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AI HL · AHL 4

Vectors

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AHL 4.9

Parametric form, intersecting lines and skew lines

AHL 4.9

Parametric form, intersecting lines and skew lines

Target: I can write the vector equation of a line as three parametric equations, and the other way round. (+4 more)

By the end of this section

Key points

Parametric form

  • Write \(\mathbf{r}\) as \(\begin{pmatrix}x\\y\\z\end{pmatrix}\) and give one equation per row: \(x=a_1+\lambda d_1\), \(y=a_2+\lambda d_2\), \(z=a_3+\lambda d_3\).
  • A zero in the direction vector means that coordinate is constant.
  • To go back, the constants form the position vector and the coefficients of the parameter form the direction vector.

Finding an intersection

  • Use a different parameter for each line: \(\mathbf{r}_1=\mathbf{a}+\lambda\mathbf{d}_1\) and \(\mathbf{r}_2=\mathbf{b}+\mu\mathbf{d}_2\). Using \(\lambda\) twice gives a wrong answer.
  • Set \(\mathbf{r}_1=\mathbf{r}_2\) and write one equation for each component.
  • Solve two of the equations for \(\lambda\) and \(\mu\). Then check the values in the third equation.
  • If all three are satisfied, put \(\lambda\) into \(\mathbf{r}_1\) to get the point.

Parallel, intersecting or skew

  • In two dimensions two lines either intersect or are parallel. In three dimensions they can also be skew: neither parallel nor intersecting.
  • If the values from two equations fail the third, the lines do not intersect.
  • First compare the direction vectors. Multiples of each other means parallel. If they are not multiples and the lines do not meet, the lines are skew.

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?

Explore: the line r = a + λd

Drag the point A, then the tip of the direction vector d (orange). Move λ and watch R slide along the line.

The points snap to whole numbers. Select one (Tab) and use the arrow keys to move it.

  1. What is λ at the point A? What is λ at the tip of d? Why?
  2. Make d point the other way along the same line. Does the line change? Does the point R at λ = 1.5 change?
  3. Make the first entry of d zero. What do the parametric equations look like, and what does the line look like?
  4. Show the other equation. Why is the same point R given by a different value of λ and μ? Is it right to compare λ with μ?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
From vector form to parametric form

Write the line \(\mathbf{r}=\begin{pmatrix}4\\-3\\0\end{pmatrix}+\lambda\begin{pmatrix}-2\\0\\6\end{pmatrix}\) in parametric form. Enter the coefficient of \(\lambda\) in the equation for \(z\).

Working
  1. Write \(\mathbf{r}\) as a column

    \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\-3\\0\end{pmatrix}+\lambda\begin{pmatrix}-2\\0\\6\end{pmatrix}\)

    Put the column of unknowns on the left.

  2. One equation per row

    \(x=4-2\lambda\), \(y=-3\), \(z=6\lambda\)

    Row 1 gives x, row 2 gives y and row 3 gives z.

  3. Zero entries

    \(y=-3\)

    A zero in the direction vector means that coordinate never changes. Here y stays at −3.

Video walkthrough coming
Where two lines meet

Find the point of intersection of \(\mathbf{r}_1=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\3\\-2\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}7\\4\\-8\end{pmatrix}+\mu\begin{pmatrix}-1\\2\\3\end{pmatrix}\). Enter the value of \(\mu\) at the intersection.

Working
  1. Equate the components

    \((1)\ \ 2+\lambda=7-\mu\); \((2)\ \ -1+3\lambda=4+2\mu\); \((3)\ \ 4-2\lambda=-8+3\mu\)

    Different letters, \(\lambda\) and \(\mu\), because the lines have different parameters.

  2. Solve two equations

    Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=5-\mu\) Put this into equation (2): \(-5\mu=-10\), so \(\mu=2,\ \lambda=3\)

    Use equations (1) and (2), then check the third.

  3. Check the third

    Check equation (3): left side \(4-2(3)=-2\), right side \(-8+3(2)=-2\). They are equal.

    The pair of values has to work in all three equations.

  4. The point

    The lines intersect. Put \(\lambda=3\) into \(\mathbf{r}_1\): \(\begin{pmatrix}2\\-1\\4\end{pmatrix}+3\begin{pmatrix}1\\3\\-2\end{pmatrix}=\begin{pmatrix}5\\8\\-2\end{pmatrix}\). The point is (5, 8, −2).

    Put \(\lambda\) into \(\mathbf{r}_1\). Putting \(\mu\) into \(\mathbf{r}_2\) gives the same point.

Video walkthrough coming
Showing two lines do not meet

Show that \(\mathbf{r}_1=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}4\\0\\6\end{pmatrix}+\mu\begin{pmatrix}1\\2\\-1\end{pmatrix}\) do not intersect. Enter the value of \(\mu\) from equations (1) and (2).

Working
  1. Equate the components

    \((1)\ \ 1+2\lambda=4+\mu\); \((2)\ \ 2-\lambda=2\mu\); \((3)\ \ 3\lambda=6-\mu\)

    Same method as for an intersection.

  2. Solve two equations

    Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=\frac{3}{2}+\frac{1}{2}\mu\) Put this into equation (2): \(-\frac{5}{2}\mu=-\frac{1}{2}\), so \(\mu=\frac{1}{5},\ \lambda=\frac{8}{5}\)

    Equations (1) and (2) always give one pair of values.

  3. Try the third

    Check equation (3): left side \(3(\frac{8}{5})=\frac{24}{5}\), right side \(6-\frac{1}{5}=\frac{29}{5}\). They are not equal.

    The pair fails the third equation.

  4. Conclusion

    The values of \(\lambda\) and \(\mu\) do not satisfy all three equations, so there is no point on both lines. The lines do not intersect.

    No pair of values works in all three, so there is no common point. The directions are not multiples, so the lines are skew.

Video walkthrough coming
Parallel lines

The lines \(\mathbf{r}_1=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\3\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}4\\-3\\8\end{pmatrix}+\mu\begin{pmatrix}-2\\4\\-6\end{pmatrix}\) have direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\). Write \(\mathbf{d}_2=k\mathbf{d}_1\) and find \(k\). Do the lines intersect?

Working
  1. Compare directions

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

    Every component of \(\mathbf{d}_2\) is −2 times the one in \(\mathbf{d}_1\). The lines are parallel.

  2. Is it the same line?

    \((4,-3,8)\text{ on }\mathbf{r}_1:\ 0+\lambda=4\Rightarrow\lambda=4,\ \ 1-2(4)=-7\neq -3\)

    Test one point of the second line in the first line.

  3. Conclusion

    The lines are parallel and distinct, so they do not intersect.

    Parallel lines meet only if they are the same line.

Video walkthrough coming