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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 A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

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

    1. (a) \(\mathbf{a}=\begin{pmatrix}4 \\ -1 \\ 3\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}12 \\ p \\ q\end{pmatrix};\)

      • \(\mathbf{b}=k\mathbf{a}\)
      • Use the components that are both known: \(12=k\times4\ \Rightarrow\ k=3\)
      • \(p=3\times(-1)=-3\)
      • \(q=3\times3=9\)

    2. (b) \(\mathbf{a}=p\mathbf{i}+4\mathbf{j}-6\mathbf{k},\ \mathbf{b}=3\mathbf{i}+q\mathbf{j}-9\mathbf{k};\)

      • \(\mathbf{b}=k\mathbf{a}\)
      • Use the components that are both known: \(-9=k\times(-6)\ \Rightarrow\ k=\frac{3}{2}\)
      • \(3=\frac{3}{2}p\ \Rightarrow\ p=2\)
      • \(q=\frac{3}{2}\times4=6\)

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

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

      • \(|\mathbf{a}|=\sqrt{(-4)^2+4^2+7^2}=\sqrt{81}=9\)
      • \(\frac{1}{9}\begin{pmatrix}-4 \\ 4 \\ 7\end{pmatrix}=\begin{pmatrix}- \frac{4}{9} \\ \frac{4}{9} \\ \frac{7}{9}\end{pmatrix}\)

    2. (b) find the vector of magnitude 20 in the direction of \(\mathbf{d}\), in column form.

      • \(|\mathbf{d}|=\sqrt{2^2+2^2+(-1)^2}=\sqrt{9}=3\)
      • \(20\times\frac{1}{3}(2\mathbf{i}+2\mathbf{j}-\mathbf{k})=\begin{pmatrix}\frac{40}{3} \\ \frac{40}{3} \\ - \frac{20}{3}\end{pmatrix}\)

  3. Points A, B and C have position vectors \(\mathbf{a}=\begin{pmatrix}4 \\ -3 \\ 1\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}6 \\ 0 \\ -2\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}-1 \\ 2 \\ 5\end{pmatrix}\). Find:

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

      • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2 \\ 3 \\ -3\end{pmatrix}\)

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

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

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

      • \(\overrightarrow{CB}=\mathbf{b}-\mathbf{c}=\begin{pmatrix}7 \\ -2 \\ -7\end{pmatrix}\)

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

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

      • \(\overrightarrow{AB}=-3\mathbf{i}+3\mathbf{j}-4\mathbf{k}\)
      • \(AB=\sqrt{(-3)^2+3^2+(-4)^2}=\sqrt{34}\)

    2. (b) \(\mathbf{a}=\begin{pmatrix}2 \\ 5 \\ -1\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}8 \\ -3 \\ -1\end{pmatrix}.\)

      • \(\overrightarrow{AB}=\begin{pmatrix}6 \\ -8 \\ 0\end{pmatrix}\)
      • \(AB=\sqrt{6^2+(-8)^2+0^2}=10\)

  5. The resultant of \(\mathbf{u}=\begin{pmatrix}8 \\ y \\ -1\end{pmatrix},\ \mathbf{v}=\begin{pmatrix}-5 \\ 4 \\ z\end{pmatrix}\text{ and }\mathbf{w}=\begin{pmatrix}x \\ -7 \\ 4\end{pmatrix}\) is \(\begin{pmatrix}2 \\ -1 \\ 10\end{pmatrix}\). Find \(x\), \(y\) and \(z\).

    • \(\text{resultant}=\mathbf{u}+\mathbf{v}+\mathbf{w}\text{, so add each component}\)
    • \(\text{Component 1: }8+(-5)+x=2\ \Rightarrow\ x=-1\)
    • \(\text{Component 2: }4+(-7)+y=-1\ \Rightarrow\ y=2\)
    • \(\text{Component 3: }(-1)+4+z=10\ \Rightarrow\ z=7\)

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

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

      • \(2\mathbf{a}-3\mathbf{b}+\mathbf{c}=2\begin{pmatrix}2 \\ -1 \\ 4\end{pmatrix}-3\begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}+\begin{pmatrix}-3 \\ 1 \\ 5\end{pmatrix}\)
      • \(=\begin{pmatrix}1 \\ -10 \\ 19\end{pmatrix}\)

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

      • \(\mathbf{a}-2\mathbf{b}=\begin{pmatrix}2 \\ -7 \\ 8\end{pmatrix}\)
      • \(|\mathbf{a}-2\mathbf{b}|=\sqrt{2^2+(-7)^2+8^2}=3 \sqrt{13}\)

  7. In each part, find the unknown.

    1. (a) Given \(\mathbf{p}=\begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix}\text{ and }\mathbf{q}=\begin{pmatrix}9 \\ -10 \\ 11\end{pmatrix}\), find the vector \(\mathbf{x}\) such that \(2\mathbf{p}+3\mathbf{x}=\mathbf{q}\).

