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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.6

Mixed practice: parallel vectors, unit vectors, displacement, distance and parameters Problem set B Worked solutions

Skills

  1. In each part the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel. Find \(p\) and \(q\).

    1. (a) \(\mathbf{a}=\begin{pmatrix}-2 \\ 3 \\ 1\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}8 \\ p \\ q\end{pmatrix};\)

      • \(\mathbf{b}=k\mathbf{a}\)
      • Use the components that are both known: \(8=k\times(-2)\ \Rightarrow\ k=-4\)
      • \(p=(-4)\times3=-12\)
      • \(q=(-4)\times1=-4\)

    2. (b) \(\mathbf{a}=p\mathbf{i}-9\mathbf{j}+3\mathbf{k},\ \mathbf{b}=4\mathbf{i}+q\mathbf{j}+\mathbf{k};\)

      • \(\mathbf{b}=k\mathbf{a}\)
      • Use the components that are both known: \(1=k\times3\ \Rightarrow\ k=\frac{1}{3}\)
      • \(4=\frac{1}{3}p\ \Rightarrow\ p=12\)
      • \(q=\frac{1}{3}\times(-9)=-3\)

  2. Given \(\mathbf{a}=\begin{pmatrix}2 \\ -3 \\ 6\end{pmatrix}\) and \(\mathbf{d}=4\mathbf{i}-4\mathbf{j}+7\mathbf{k}\):

    1. (a) find the unit vector in the same direction as \(\mathbf{a}\);

      • \(|\mathbf{a}|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7\)
      • \(\frac{1}{7}\begin{pmatrix}2 \\ -3 \\ 6\end{pmatrix}=\begin{pmatrix}\frac{2}{7} \\ - \frac{3}{7} \\ \frac{6}{7}\end{pmatrix}\)

    2. (b) find the vector of magnitude 15 in the direction of \(\mathbf{d}\), in column form.

      • \(|\mathbf{d}|=\sqrt{4^2+(-4)^2+7^2}=\sqrt{81}=9\)
      • \(15\times\frac{1}{9}(4\mathbf{i}-4\mathbf{j}+7\mathbf{k})=\begin{pmatrix}\frac{20}{3} \\ - \frac{20}{3} \\ \frac{35}{3}\end{pmatrix}\)

  3. Points A, B and C have position vectors \(\mathbf{a}=\begin{pmatrix}-2 \\ 5 \\ 3\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}1 \\ -4 \\ 6\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}3 \\ 1 \\ -5\end{pmatrix}\). Find:

    1. (a) \(\overrightarrow{AB}\);

      • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}3 \\ -9 \\ 3\end{pmatrix}\)

    2. (b) \(\overrightarrow{AC}\);

      • \(\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=\begin{pmatrix}5 \\ -4 \\ -8\end{pmatrix}\)

    3. (c) \(\overrightarrow{CB}\).

      • \(\overrightarrow{CB}=\mathbf{b}-\mathbf{c}=\begin{pmatrix}-2 \\ -5 \\ 11\end{pmatrix}\)

  4. Find the exact distance \(AB\) between the points with position vectors:

    1. (a) \(\mathbf{a}=2\mathbf{i}+3\mathbf{j}-5\mathbf{k},\ \mathbf{b}=-2\mathbf{i}+\mathbf{j}-3\mathbf{k};\)

      • \(\overrightarrow{AB}=-4\mathbf{i}-2\mathbf{j}+2\mathbf{k}\)
      • \(AB=\sqrt{(-4)^2+(-2)^2+2^2}=2 \sqrt{6}\)

    2. (b) \(\mathbf{a}=\begin{pmatrix}-1 \\ 0 \\ 4\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}3 \\ -3 \\ 16\end{pmatrix}.\)

      • \(\overrightarrow{AB}=\begin{pmatrix}4 \\ -3 \\ 12\end{pmatrix}\)
      • \(AB=\sqrt{4^2+(-3)^2+12^2}=13\)

  5. The resultant of \(\mathbf{u}=\begin{pmatrix}x \\ 3 \\ -6\end{pmatrix},\ \mathbf{v}=\begin{pmatrix}5 \\ y \\ 2\end{pmatrix}\text{ and }\mathbf{w}=\begin{pmatrix}-9 \\ -8 \\ z\end{pmatrix}\) is \(\begin{pmatrix}-3 \\ -2 \\ 1\end{pmatrix}\). Find \(x\), \(y\) and \(z\).

