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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.10

Mixed practice: lines in 3D, parametric form and intersections Problem set A Worked solutions

Skills

  1. The points \(A(2, -3, 1)\), \(B(5, -1, -3)\) and \(C(-4, -7, 9)\) are in three dimensions.

    1. (a) Find a vector equation of the line \(AB\).

      • Direction: \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)
      • Use \(A\) as the point: \(\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)

    2. (b) Determine whether \(C\) lies on the line \(AB\).

      • Put the coordinates of \(C\) into the line and find \(\lambda\) from each: \(x: 2 + 3\lambda = -4 \Rightarrow \lambda = -2\); \(y: -3 + 2\lambda = -7 \Rightarrow \lambda = -2\); \(z: 1 - 4\lambda = 9 \Rightarrow \lambda = -2\)
      • Every coordinate gives the same \(\lambda\), so \(C\) lies on the line.

  2. Write down a vector equation of each line.

    1. (a) The line through the point \((3, -2, 4)\) with direction \(\begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix}\).

      • Use \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), with \(\mathbf{a}\) the point and \(\mathbf{b}\) the direction.
      • \(\mathbf{r} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix}\)

    2. (b) The line through the point \((-1, 4, 2)\) with direction \(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\).

      • The point has position vector \(-\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\).
      • \(\mathbf{r} = (-\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) + \lambda(3\mathbf{i} - 2\mathbf{j} + \mathbf{k})\)

    3. (c) The line through the point \((2, 6)\) with direction \(\begin{pmatrix} -3 \\ 5 \end{pmatrix}\).

      • The same in two dimensions.
      • \(\mathbf{r} = \begin{pmatrix} 2 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

  3. Write each line in the form asked.

    1. (a) Write \(\mathbf{r} = \begin{pmatrix} 2 \\ -4 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ 3 \\ -2 \end{pmatrix}\) in parametric form.

      • Each row of the vector equation gives one coordinate.
      • \(x = 2 + 5\lambda, y = -4 + 3\lambda, z = 7 - 2\lambda\)

    2. (b) Write \(\mathbf{r} = (3\mathbf{i} - \mathbf{j}) + \lambda(2\mathbf{j} - 5\mathbf{k})\) in parametric form.

      • A missing \(\mathbf{i}\), \(\mathbf{j}\) or \(\mathbf{k}\) means a zero coordinate.
      • \(x = 3, y = -1 + 2\lambda, z = -5\lambda\)

    3. (c) A line has parametric equations \(x = 6 - 4\lambda\), \(y = -2\), \(z = \frac{3\lambda + 1}{2}\). Write it as a vector equation.

      • Constant parts form the point; the multiples of \(\lambda\) form the direction.
      • Split \(z\): \(z = \frac{1}{2} + \frac{3}{2}\lambda\)
      • \(\mathbf{r} = \begin{pmatrix} 6 \\ -2 \\ \frac{1}{2} \end{pmatrix} + \lambda\begin{pmatrix} -4 \\ 0 \\ \frac{3}{2} \end{pmatrix}\)

  4. A line \(l\) has equation \(\mathbf{r} = (2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\). Find parametric equations of the line parallel to \(l\) that passes through the point \((4, -1, 3)\).

    • Parallel lines have the same direction vector: \(\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}\).
    • Use the point \((4, -1, 3)\) and this direction.
    • \(x = 4 + 3\lambda, y = -1 - \lambda, z = 3 + 2\lambda\)

  5. A line has parametric equations \(x = 4 - \lambda, y = 2 + 3\lambda, z = -5 + 2\lambda\).

    1. (a) Determine whether the point \((6, -4, -9)\) lies on the line.

      • Find \(\lambda\) from each coordinate: \(x: 4 - \lambda = 6 \Rightarrow \lambda = -2\); \(y: 2 + 3\lambda = -4 \Rightarrow \lambda = -2\); \(z: -5 + 2\lambda = -9 \Rightarrow \lambda = -2\)
      • The same \(\lambda\) works in every coordinate, so the point is on the line.

    2. (b) The point \((1, p, q)\) lies on the line. Find \(p\) and \(q\).