      • \(2\mathbf{p}+3\mathbf{x}=\mathbf{q}\ \Rightarrow\ 3\mathbf{x}=\mathbf{q}-2\mathbf{p}\)
      • \(3\mathbf{x}=\begin{pmatrix}9 \\ -10 \\ 11\end{pmatrix}-2\begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix}=\begin{pmatrix}3 \\ -6 \\ 9\end{pmatrix}\)
      • \(\mathbf{x}=\begin{pmatrix}1 \\ -2 \\ 3\end{pmatrix}\)

    2. (b) Given \(\mathbf{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k},\ \mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\text{ and }\mathbf{c}=-5\mathbf{j}+5\mathbf{k}\), find the scalar \(t\) such that \(\mathbf{a}+t\mathbf{b}=\mathbf{c}\).

      • \(\mathbf{a}+t\mathbf{b}=\begin{pmatrix}2+t \\ -1+2t \\ 3-t\end{pmatrix}=\begin{pmatrix}0 \\ -5 \\ 5\end{pmatrix}\)
      • Equate one component, then check another.
      • \(t=-2\)

  8. The vector \((4+2t)\mathbf{i}+(-2-t)\mathbf{j}+(1-3t)\mathbf{k}\) has magnitude 7. Find the possible values of \(t\).

    • \(\sqrt{(4+2t)^2+(-2-t)^2+(1-3t)^2}=7\)
    • \(14t^2+14t+21=49\ \Rightarrow\ 14t^2+14t-28=0\)
    • \(t=-2\ \text{or}\ t=1\)

In context

  1. Three forces act on a boat: \(\mathbf{F}_1=\begin{pmatrix}8 \\ -3 \\ 5\end{pmatrix},\ \mathbf{F}_2=\begin{pmatrix}-2 \\ 7 \\ -9\end{pmatrix}\text{ and }\mathbf{F}_3=\begin{pmatrix}4 \\ -1 \\ 3\end{pmatrix}\), in newtons. Find the magnitude of the resultant force, to 3 significant figures.

    • \(\mathbf{R}=\begin{pmatrix}8 \\ -3 \\ 5\end{pmatrix}+\begin{pmatrix}-2 \\ 7 \\ -9\end{pmatrix}+\begin{pmatrix}4 \\ -1 \\ 3\end{pmatrix}=\begin{pmatrix}10 \\ 3 \\ -1\end{pmatrix}\)
    • \(|\mathbf{R}|=\sqrt{10^2+3^2+(-1)^2}=\sqrt{110}\)
    • To 3 significant figures: 10.5 N.

  2. The diagram shows a rectangular tile PQRS. M is the midpoint of QR, N is the midpoint of RS and T is the point on PQ extended beyond Q such that \(QT=\frac{1}{2}PQ\). Let \(\mathbf{a}=\overrightarrow{PQ}\) and \(\mathbf{b}=\overrightarrow{PS}\). Express in terms of \(\mathbf{a}\) and \(\mathbf{b}\):

    abPQRSMNT

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

      • \(\overrightarrow{PQ}=\mathbf{a},\ \overrightarrow{PS}=\mathbf{b},\ \text{so}\ \overrightarrow{PR}=\mathbf{a}+\mathbf{b}\)
      • \(\overrightarrow{PM}=\mathbf{a}+\frac{1}{2}\mathbf{b},\ \overrightarrow{PN}=\frac{1}{2}\mathbf{a}+\mathbf{b},\ \overrightarrow{PT}=\frac{3}{2}\mathbf{a}\)
      • \(\overrightarrow{PM}=\mathbf{a}+\frac{1}{2}\mathbf{b}\)

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

      • \(\overrightarrow{MN}=\overrightarrow{PN}-\overrightarrow{PM}=(\frac{1}{2}\mathbf{a}+\mathbf{b})-(\mathbf{a}+\frac{1}{2}\mathbf{b})=- \frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}\)

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

      • \(\overrightarrow{ST}=\overrightarrow{PT}-\overrightarrow{PS}=(\frac{3}{2}\mathbf{a})-(\mathbf{b})=\frac{3}{2}\mathbf{a}-\mathbf{b}\)

  3. In triangle OAB, \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). M is the midpoint of OB. P is the point on AB such that \(AP:PB=2:3\). Express in terms of \(\mathbf{a}\) and \(\mathbf{b}\):

    aOABMP

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

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

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

      • \(AP:PB=2:3\ \Rightarrow\ \overrightarrow{AP}=\frac{2}{5}\overrightarrow{AB}=\frac{2}{5}(\mathbf{b}-\mathbf{a}),\quad \overrightarrow{OM}=\frac{1}{2}\mathbf{b}\)
      • \(\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\frac{3}{5}\mathbf{a}+\frac{2}{5}\mathbf{b}\)
      • \(\overrightarrow{OP}=\frac{3}{5}\mathbf{a}+\frac{2}{5}\mathbf{b}\)

