Unit 20 Review Exercises
Vectors I
These questions mix the ideas from the whole unit. All of them are HL content. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 AHL The vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are non-zero and non-parallel. Let \(\boldsymbol{p} = 2\boldsymbol{a} - 3\boldsymbol{b}\) and \(\boldsymbol{q} = \boldsymbol{a} + \boldsymbol{b}\).
(a) Express \(\boldsymbol{p} - 4\boldsymbol{q}\) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(b) Find the value of \(k\) for which \(\boldsymbol{p} + k\boldsymbol{q}\) is parallel to \(\boldsymbol{a}\).
(a) \(2\boldsymbol{a} - 3\boldsymbol{b} - 4\boldsymbol{a} - 4\boldsymbol{b} = -2\boldsymbol{a} - 7\boldsymbol{b}\).
(b) \(\boldsymbol{p} + k\boldsymbol{q} = (2 + k)\boldsymbol{a} + (k - 3)\boldsymbol{b}\). This is a multiple of \(\boldsymbol{a}\) only when the \(\boldsymbol{b}\) coefficient is \(0\): \(k = 3\) (then \(\boldsymbol{p} + 3\boldsymbol{q} = 5\boldsymbol{a}\)). See 20.1.
2 Paper 1 AHL The points \(A(2, -1, 4)\) and \(B(5, 3, 4)\) are given.
(a) Find \(\big|\overrightarrow{AB}\big|\) and the unit vector in the direction of \(\overrightarrow{AB}\).
(b) The point \(C\) satisfies \(\overrightarrow{AC} = 3\overrightarrow{AB}\). Find the coordinates of \(C\).
(a) \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}\), \(\big|\overrightarrow{AB}\big| = 5\), unit vector \(\dfrac{1}{5}\begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}\).
(b) \(\boldsymbol{c} = \boldsymbol{a} + 3\overrightarrow{AB} = \begin{pmatrix} 2 + 9 \\ -1 + 12 \\ 4 + 0 \end{pmatrix}\), so \(C\) is \((11, 11, 4)\). See 20.2.
3 Paper 1 AHL \(A\) and \(B\) have position vectors \(\begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 8 \\ 0 \\ 6 \end{pmatrix}\). The point \(P\) divides \(AB\) internally in the ratio \(3 : 2\). Find the position vector of \(P\).
\(\boldsymbol{p} = \dfrac{2\boldsymbol{a} + 3\boldsymbol{b}}{5} = \dfrac{1}{5}\begin{pmatrix} -4 + 24 \\ 10 + 0 \\ 2 + 18 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}\). See 20.3.
4 Paper 1 AHL The points \(A(1, 2, -1)\), \(B(3, k, 1)\) and \(C(7, 11, m)\) are collinear. Find \(k\) and \(m\).
\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ k - 2 \\ 2 \end{pmatrix}\) and \(\overrightarrow{AC} = \begin{pmatrix} 6 \\ 9 \\ m + 1 \end{pmatrix} = t\,\overrightarrow{AB}\).
\(x\): \(6 = 2t\), so \(t = 3\). Then \(9 = 3(k - 2)\), so \(k = 5\); and \(m + 1 = 6\), so \(m = 5\). See 20.3.
5 Paper 1 AHL \(|\boldsymbol{a}| = 2\), \(|\boldsymbol{b}| = 5\) and \(\boldsymbol{a} \cdot \boldsymbol{b} = 5\).
(a) Find the angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(b) Find \(|\boldsymbol{a} - \boldsymbol{b}|\).
(c) Find the value of \(\mu\) for which \(\boldsymbol{a} + \mu\boldsymbol{b}\) is perpendicular to \(\boldsymbol{a}\).
(a) \(\cos\theta = \dfrac{5}{10} = \dfrac{1}{2}\), so \(\theta = \dfrac{\pi}{3}\) \((60^\circ)\).
(b) \(|\boldsymbol{a} - \boldsymbol{b}|^2 = 4 - 2(5) + 25 = 19\), so \(|\boldsymbol{a} - \boldsymbol{b}| = \sqrt{19}\).
(c) \((\boldsymbol{a} + \mu\boldsymbol{b}) \cdot \boldsymbol{a} = 4 + 5\mu = 0\), so \(\mu = -\dfrac{4}{5}\). See 20.4.
6 Paper 1 AHL The points \(A(1, 0, 3)\), \(B(2, 2, 1)\) and \(C(4, -1, 3)\) are given.
(a) Find the exact value of \(\cos\angle BAC\).
(b) Find \(\overrightarrow{AB} \times \overrightarrow{AC}\), and hence the exact area of triangle \(ABC\).