    • \(\text{resultant}=\mathbf{u}+\mathbf{v}+\mathbf{w}\text{, so add each component}\)
    • \(\text{Component 1: }5+(-9)+x=-3\ \Rightarrow\ x=1\)
    • \(\text{Component 2: }3+(-8)+y=-2\ \Rightarrow\ y=3\)
    • \(\text{Component 3: }(-6)+2+z=1\ \Rightarrow\ z=5\)

  6. Given \(\mathbf{a}=\begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}4 \\ 0 \\ -3\end{pmatrix}\text{ and }\mathbf{c}=\begin{pmatrix}2 \\ -5 \\ 1\end{pmatrix}\), find:

    1. (a) \(3\mathbf{a}+2\mathbf{b}-2\mathbf{c}\);

      • \(3\mathbf{a}+2\mathbf{b}-2\mathbf{c}=3\begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}+2\begin{pmatrix}4 \\ 0 \\ -3\end{pmatrix}-2\begin{pmatrix}2 \\ -5 \\ 1\end{pmatrix}\)
      • \(=\begin{pmatrix}1 \\ 16 \\ 1\end{pmatrix}\)

    2. (b) the exact value of \(|-\mathbf{a}+3\mathbf{b}|\).

      • \(-\mathbf{a}+3\mathbf{b}=\begin{pmatrix}13 \\ -2 \\ -12\end{pmatrix}\)
      • \(|-\mathbf{a}+3\mathbf{b}|=\sqrt{13^2+(-2)^2+(-12)^2}=\sqrt{317}\)

  7. In each part, find the unknown.

    1. (a) Given \(\mathbf{p}=\begin{pmatrix}2 \\ 0 \\ -1\end{pmatrix}\text{ and }\mathbf{q}=\begin{pmatrix}4 \\ 4 \\ -13\end{pmatrix}\), find the vector \(\mathbf{x}\) such that \(5\mathbf{p}-2\mathbf{x}=\mathbf{q}\).

      • \(5\mathbf{p}-2\mathbf{x}=\mathbf{q}\ \Rightarrow\ -2\mathbf{x}=\mathbf{q}-5\mathbf{p}\)
      • \(-2\mathbf{x}=\begin{pmatrix}4 \\ 4 \\ -13\end{pmatrix}-5\begin{pmatrix}2 \\ 0 \\ -1\end{pmatrix}=\begin{pmatrix}-6 \\ 4 \\ -8\end{pmatrix}\)
      • \(\mathbf{x}=\begin{pmatrix}3 \\ -2 \\ 4\end{pmatrix}\)

    2. (b) Given \(\mathbf{a}=\mathbf{i}+4\mathbf{j}-3\mathbf{k},\ \mathbf{b}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\text{ and }\mathbf{c}=7\mathbf{i}+\mathbf{j}\), find the scalar \(t\) such that \(\mathbf{a}+t\mathbf{b}=\mathbf{c}\).

      • \(\mathbf{a}+t\mathbf{b}=\begin{pmatrix}1+2t \\ 4-t \\ -3+t\end{pmatrix}=\begin{pmatrix}7 \\ 1 \\ 0\end{pmatrix}\)
      • Equate one component, then check another.
      • \(t=3\)

  8. The vector \((2-t)\mathbf{i}+(1+t)\mathbf{j}+(-4+2t)\mathbf{k}\) has magnitude 9. Find the possible values of \(t\).

    • \(\sqrt{(2-t)^2+(1+t)^2+(-4+2t)^2}=9\)
    • \(6t^2-18t+21=81\ \Rightarrow\ 6t^2-18t-60=0\)
    • \(t=-2\ \text{or}\ t=5\)

In context

  1. Three tugs pull a ship with the forces \(\mathbf{F}_1=\begin{pmatrix}-6 \\ 4 \\ 2\end{pmatrix},\ \mathbf{F}_2=\begin{pmatrix}9 \\ -8 \\ 5\end{pmatrix}\text{ and }\mathbf{F}_3=\begin{pmatrix}1 \\ 6 \\ -7\end{pmatrix}\), in kilonewtons. Find the magnitude of the resultant force, to 3 significant figures.