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

      • \(\overrightarrow{MP}=\overrightarrow{OP}-\overrightarrow{OM}=(\frac{3}{5}\mathbf{a}+\frac{2}{5}\mathbf{b})-(\frac{1}{2}\mathbf{b})=\frac{3}{5}\mathbf{a}- \frac{1}{10}\mathbf{b}\)

Stretch

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

    1. (a) Given \(\mathbf{a}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}\text{ and }\mathbf{b}=\begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix}\), find \(p\) such that \(\mathbf{a}+p\mathbf{b}\) is parallel to \(\begin{pmatrix}-1 \\ 5 \\ -4\end{pmatrix}\).

      • \(\mathbf{a}+p\mathbf{b}=\begin{pmatrix}3+p \\ -1+2p \\ 2-p\end{pmatrix}=\mu\begin{pmatrix}-1 \\ 5 \\ -4\end{pmatrix}\)
      • \(3+p=-\mu,\quad -1+2p=5\mu,\quad 2-p=-4\mu\)
      • Solve two of them for \(p\) and \(\mu\): \(p=-2,\ \mu=-1\). Check the third.

    2. (b) Given \(\mathbf{x}=\begin{pmatrix}3 \\ 5 \\ -2\end{pmatrix}\text{ and }\mathbf{y}=\begin{pmatrix}-6 \\ 1 \\ 4\end{pmatrix}\), find \(\lambda\) and \(k\) such that \(\lambda\mathbf{x}+\mathbf{y}=k\mathbf{j}\).

      • \(\lambda\mathbf{x}+\mathbf{y}=\begin{pmatrix}-6+3\lambda \\ 1+5\lambda \\ 4-2\lambda\end{pmatrix}=k\mathbf{j}\)
      • A vector parallel to \(\mathbf{j}\) has zero components in the other two directions: \(-6+3\lambda=0\ \Rightarrow\ \lambda=2\), and \(4-2\lambda=0\) agrees.
      • \(k=1+5\lambda=11\)

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

    1. (a) Find the values of \(\lambda\) such that \(|\mathbf{a}+\lambda\mathbf{b}|=6\), where \(\mathbf{a}=\begin{pmatrix}1 \\ -3 \\ 2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}2 \\ 1 \\ -2\end{pmatrix}\).

      • \(\mathbf{a}+\lambda\mathbf{b}=\begin{pmatrix}1+2\lambda \\ -3+\lambda \\ 2-2\lambda\end{pmatrix}\)
      • \(\sqrt{(1+2\lambda)^2+(-3+\lambda)^2+(2-2\lambda)^2}=6\)
      • From the calculator: nSolve with a guess above 0 and one below 0: λ = 2.21 and λ = −1.10.

    2. (b) Points A and B are such that \(\overrightarrow{OA}=\begin{pmatrix}-2 \\ 3 \\ 5\end{pmatrix}\) and \(\overrightarrow{OB}=\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+t\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\), where O is the origin. Find the values of \(t\) such that \(AB=6\).

      • \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}3+2t \\ -3-t \\ -3+t\end{pmatrix}\)
      • \(AB=\sqrt{(3+2t)^2+(-3-t)^2+(-3+t)^2}=6\)
      • From the calculator: nSolve with two different guesses: t = −2.58 and t = 0.581.

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

      • \(|\mathbf{v}|=\sqrt{(2+t)^2+(-1+2t)^2+(2-t)^2}\)
      • From the calculator: graph this and find the minimum.
      • The minimum is at t = 0.333, where the magnitude is 2.89.

  3. Points P and Q have position vectors \(\mathbf{p}=4\mathbf{i}-\mathbf{j}+6\mathbf{k}\) and \(\mathbf{q}=-2\mathbf{i}+5\mathbf{j}+2\mathbf{k}\).

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

      • \(\overrightarrow{OM}=\frac{1}{2}(\mathbf{p}+\mathbf{q})=\frac{1}{2}(2\mathbf{i}+4\mathbf{j}+8\mathbf{k})=\mathbf{i}+2\mathbf{j}+4\mathbf{k}\)

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

      • \(QM=\frac{1}{2}PQ,\ \text{so}\ \overrightarrow{QR}=\frac{1}{2}\overrightarrow{PQ}=\frac{1}{2}(-6\mathbf{i}+6\mathbf{j}-4\mathbf{k})=-3\mathbf{i}+3\mathbf{j}-2\mathbf{k}\)
      • \(\overrightarrow{OR}=\mathbf{q}+\overrightarrow{QR}=-5\mathbf{i}+8\mathbf{j}\)
      • 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