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AI HL · AHL 4

Vectors

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AHL 4.10

Mixed practice: lines in 3D, parametric form and intersections

AHL 4.10

Mixed practice: lines in 3D, parametric form and intersections

Target: I can write a vector equation of a line from a point and a direction, or from two points, in column or \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) form. (+5 more)

By the end of this section

Key points

Writing a line

  • A line is \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\): \(\mathbf{a}\) is the position vector of a point on it, \(\mathbf{b}\) is the direction vector.
  • Through two points \(A\) and \(B\): the direction is \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) (end minus start). Use either point as \(\mathbf{a}\).
  • Many equations describe the same line: any point on it and any multiple of the direction will do.
  • Parametric form lists one equation for each coordinate: \(x = a_1 + \lambda b_1,\ y = a_2 + \lambda b_2,\ z = a_3 + \lambda b_3\). A missing \(\mathbf{i}\), \(\mathbf{j}\) or \(\mathbf{k}\) means a zero.

Testing points and finding unknowns

  • A point is on the line only if one value of \(\lambda\) gives all its coordinates. Solve each coordinate and compare the values.
  • A coordinate with a zero in the direction never changes: it must equal the point's coordinate exactly.
  • For an unknown coordinate, use the known one to find \(\lambda\), then put \(\lambda\) into the other equations.

Two lines

  • Use different letters for the two parameters: \(\lambda\) for one line, \(\mu\) for the other.
  • Equate the coordinates to get three equations. Solve two of them for \(\lambda\) and \(\mu\), then check the third.
  • If the third equation fails, the lines do not intersect. In 3D two lines can miss without being parallel (skew lines).
  • Parallel lines have directions that are multiples of each other. Parallel lines that share a point are the same line.

Distances and axes

  • A step of 1 in \(\lambda\) moves \(|\mathbf{b}|\) along the line, so a distance \(D\) from the point \(\mathbf{a}\) needs \(|\lambda| = D \div |\mathbf{b}|\). There are two points, one each side.
  • \(PR = 3PQ\) means \(\lambda = \pm 3\) when the direction is \(\overrightarrow{PQ}\) and the point is \(P\).
  • A line meets the \(y\)-axis where \(x = 0\) and \(z = 0\) together. If the two coordinates give different values of \(\lambda\), it does not meet that axis.
  • For an unknown in a direction, find \(\lambda\) and \(\mu\) from the equations without it, then use the remaining equation to find the unknown.

On the calculator

Solving two equations for λ and μ

  • Take two of the three equations and write each as \(a\lambda + b\mu = c\).
  • Use the calculator to solve the pair for λ and μ.
  • Put the values into the third equation to check, or to find an unknown.

Screenshot and video to come

Explore: where two lines meet

Drag the points A and B and the tips of the direction arrows d₁ and d₂, or type whole numbers. Before you read the working, say whether the lines meet, are parallel or (in 3D) are skew.

xy
Line 1 (black): r₁ = a + λd₁
a
d₁
Line 2 (teal): r₂ = b + μd₂
b
d₂

  1. Drag only the tip of d₂ until the lines are parallel. What must be true about d₁ and d₂? How does the working show it?
  2. Two parallel lines: drag only B to turn them into the same line. What is special about b − a?
  3. Use the same letter λ for both lines in your head. Why would that give a wrong answer when the lines meet at different parameter values?
  4. In 3D, press They meet, then change only the z of b. Seen from above the lines still cross. Why do they no longer meet, and which equation shows it?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Line through two points, then test a third

The points are \(A(3, -2, 5)\), \(B(1, 1, -1)\) and \(C(-1, 4, -7)\). Find a vector equation of the line \(AB\) and show that \(C\) lies on it.

Working
  1. Direction

    \(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \\ -6 \end{pmatrix}\)

    End point minus start point: B minus A.

  2. Vector equation

    \(\mathbf{r} = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} + \lambda\begin{pmatrix} -2 \\ 3 \\ -6 \end{pmatrix}\)

    Use A as the point. Using B as the point is also correct.

  3. Put C in

    \(x: 3 - 2\lambda = -1 \Rightarrow \lambda = 2\); \(y: -2 + 3\lambda = 4 \Rightarrow \lambda = 2\); \(z: 5 - 6\lambda = -7 \Rightarrow \lambda = 2\)

    Solve one coordinate at a time.

  4. Conclude

    \(\lambda = 2\) in all three

    The same value of λ works in every coordinate, so C is on the line. If one disagrees, C is not on it.

Video walkthrough coming
Point and direction in \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) form

A line passes through the point \((2, 3, -1)\) and is parallel to \(\mathbf{i} - 3\mathbf{k}\). Write down a vector equation of the line and then its parametric equations. Find \(z\) when \(\lambda = 4\).

Working
  1. Vector equation

    \(\mathbf{r} = (2\mathbf{i} + 3\mathbf{j} - \mathbf{k}) + \lambda(\mathbf{i} - 3\mathbf{k})\)

    The point is a position vector; the direction goes inside the bracket with λ.

