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

Vectors: exam review Exam review: second paper Answers

Section A: short questions

    1. (a) \(\overrightarrow{PQ} = \begin{pmatrix} 5 \\ -4 \\ -4 \end{pmatrix}\)
    2. (b) \(\sqrt{57}\) metres
    3. (c) \(\begin{pmatrix} 7.95 \\ -6.36 \\ -6.36 \end{pmatrix}\)
    1. (a) \(\overrightarrow{CD} = \mathbf{d} - \mathbf{c}\)
    2. (b) Shown.
    3. (c) \(\overrightarrow{NS} = 3\overrightarrow{NR}\), so N, R, S are collinear.
    1. (a) \(p = 3\), \(q = -6\)
    2. (b) \(k = -3.49\) or \(k = 3.09\)
    1. (a) \(\mathbf{r} = \begin{pmatrix} -3 \\ 4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}\)
    2. (b) No: R does not lie on \(l\)
    3. (c) \(x = 4 \lambda - 3,\ y = 4 - 2 \lambda,\ z = 5 \lambda + 1\)
    1. (a) \(\sqrt{74}\) = 8.60 km/h
    2. (b) 306°
    3. (c) 24.1 km
    1. (a) \(\mathbf{a} \cdot \mathbf{b} = -15.4\)
    2. (b) 113° (obtuse, because the scalar product is negative)
    3. (c) \(t = \frac{17}{2}\)
    1. (a) \(|\mathbf{a} \times \mathbf{b}| = 36.3\)
    2. (b) \(\mathbf{p} \times \mathbf{q} = \begin{pmatrix} -8 \\ 10 \\ -7 \end{pmatrix}\) (\(-8\mathbf{i} + 10\mathbf{j} - 7\mathbf{k}\))
    3. (c) -2.41
    1. (a) \(1.96\text{ m s}^{-2}\)
    2. (b) \(26.1\) metres
    3. (c) \(5.76\text{ m s}^{-2}\)
    1. (a) \(\begin{pmatrix} 36 \\ 4 \end{pmatrix}\text{ m s}^{-2}\)
    2. (b) \(\mathbf{r} = \begin{pmatrix} 2 t^{3} - 2 t - 3 \\ 2 t^{2} - 7 t + 4 \end{pmatrix}\) metres
    3. (c) \(22.0\) m s⁻¹
    1. (a) \(\mathbf{v} = \begin{pmatrix} 11 \\ 12 - 9.8t \end{pmatrix}\)
    2. (b) 7.35 metres
    3. (c) 26.9 metres
    1. (a) Centre \((-2, 5)\), radius \(3\) metres
    2. (b) Clockwise, 4.19 seconds
    3. (c) \(\mathbf{r} = \begin{pmatrix} -2 + 3\sin(\dfrac{2\pi t}{5}) \\ 5 - 3\cos(\dfrac{2\pi t}{5}) \end{pmatrix}\) metres

Section B: long questions

    1. (a) \((0, 7, -3)\) (with \(\lambda = 2\), \(\mu = 1\))
    2. (b) 63.6°
    3. (c) 1.41
    4. (d) No solution for the third component, so no intersection (and the lines are not parallel).
    1. (a) 11.6 m/s
    2. (b) Shown.
    3. (c) \(\mathbf{r}_{K} = \begin{pmatrix} -80 \\ 60 \\ 25 \end{pmatrix} + t\begin{pmatrix} 12 \\ -6 \\ 4 \end{pmatrix}\)
    4. (d) 33.5 metres
    5. (e) The same value of t does not satisfy all three components.
    6. (f) t = 14.2 seconds, distance 24.9 metres
    1. (a) \(\overrightarrow{DE} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\), \(\overrightarrow{DF} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\)
    2. (b) 63°
    3. (c) \(\begin{pmatrix} 11 \\ -10 \\ 13 \end{pmatrix}\)
    4. (d) 9.87 m²
    5. (e) \(\pm\dfrac{1}{\sqrt{390}}\begin{pmatrix} 11 \\ -10 \\ 13 \end{pmatrix}\)
    6. (f) 5.28 metres
    1. (a) Radius \(6\) metres
    2. (b) \(\mathbf{v}_P = \begin{pmatrix} -0.96t\sin(0.08t^2) \\ 0.96t\cos(0.08t^2) \end{pmatrix}\) m s⁻¹
    3. (c) \(2.88\text{ m s}^{-1}\)
    4. (d) 2 complete revolutions
    5. (e) \(\mathbf{r}_Q = \begin{pmatrix} 6\cos(0.08(t - 4)^2) \\ 6\sin(0.08(t - 4)^2) \end{pmatrix}\) metres
    6. (f) 11.3 metres
    1. (a) At rest when \(t = 2\) and \(t = 8\); it changes direction at both times
    2. (b) 65.3 metres
    3. (c) 4.63 seconds
    1. (a) \(\mathbf{r} = \begin{pmatrix} 10.7t \\ 1.8 + 9.00t - 4.9t^2 \end{pmatrix}\)
    2. (b) 2.02 seconds; 21.6 metres
    3. (c) 5.93 metres
    4. (d) 15.2 m s⁻¹; 45.2° below the horizontal
    5. (e) No, it hits the wall, 0.569 m below the top

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