    • \(\mathbf{R}=\begin{pmatrix}-6 \\ 4 \\ 2\end{pmatrix}+\begin{pmatrix}9 \\ -8 \\ 5\end{pmatrix}+\begin{pmatrix}1 \\ 6 \\ -7\end{pmatrix}=\begin{pmatrix}4 \\ 2 \\ 0\end{pmatrix}\)
    • \(|\mathbf{R}|=\sqrt{4^2+2^2+0^2}=\sqrt{20}\)
    • To 3 significant figures: 4.47 kN.

  2. The diagram shows a rectangular lawn EFGH. J is the midpoint of FG, K is the midpoint of GH and L is the point on EF extended beyond F such that \(FL=\frac{1}{4}EF\). Let \(\mathbf{a}=\overrightarrow{EF}\) and \(\mathbf{b}=\overrightarrow{EH}\). Express in terms of \(\mathbf{a}\) and \(\mathbf{b}\):

    abEFGHJKL

    1. (a) \(\overrightarrow{EJ}\)

      • \(\overrightarrow{EF}=\mathbf{a},\ \overrightarrow{EH}=\mathbf{b},\ \text{so}\ \overrightarrow{EG}=\mathbf{a}+\mathbf{b}\)
      • \(\overrightarrow{EJ}=\mathbf{a}+\frac{1}{2}\mathbf{b},\ \overrightarrow{EK}=\frac{1}{2}\mathbf{a}+\mathbf{b},\ \overrightarrow{EL}=\frac{5}{4}\mathbf{a}\)
      • \(\overrightarrow{EJ}=\mathbf{a}+\frac{1}{2}\mathbf{b}\)

    2. (b) \(\overrightarrow{JK}\)

      • \(\overrightarrow{JK}=\overrightarrow{EK}-\overrightarrow{EJ}=(\frac{1}{2}\mathbf{a}+\mathbf{b})-(\mathbf{a}+\frac{1}{2}\mathbf{b})=- \frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\)

    3. (c) \(\overrightarrow{KL}\)

      • \(\overrightarrow{KL}=\overrightarrow{EL}-\overrightarrow{EK}=(\frac{5}{4}\mathbf{a})-(\frac{1}{2}\mathbf{a}+\mathbf{b})=\frac{3}{4}\mathbf{a}-\mathbf{b}\)

  3. In triangle OAB, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). M is the midpoint of OA. N is the point on AB such that \(BN:NA=1:3\). Express in terms of \(\mathbf{a}\) and \(\mathbf{b}\):

    bOABMN

    1. (a) \(\overrightarrow{BA}\)

      • \(\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=(\mathbf{a})-(\mathbf{b})=\mathbf{a}-\mathbf{b}\)

    2. (b) \(\overrightarrow{ON}\)

      • \(BN:NA=1:3\ \Rightarrow\ \overrightarrow{BN}=\frac{1}{4}\overrightarrow{BA},\quad \overrightarrow{OM}=\frac{1}{2}\mathbf{a}\)
      • \(\overrightarrow{ON}=\overrightarrow{OB}+\overrightarrow{BN}=\frac{1}{4}\mathbf{a}+\frac{3}{4}\mathbf{b}\)
      • \(\overrightarrow{ON}=\frac{1}{4}\mathbf{a}+\frac{3}{4}\mathbf{b}\)

    3. (c) \(\overrightarrow{MN}\)

      • \(\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=(\frac{1}{4}\mathbf{a}+\frac{3}{4}\mathbf{b})-(\frac{1}{2}\mathbf{a})=- \frac{1}{4}\mathbf{a}+\frac{3}{4}\mathbf{b}\)

Stretch

  1. In each part, find the scalar or scalars asked for.

    1. (a) Given \(\mathbf{a}=\begin{pmatrix}1 \\ -1 \\ 4\end{pmatrix}\text{ and }\mathbf{b}=\begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}\), find \(p\) such that \(\mathbf{a}+p\mathbf{b}\) is parallel to \(\begin{pmatrix}14 \\ 4 \\ 2\end{pmatrix}\).

      • \(\mathbf{a}+p\mathbf{b}=\begin{pmatrix}1+2p \\ -1+p \\ 4-p\end{pmatrix}=\mu\begin{pmatrix}14 \\ 4 \\ 2\end{pmatrix}\)
      • \(1+2p=14\mu,\quad -1+p=4\mu,\quad 4-p=2\mu\)
      • Solve two of them for \(p\) and \(\mu\): \(p=3,\ \mu=\frac{1}{2}\). Check the third.