      • The known coordinate gives \(\lambda\): \(4 - \lambda = 1 \Rightarrow \lambda = 3\)
      • Put this into the other two equations.
      • \(p = 11, q = 1\)

  6. Find the point of intersection of the lines \(\mathbf{r} = \begin{pmatrix} 1 \\ -8 \\ 9 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}\) and \(\mathbf{r} = (5\mathbf{i} - 3\mathbf{j} + 9\mathbf{k}) + \mu(2\mathbf{i} - \mathbf{j} + 4\mathbf{k})\).

    • Equate the coordinates: \(1 + \lambda = 5 + 2\mu\) (1), \(-8 + 3\lambda = -3 - \mu\) (2), \(9 - 2\lambda = 9 + 4\mu\) (3)
    • Solve (1) and (2) together. From the calculator: \(\lambda = 2, \mu = -1\)
    • Check (3): \(9 - 2\lambda = 5\) and \(9 + 4\mu = 5\). They are equal, so the lines meet.
    • Put \(\lambda = 2\) into the first line: \((3, -2, 5)\)

  7. Decide about the lines in each part.

    1. (a) Show that the lines \(\mathbf{r} = \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} 3 \\ -2 \\ -3 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}\) do not intersect.

      • Equate the coordinates: \(-1 + 2\lambda = 3 - \mu\) (1), \(3 + \lambda = -2 + 3\mu\) (2), \(1 - 3\lambda = -3 + 2\mu\) (3)
      • Solve (1) and (2) together. From the calculator: \(\lambda = 1, \mu = 2\)
      • Check (3): left side \(-2\), right side \(1\). They are not equal, so the lines do not intersect.

    2. (b) The lines \(l_1\) and \(l_2\) have parametric equations \(l_1: x = -1 + 3\lambda, y = -3 + 2\lambda, z = 7 - \lambda\) and \(l_2: x = 4 + \mu, y = -3 - \mu, z = 14 + 4\mu\). Show that they intersect and find the point of intersection.

      • Equate the coordinates: \(-1 + 3\lambda = 4 + \mu\) (1), \(-3 + 2\lambda = -3 - \mu\) (2), \(7 - \lambda = 14 + 4\mu\) (3)
      • Solve (1) and (2) together. From the calculator: \(\lambda = 1, \mu = -2\)
      • Check (3): \(7 - \lambda = 6\) and \(14 + 4\mu = 6\). They are equal, so the lines meet.
      • Put \(\lambda = 1\) into the first line: \((2, -1, 6)\)

  8. The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\).

    1. (a) Show that the equations \(x = 5 - 2\mu, y = -4 + 4\mu, z = 2 - 6\mu\) describe the same line as \(l\).

      • The direction \(\begin{pmatrix} -2 \\ 4 \\ -6 \end{pmatrix}\) is \(-2\) times \(\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\), so the lines are parallel.
      • Is the point \((5, -4, 2)\) on \(l\)? \(x: 3 + \lambda = 5 \Rightarrow \lambda = 2\); \(y: -2\lambda = -4 \Rightarrow \lambda = 2\); \(z: -4 + 3\lambda = 2 \Rightarrow \lambda = 2\)
      • It is on \(l\), so the lines are the same.

    2. (b) Show that \(\mathbf{r} = (6\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} + 6\mathbf{k})\) describes a different line from \(l\).

      • The direction \(2\mathbf{i} - 4\mathbf{j} + 6\mathbf{k}\) is \(2\) times \(\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\), so the lines are parallel.
      • Is the point \((6, -6, 7)\) on \(l\)? \(x: 3 + \lambda = 6 \Rightarrow \lambda = 3\); \(y: -2\lambda = -6 \Rightarrow \lambda = 3\); \(z: -4 + 3\lambda = 7 \Rightarrow \lambda = \frac{11}{3}\)
      • It is not on \(l\), so this is a different (parallel) line.

In context

  1. A laser show uses two straight beams. One beam passes through the points \(A\) and \(B\), the other through \(C\) and \(D\). Coordinates are in metres. \(A(0, 4, -7)\), \(B(2, 3, -4)\), \(C(7, 8, -7)\), \(D(8, 10, -9)\). Find the coordinates of the point where the two lines cross.

