Parametric form, intersecting lines and skew lines Problem set A Worked solutions

Write the parametric equations of the line \(\mathbf{r}=\begin{pmatrix}-2\\5\\1\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\-4\end{pmatrix}\).

• \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-2\\5\\1\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\-4\end{pmatrix}\)
• One equation for each component: \(x=-2+3\lambda\), \(y=5\), \(z=1-4\lambda\)

A line has parametric equations \(x=4-2t\), \(y=1+5t\), \(z=3\).

1. (a) Write down a vector equation of the line.

   • The constants give the position vector and the coefficients of \(t\) give the direction: \(\mathbf{r}=\begin{pmatrix}4\\1\\3\end{pmatrix}+t\begin{pmatrix}-2\\5\\0\end{pmatrix}\)

2. (b) Find the coordinates of the point on the line where \(t=2\).

   • \(x=4-2(2)=0,\ y=1+5(2)=11,\ z=3\)
   • The point is (0, 11, 3).

Find the coordinates of the point of intersection of the lines \(\mathbf{r}_1=\begin{pmatrix}-3\\5\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}2\\5\\0\end{pmatrix}+\mu\begin{pmatrix}1\\-2\\3\end{pmatrix}\).

• Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ -3+2\lambda=2+\mu\); \((2)\ \ 5+\lambda=5-2\mu\); \((3)\ \ -1-\lambda=3\mu\)
• Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=\frac{5}{2}+\frac{1}{2}\mu\)
• Put this into equation (2): \(\frac{5}{2}\mu=-\frac{5}{2}\), so \(\mu=-1,\ \lambda=2\)
• Check equation (3): left side \(-1-2=-3\), right side \(3(-1)=-3\). They are equal.
• The lines intersect. Put \(\lambda=2\) into \(\mathbf{r}_1\): \(\begin{pmatrix}-3\\5\\-1\end{pmatrix}+2\begin{pmatrix}2\\1\\-1\end{pmatrix}=\begin{pmatrix}1\\7\\-3\end{pmatrix}\). The point is (1, 7, −3).

Show that the lines \(\mathbf{r}_1=\begin{pmatrix}0\\3\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}5\\-2\\4\end{pmatrix}+\mu\begin{pmatrix}2\\-1\\1\end{pmatrix}\) do not intersect.

• Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ \lambda=5+2\mu\); \((2)\ \ 3+2\lambda=-2-\mu\); \((3)\ \ -1+2\lambda=4+\mu\)
• Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=5+2\mu\)
• Put this into equation (2): \(5\mu=-15\), so \(\mu=-3,\ \lambda=-1\)
• Check equation (3): left side \(-1+2(-1)=-3\), right side \(4+(-3)=1\). They are not equal.
• 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.

Two drones fly along straight paths with equations \(\mathbf{r}_1=\begin{pmatrix}10\\42\\24\end{pmatrix}+\lambda\begin{pmatrix}5\\2\\-1\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}45\\30\\15\end{pmatrix}+\mu\begin{pmatrix}-3\\4\\1\end{pmatrix}\), with coordinates in metres.

1. (a) Show that the two paths cross.

   • Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ 10+5\lambda=45-3\mu\); \((2)\ \ 42+2\lambda=30+4\mu\); \((3)\ \ 24-\lambda=15+\mu\)
   • Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=7-\frac{3}{5}\mu\)
   • Put this into equation (2): \(-\frac{26}{5}\mu=-26\), so \(\mu=5,\ \lambda=4\)
   • Check equation (3): left side \(24-4=20\), right side \(15+5=20\). They are equal.

2. (b) Find the coordinates of the crossing point.

   • The lines intersect. Put \(\lambda=4\) into \(\mathbf{r}_1\): \(\begin{pmatrix}10\\42\\24\end{pmatrix}+4\begin{pmatrix}5\\2\\-1\end{pmatrix}=\begin{pmatrix}30\\50\\20\end{pmatrix}\). The point is (30, 50, 20).

The flight paths of two aircraft are modelled by the lines \(\mathbf{r}_1=\begin{pmatrix}0\\0\\9\end{pmatrix}+\lambda\begin{pmatrix}4\\3\\-1\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}10\\-5\\7\end{pmatrix}+\mu\begin{pmatrix}1\\2\\0\end{pmatrix}\), with coordinates in kilometres.

