) \(k = -4.87\) or \(k = 2.87\)
    1. (a) \(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)
    2. (b) Yes: C lies on \(l\) (\(\lambda = -1\))
    3. (c) \(x = 3 \lambda + 2,\ y = 2 \lambda - 1,\ z = 3 - 4 \lambda\)
    1. (a) \(\sqrt{97}\) = 9.85 km/h
    2. (b) 114°
    3. (c) 18.7 km
    1. (a) \(\mathbf{a} \cdot \mathbf{b} = -21.4\)
    2. (b) 122° (obtuse, because the scalar product is negative)
    3. (c) \(t = \frac{13}{4}\)
    1. (a) \(|\mathbf{a} \times \mathbf{b}| = 34.4\)
    2. (b) \(\mathbf{p} \times \mathbf{q} = \begin{pmatrix} -11 \\ 5 \\ 9 \end{pmatrix}\) (\(-11\mathbf{i} + 5\mathbf{j} + 9\mathbf{k}\))
    3. (c) -1.18
    1. (a) \(-5.60\text{ m s}^{-2}\)
    2. (b) \(5.91\) metres
    3. (c) \(6.24\text{ m s}^{-2}\)
    1. (a) \(\begin{pmatrix} 12 \\ 6 \end{pmatrix}\text{ m s}^{-2}\)
    2. (b) \(\mathbf{r} = \begin{pmatrix} t^{3} - 4 t + 2 \\ 3 t^{2} - 5 t - 1 \end{pmatrix}\) metres
    3. (c) \(10.6\) m s⁻¹
    1. (a) \(\mathbf{v} = \begin{pmatrix} 9 \\ 14 - 9.8t \end{pmatrix}\)
    2. (b) 10.0 metres
    3. (c) 25.7 metres
    1. (a) Centre \((6, -3)\), radius \(4\) metres
    2. (b) Anticlockwise, 12.6 seconds
    3. (c) \(\mathbf{r} = \begin{pmatrix} 6 + 4\sin(\dfrac{\pi t}{4}) \\ -3 + 4\cos(\dfrac{\pi t}{4}) \end{pmatrix}\) metres

Section B: long questions

    1. (a) \((5, 4, 5)\) (with \(\lambda = 2\), \(\mu = 1\))
    2. (b) 57.7°
    3. (c) 2.24
    4. (d) No solution for the third component, so no intersection (and the lines are not parallel).
    1. (a) 4.58 m/s
    2. (b) Shown.
    3. (c) \(\mathbf{r}_{B} = \begin{pmatrix} -8 \\ 32 \\ 12 \end{pmatrix} + t\begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\)
    4. (d) 26.3 metres
    5. (e) The same value of t does not satisfy all three components.
    6. (f) t = 2.47 seconds, distance 26.2 metres
    1. (a) \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix}\)
    2. (b) 26°
    3. (c) \(\begin{pmatrix} -2 \\ -5 \\ 8 \end{pmatrix}\)
    4. (d) 4.82 m²
    5. (e) \(\pm\dfrac{1}{\sqrt{93}}\begin{pmatrix} -2 \\ -5 \\ 8 \end{pmatrix}\)
    6. (f) 2.34 metres
    1. (a) Radius \(8\) metres
    2. (b) \(\mathbf{v}_C = \begin{pmatrix} -0.8t\sin(0.05t^2) \\ 0.8t\cos(0.05t^2) \end{pmatrix}\) m s⁻¹
    3. (c) \(3.20\text{ m s}^{-1}\)
    4. (d) 3 complete revolutions
    5. (e) \(\mathbf{r}_D = \begin{pmatrix} 8\cos(0.05(t - 5)^2) \\ 8\sin(0.05(t - 5)^2) \end{pmatrix}\) metres
    6. (f) 16.0 metres
    1. (a) At rest when \(t = 2\) and \(t = 7\); it changes direction at both times
    2. (b) 36.3 metres
    3. (c) 4.86 seconds
    1. (a) \(\mathbf{r} = \begin{pmatrix} 13.1t \\ 1.5 + 9.18t - 4.9t^2 \end{pmatrix}\)
    2. (b) 2.02 seconds; 26.5 metres
    3. (c) 5.80 metres
    4. (d) 16.9 m s⁻¹; 39.1° below the horizontal
    5. (e) Yes, it clears the fence by 0.585 m

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