    • Directions: \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\overrightarrow{CD} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\).
    • Lines: \(\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ -7 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} 7 \\ 8 \\ -7 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\).
    • Equate the coordinates: \(2\lambda = 7 + \mu\) (1), \(4 - \lambda = 8 + 2\mu\) (2), \(-7 + 3\lambda = -7 - 2\mu\) (3)
    • Solve (1) and (2) together. From the calculator: \(\lambda = 2, \mu = -3\)
    • Check (3): \(-7 + 3\lambda = -1\) and \(-7 - 2\mu = -1\). They are equal, so the lines meet.
    • Put \(\lambda = 2\) into the first line: \((4, 2, -1)\)

  2. A zip wire runs in a straight line from post \(P\) to post \(Q\). Coordinates are in metres. A rest platform \(R\) is on the line through \(P\) and \(Q\), with \(PR = 3PQ\). The posts are at \(P(2, -1, 4)\) and \(Q(5, 1, 0)\).

    1. (a) Find a vector equation of the line \(PQ\).

      • Direction: \(\overrightarrow{PQ} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)
      • \(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)

    2. (b) Find the possible coordinates of \(R\).

      • \(PR = 3PQ\) means \(R\) is 3 lots of \(\overrightarrow{PQ}\) from \(P\), in either direction: \(\lambda = 3\) or \(\lambda = -3\).
      • \(\lambda = 3: (11, 5, -8)\)
      • \(\lambda = -3: (-7, -7, 16)\)

Stretch

  1. A line has equation \(\mathbf{r} = \begin{pmatrix} -6 \\ -16 \\ 9 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 7 \\ -3 \end{pmatrix}\).

    1. (a) Find the coordinates of the point where the line meets the \(y\)-axis.

      • On the \(y\)-axis, \(x\) and \(z\) are both 0.
      • \(x: -6 + 2\lambda = 0 \Rightarrow \lambda = 3\); \(z: 9 - 3\lambda = 0 \Rightarrow \lambda = 3\)
      • Both give \(\lambda = 3\). Put it into the line: \((0, 5, 0)\)

    2. (b) Show that the line does not meet the \(z\)-axis.

      • On the \(z\)-axis, \(x\) and \(y\) are both 0.
      • \(x: -6 + 2\lambda = 0 \Rightarrow \lambda = 3\); \(y: -16 + 7\lambda = 0 \Rightarrow \lambda = \frac{16}{7}\)
      • The values of \(\lambda\) are different, so no point of the line has both coordinates 0.

  2. The lines \(\mathbf{r} = \begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} 7 \\ -5 \\ -6 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 3 \\ p \end{pmatrix}\) intersect. Find the value of \(p\) and the coordinates of the point of intersection.

    • Equate the coordinates: \(3 + \lambda = 7 - \mu\) (1), \(5 - 2\lambda = -5 + 3\mu\) (2), \(-4 + 3\lambda = -6 + \mu\,p\) (3)
    • Equations (1) and (2) do not contain \(p\). From the calculator: \(\lambda = 2, \mu = 2\)
    • Put these into (3): \(2 = -6 + 2p\), so \(p = 4\)
    • Put \(\lambda = 2\) into the first line: \((5, 1, 2)\)

  3. A cable \(l\) passes through the point \(A\) and is parallel to the vector \(2\mathbf{i} + 6\mathbf{j} - 9\mathbf{k}\). Distances are in metres. The point \(A\) has coordinates \((3, -1, 2)\).

    1. (a) Calculate the magnitude of \(2\mathbf{i} + 6\mathbf{j} - 9\mathbf{k}\).

      • \(|2\mathbf{i} + 6\mathbf{j} - 9\mathbf{k}| = \sqrt{4 + 36 + 81} = \sqrt{121} = 11\)

    2. (b) Find the possible coordinates of the point \(P\) on \(l\) such that \(AP = 44\).

      • The line is \(\mathbf{r} = (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \lambda(2\mathbf{i} + 6\mathbf{j} - 9\mathbf{k})\).
      • A step of \(\lambda\) along \(l\) moves a distance \(|\lambda|\times 11\). Set this equal to 44: \(|\lambda| = 44 \div 11 = 4\)
      • \(\lambda = 4: (11, 23, -34)\), \(\lambda = -4: (-5, -25, 38)\)

jmaths.xyz · AI HL · AHL 4.10 Mixed practice: lines in 3D, parametric form and intersections