    2. (b) Given \(\mathbf{x}=\begin{pmatrix}2 \\ -3 \\ 4\end{pmatrix}\text{ and }\mathbf{y}=\begin{pmatrix}-3 \\ 12 \\ -16\end{pmatrix}\), find \(\lambda\) and \(k\) such that \(\lambda\mathbf{x}+\mathbf{y}=k\mathbf{i}\).

      • \(\lambda\mathbf{x}+\mathbf{y}=\begin{pmatrix}-3+2\lambda \\ 12-3\lambda \\ -16+4\lambda\end{pmatrix}=k\mathbf{i}\)
      • A vector parallel to \(\mathbf{i}\) has zero components in the other two directions: \(12-3\lambda=0\ \Rightarrow\ \lambda=4\), and \(-16+4\lambda=0\) agrees.
      • \(k=-3+2\lambda=5\)

  2. Use your calculator where it helps. Give answers to 3 significant figures.

    1. (a) Find the values of \(\lambda\) such that \(|\mathbf{a}+\lambda\mathbf{b}|=5\), where \(\mathbf{a}=\begin{pmatrix}-1 \\ 2 \\ 2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}3 \\ -1 \\ 1\end{pmatrix}\).

      • \(\mathbf{a}+\lambda\mathbf{b}=\begin{pmatrix}-1+3\lambda \\ 2-\lambda \\ 2+\lambda\end{pmatrix}\)
      • \(\sqrt{(-1+3\lambda)^2+(2-\lambda)^2+(2+\lambda)^2}=5\)
      • From the calculator: nSolve with a guess above 0 and one below 0: λ = 1.51 and λ = −0.964.

    2. (b) Points A and B are such that \(\overrightarrow{OA}=\begin{pmatrix}4 \\ -1 \\ 2\end{pmatrix}\) and \(\overrightarrow{OB}=\begin{pmatrix}0 \\ 3 \\ -1\end{pmatrix}+t\begin{pmatrix}1 \\ 2 \\ 2\end{pmatrix}\), where O is the origin. Find the values of \(t\) such that \(AB=8\).

      • \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}-4+t \\ 4+2t \\ -3+2t\end{pmatrix}\)
      • \(AB=\sqrt{(-4+t)^2+(4+2t)^2+(-3+2t)^2}=8\)
      • From the calculator: nSolve with two different guesses: t = −1.39 and t = 1.84.

    3. (c) Find the smallest possible magnitude of the vector \((-3+2t)\mathbf{i}+(2+t)\mathbf{j}+(-1+3t)\mathbf{k}\), where \(t\) is a real number.

      • \(|\mathbf{v}|=\sqrt{(-3+2t)^2+(2+t)^2+(-1+3t)^2}\)
      • From the calculator: graph this and find the minimum.
      • The minimum is at t = 0.500, where the magnitude is 3.24.

  3. Points P and Q have position vectors \(\mathbf{p}=-3\mathbf{i}+2\mathbf{j}+5\mathbf{k}\) and \(\mathbf{q}=7\mathbf{i}-4\mathbf{j}-\mathbf{k}\).

    1. (a) Find the position vector of the midpoint M of PQ.

      • \(\overrightarrow{OM}=\frac{1}{2}(\mathbf{p}+\mathbf{q})=\frac{1}{2}(4\mathbf{i}-2\mathbf{j}+4\mathbf{k})=2\mathbf{i}-\mathbf{j}+2\mathbf{k}\)

    2. (b) The point R lies on the line PQ, beyond Q, so that \(QR=QM\). Find the coordinates of R.

      • \(QM=\frac{1}{2}PQ,\ \text{so}\ \overrightarrow{QR}=\frac{1}{2}\overrightarrow{PQ}=\frac{1}{2}(10\mathbf{i}-6\mathbf{j}-6\mathbf{k})=5\mathbf{i}-3\mathbf{j}-3\mathbf{k}\)
      • \(\overrightarrow{OR}=\mathbf{q}+\overrightarrow{QR}=12\mathbf{i}-7\mathbf{j}-4\mathbf{k}\)
      • R is beyond Q, on the opposite side from P.

jmaths.xyz · AI HL · AHL 4.6 Mixed practice: parallel vectors, unit vectors, displacement, distance and parameters