(a) \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}\). Dot product \(= 3 - 2 + 0 = 1\); magnitudes \(3\) and \(\sqrt{10}\). \(\;\cos\angle BAC = \dfrac{1}{3\sqrt{10}}\).
(b) \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} 0 - 2 \\ -6 - 0 \\ -1 - 6 \end{pmatrix} = \begin{pmatrix} -2 \\ -6 \\ -7 \end{pmatrix}\), with magnitude \(\sqrt{4 + 36 + 49} = \sqrt{89}\).
7 Paper 2 AHL Find the angle between \(\boldsymbol{u} = 3\boldsymbol{i} - \boldsymbol{j} + 2\boldsymbol{k}\) and \(\boldsymbol{v} = \boldsymbol{i} + 4\boldsymbol{j} - 2\boldsymbol{k}\), in degrees to 1 decimal place.
\(\boldsymbol{u} \cdot \boldsymbol{v} = 3 - 4 - 4 = -5\), \(|\boldsymbol{u}| = \sqrt{14}\), \(|\boldsymbol{v}| = \sqrt{21}\).
\(\cos\theta = \dfrac{-5}{\sqrt{14}\sqrt{21}} = \dfrac{-5}{\sqrt{294}}\), so \(\theta = 107.0^\circ\) (1 d.p.). (On the GDC: cos⁻¹(dotP(u,v)/(norm(u)·norm(v))) in Degree mode.) See 20.4.
8 Paper 1 AHL \(OABC\) is a parallelogram with \(\overrightarrow{OA} = \boldsymbol{a}\) and \(\overrightarrow{OC} = \boldsymbol{c}\). Use vectors to prove that the diagonals \(OB\) and \(AC\) bisect each other.
\(\overrightarrow{OB} = \boldsymbol{a} + \boldsymbol{c}\), so the midpoint of \(OB\) has position vector \(\frac{1}{2}(\boldsymbol{a} + \boldsymbol{c})\).
The midpoint of \(AC\) has position vector \(\frac{1}{2}(\boldsymbol{a} + \boldsymbol{c})\) too (ratio theorem with \(\lambda = \mu = 1\)).
The two midpoints are the same point, so each diagonal passes through the midpoint of the other: the diagonals bisect each other. \(\;\blacksquare\) See 20.1 and 20.3.
9 Paper 1 AHL Let \(\boldsymbol{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and \(\boldsymbol{b} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}\).
(a) Find \(\boldsymbol{a} \times \boldsymbol{b}\).
(b) Write down a unit vector perpendicular to both \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(c) Find the area of the triangle with two sides represented by \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(a) \(\boldsymbol{a} \times \boldsymbol{b} = \begin{pmatrix} 2(-1) - 3(0) \\ 3(2) - 1(-1) \\ 1(0) - 2(2) \end{pmatrix} = \begin{pmatrix} -2 \\ 7 \\ -4 \end{pmatrix}\).
(b) \(|\boldsymbol{a} \times \boldsymbol{b}| = \sqrt{4 + 49 + 16} = \sqrt{69}\), so \(\dfrac{1}{\sqrt{69}}\begin{pmatrix} -2 \\ 7 \\ -4 \end{pmatrix}\) (or its negative).
(c) \(\dfrac{\sqrt{69}}{2}\). See 20.5.
10 Paper 2 AHL \(ABCD\) is a parallelogram with \(A(2, 1, -1)\), \(B(5, 3, 0)\) and \(C(4, -2, 1)\).
(a) Find the coordinates of \(D\).
(b) Find the size of angle \(DAB\).
(c) Find the area of \(ABCD\).
(a) \(\overrightarrow{AD} = \overrightarrow{BC}\), so \(\boldsymbol{d} = \boldsymbol{a} + \boldsymbol{c} - \boldsymbol{b} = \begin{pmatrix} 1 \\ -4 \\ 0 \end{pmatrix}\). \(D\) is \((1, -4, 0)\).
(b) \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\), \(\overrightarrow{AD} = \begin{pmatrix} -1 \\ -5 \\ 1 \end{pmatrix}\). Dot product \(= -3 - 10 + 1 = -12\); magnitudes \(\sqrt{14}\) and \(\sqrt{27}\). \(\cos\angle DAB = \dfrac{-12}{\sqrt{14}\sqrt{27}}\), so \(\angle DAB = 128.1^\circ\) (1 d.p.).
(c) \(\overrightarrow{AB} \times \overrightarrow{AD} = \begin{pmatrix} 2 + 5 \\ -1 - 3 \\ -15 + 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -4 \\ -13 \end{pmatrix}\). Area \(= \sqrt{49 + 16 + 169} = \sqrt{234} = 15.3\) (3 s.f.). See 20.2, 20.4 and 20.5.