  2. Parametric form

    \(x = 2 + \lambda, y = 3, z = -1 - 3\lambda\)

    No \(\mathbf{j}\) in the direction, so \(y\) stays 3.

  3. z at λ = 4

    \(z = -1 - 3(4) = -13\)

    Put λ = 4 into the \(z\) equation only.

Video walkthrough coming
Parametric form to vector equation, and an unknown coordinate

A line has parametric equations \(x = 7 - 2\lambda\), \(y = 3\lambda\), \(z = \dfrac{\lambda + 4}{2}\). Write it as a vector equation. The point \((1, p, q)\) lies on the line. Find \(p\) and \(q\).

Working
  1. Split z

    \(z = 2 + \frac{1}{2}\lambda\)

    Divide each part of the top by 2.

  2. Vector equation

    \(\mathbf{r} = \begin{pmatrix} 7 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} -2 \\ 3 \\ \frac{1}{2} \end{pmatrix}\)

    Constants make the point; the numbers multiplying λ make the direction.

  3. Find λ

    \(7 - 2\lambda = 1 \Rightarrow \lambda = 3\)

    The known coordinate gives λ.

  4. Find p and q

    \(p = 3(3) = 9, \quad q = \frac{3 + 4}{2} = \frac{7}{2}\)

    Put λ into the other two equations.

Video walkthrough coming
Where two lines meet

Find the point of intersection of the lines \(\mathbf{r} = \begin{pmatrix} -1 \\ -7 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\mathbf{r} = (6\mathbf{i} - 3\mathbf{j} + 10\mathbf{k}) + \mu(2\mathbf{i} - \mathbf{j} + 3\mathbf{k})\).

Working
  1. Equate

    \(-1 + \lambda = 6 + 2\mu\) (1), \(-7 + 2\lambda = -3 - \mu\) (2), \(7 - \lambda = 10 + 3\mu\) (3)

    Different letters for the two lines: λ and μ. Each coordinate gives an equation.

  2. Solve two

    \(\lambda = 3, \mu = -2\)

    From the calculator, solving the first two equations together.

  3. Check the third

    \(7 - \lambda = 4\) and \(10 + 3\mu = 4\). They are equal, so the lines meet.

    In three dimensions the third equation must also hold.

  4. The point

    \((2, -1, 4)\)

    Put λ = 3 into the first line (or μ = −2 into the second: it gives the same point).

Video walkthrough coming
Showing two lines do not intersect

Show that the lines \(\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ -6 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) do not intersect.

Working
  1. Equate

    \(1 + \lambda = 5 + 2\mu\) (1), \(5 - 2\lambda = 2 + \mu\) (2), \(-6 + 3\lambda = -3 - \mu\) (3)

    Treat the lines as meeting and see whether that is possible.

  2. Solve two

    \(\lambda = 2, \mu = -1\)

    From the calculator, using the first two equations.

  3. Test the third

    left side \(0\), right side \(-2\). They are not equal, so the lines do not intersect.

    The two sides differ, so no values of λ and μ satisfy all three.

  4. Conclude

    The lines do not intersect.

    They are not parallel either (the directions are not multiples), so the lines are skew.

Video walkthrough coming
Parallel lines and the same line

The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\). (a) Find parametric equations of the line through \((5, 0, 2)\) parallel to \(l\). (b) Show that \(x = 7 - 4\mu, y = 1 - 2\mu, z = -5 + 6\mu\) describes the same line as \(l\).

Working
  1. (a) Same direction

    \(x = 5 + 2\lambda, y = \lambda, z = 2 - 3\lambda\)

    Parallel lines share a direction vector. Use the new point.

  2. (b) Direction

    \(\begin{pmatrix} -4 \\ -2 \\ 6 \end{pmatrix} = -2\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\)

    A multiple of the direction means the lines are parallel.

  3. (b) A point on l

    \(\text{Is }(7, 1, -5)\text{ on } l?\) \(x: 1 + 2\lambda = 7 \Rightarrow \lambda = 3\); \(y: -2 + \lambda = 1 \Rightarrow \lambda = 3\); \(z: 4 - 3\lambda = -5 \Rightarrow \lambda = 3\)

    Parallel lines are the same line only if they share a point.

  4. Conclude

    Same direction and a common point: the same line.

    If the point were not on l, the lines would be parallel but different.

Video walkthrough coming
Points a given distance along a line

The line \(l\) passes through \(A(1, -2, 3)\) and is parallel to \(4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\). Find the possible coordinates of the point \(P\) on \(l\) such that \(AP = 45\).

Working
  1. Magnitude

    \(|4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}| = \sqrt{16 + 16 + 49} = 9\)

    One step of λ moves 9 units along the line.

  2. The line

    \(\mathbf{r} = (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) + \lambda(4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k})\)

    P is the point with parameter λ.

  3. Set the distance

    \(|\lambda| \times 9 = 45 \Rightarrow \lambda = \pm 5\)

    Two values: P can be either side of A.

  4. The points

    \((21, -22, 38)\) or \((-19, 18, -32)\)

    Put λ = 5 and λ = −5 into the line.

Video walkthrough coming