1. (a) Show that the two flight paths do not cross.

   • Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ 4\lambda=10+\mu\); \((2)\ \ 3\lambda=-5+2\mu\); \((3)\ \ 9-\lambda=7\)
   • Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=\frac{5}{2}+\frac{1}{4}\mu\)
   • Put this into equation (2): \(-\frac{5}{4}\mu=-\frac{25}{2}\), so \(\mu=10,\ \lambda=5\)
   • Check equation (3): left side \(9-5=4\), right side \(7=7\). They are not equal.
   • 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.

2. (b) Explain why this does not mean they are parallel.

   • The direction vectors \(\begin{pmatrix}4\\3\\-1\end{pmatrix}\) and \(\begin{pmatrix}1\\2\\0\end{pmatrix}\) are not multiples of each other.
   • So the flight paths are not parallel. They are skew: neither parallel nor intersecting.

On a map, two straight roads have equations \(\mathbf{r}_1=\begin{pmatrix}1\\2\end{pmatrix}+\lambda\begin{pmatrix}3\\1\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}10\\-2\end{pmatrix}+\mu\begin{pmatrix}-1\\2\end{pmatrix}\), with distances in kilometres. Find the coordinates of the point where the roads meet.

• Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ 1+3\lambda=10-\mu\); \((2)\ \ 2+\lambda=-2+2\mu\)
• Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=3-\frac{1}{3}\mu\)
• Put this into equation (2): \(-\frac{7}{3}\mu=-7\), so \(\mu=3,\ \lambda=2\)
• The lines intersect. Put \(\lambda=2\) into \(\mathbf{r}_1\): \(\begin{pmatrix}1\\2\end{pmatrix}+2\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}7\\4\end{pmatrix}\). The point is (7, 4).

The lines \(\mathbf{r}_1=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(\mathbf{r}_2=\begin{pmatrix}3\\-1\\0\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\2\end{pmatrix}\) are in three dimensions.

1. (a) Show that the lines are not parallel.

   • If \(\begin{pmatrix}2\\1\\-1\end{pmatrix}=k\begin{pmatrix}1\\-1\\2\end{pmatrix}\) then the first components give \(k=2\).
   • The component number 2 would need \(2(-1)=-2\), but it is \(1\).
   • So the direction vectors are not multiples and the lines are not parallel.

2. (b) Show that the lines do not intersect.

   • Put \(\mathbf{r}_1=\mathbf{r}_2\) and use different letters for the two parameters. Equate the components: \((1)\ \ 1+2\lambda=3+\mu\); \((2)\ \ \lambda=-1-\mu\); \((3)\ \ 2-\lambda=2\mu\)
   • Take equation (1) and write \(\lambda\) in terms of \(\mu\): \(\lambda=1+\frac{1}{2}\mu\)
   • Put this into equation (2): \(\frac{3}{2}\mu=-2\), so \(\mu=-\frac{4}{3},\ \lambda=\frac{1}{3}\)
   • Check equation (3): left side \(2-\frac{1}{3}=\frac{5}{3}\), right side \(2(-\frac{4}{3})=-\frac{8}{3}\). They are not equal.
   • 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.

3. (c) Write down the word for two lines that are neither parallel nor intersecting.

   • Lines that are neither parallel nor intersecting are skew.

The lines \(\mathbf{r}_1=\begin{pmatrix}1\\-1\\k\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}\quad\text{and}\quad \mathbf{r}_2=\begin{pmatrix}5\\4\\5\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\2\end{pmatrix}\) intersect. Find the value of \(k\) and the coordinates of the point of intersection.

• Equate components: \((1)\ \ 1+2\lambda=5+\mu\); \((2)\ \ -1+\lambda=4-\mu\); \((3)\ \ k-\lambda=5+2\mu\)
• Equations (1) and (2) do not contain \(k\). Solve them: \(\lambda=3,\ \mu=2\)
• Put these into equation (3): \(k-3=9\), so \(k=12\)
• The point of intersection is (7, 2, 9).