AS & A Level · Practice questions · Skew lines: shortest distance and common perpendicular

Cross, dot, divide.

Six original Cambridge-style questions on skew lines: showing lines are skew, the shortest distance from a point to a line and between two lines, the end points and equation of the common perpendicular, and why the perpendicular segment is the shortest.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Check, find \(\mathbf{n}\), then project onto it.

Skew? Directions not multiples, and the three component equations have no common solution: solve two, test the third.
Normal: \(\mathbf{n} = \mathbf{b}_{1} \times \mathbf{b}_{2}\); check \(\mathbf{n} \cdot \mathbf{b}_{1} = \mathbf{n} \cdot \mathbf{b}_{2} = 0\) and take out common factors.
Distance: \(d = \dfrac{|(\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n}|}{|\mathbf{n}|}\); \(d = 0\) means the lines meet.
End points: general \(P\) and \(Q\), then \(\overrightarrow{PQ} \cdot \mathbf{b}_{1} = 0\) and \(\overrightarrow{PQ} \cdot \mathbf{b}_{2} = 0\); the common perpendicular is \(\mathbf{r} = \overrightarrow{OP} + t\,\overrightarrow{PQ}\).
Parallel lines: find the foot of the perpendicular from a point of one line to the other, or use \(\dfrac{|(\mathbf{a}_{2} - \mathbf{a}_{1}) \times \mathbf{b}|}{|\mathbf{b}|}\).

Recall

[3 marks]

Show that the lines \[l_{1}: \mathbf{r} = \begin{pmatrix}3\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\end{pmatrix} \quad\text{and}\quad l_{2}: \mathbf{r} = \begin{pmatrix}0\\2\\1\end{pmatrix} + \mu\begin{pmatrix}1\\-1\\2\end{pmatrix}\] are skew.

\(\begin{pmatrix}1\\1\\1\end{pmatrix}\) is not a multiple of \(\begin{pmatrix}1\\-1\\2\end{pmatrix}\), so the lines are not parallel. ✓
\(x\): \(3 + \lambda = \mu\); \(y\): \(1 + \lambda = 2 - \mu\). Substituting, \(1 + \lambda = -1 - \lambda\), so \(\lambda = -1\) and \(\mu = 2\). ✓
\(z\): \(l_{1}\) gives \(\lambda = -1\), \(l_{2}\) gives \(1 + 2\mu = 5\). Since \(-1 \ne 5\) the lines do not meet, so they are skew. ✓

Calculation

[4 marks]

Find the exact shortest distance between the lines \[\mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\-1\end{pmatrix} \quad\text{and}\quad \mathbf{r} = \begin{pmatrix}1\\1\\0\end{pmatrix} + \mu\begin{pmatrix}3\\0\\1\end{pmatrix}.\]

\(\begin{pmatrix}1\\2\\-1\end{pmatrix} \times \begin{pmatrix}3\\0\\1\end{pmatrix} = \begin{pmatrix}2(1) - (-1)(0)\\ -1(3) - 1(1)\\1(0) - 2(3)\end{pmatrix} = \begin{pmatrix}2\\-4\\-6\end{pmatrix}\), so take \(\mathbf{n} = \begin{pmatrix}1\\-2\\-3\end{pmatrix}\), \(|\mathbf{n}| = \sqrt{14}\). ✓
\(\mathbf{a}_{2} - \mathbf{a}_{1} = \begin{pmatrix}-1\\2\\-3\end{pmatrix}\), so \((\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n} = -1 - 4 + 9 = 4\). ✓
\(d = \dfrac{4}{\sqrt{14}} = \dfrac{2\sqrt{14}}{7} \approx 1.07\). ✓

Calculation

[5 marks]

The lines \(l_{1}\) and \(l_{2}\) have equations \[l_{1}: \mathbf{r} = \begin{pmatrix}0\\1\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}, \qquad l_{2}: \mathbf{r} = \begin{pmatrix}3\\2\\1\end{pmatrix} + \mu\begin{pmatrix}-2\\4\\-4\end{pmatrix}.\]

(a) Show that \(l_{1}\) and \(l_{2}\) are parallel.

(b) Find the foot of the perpendicular from \(A(3, 2, 1)\) to \(l_{1}\), and hence the exact distance between the lines.

(a) \(\begin{pmatrix}-2\\4\\-4\end{pmatrix} = -2\begin{pmatrix}1\\-2\\2\end{pmatrix}\), so the directions are parallel. \(A\) is not on \(l_{1}\) (from \(x\), \(\lambda = 3\), giving \(y = -5 \ne 2\)), so they are distinct parallel lines. ✓
(b) \(F = (\lambda,\; 1 - 2\lambda,\; 3 + 2\lambda)\), \(\overrightarrow{AF} = \begin{pmatrix}\lambda - 3\\-1 - 2\lambda\\2 + 2\lambda\end{pmatrix}\). ✓
\(\overrightarrow{AF} \cdot \begin{pmatrix}1\\-2\\2\end{pmatrix} = (\lambda - 3) + (2 + 4\lambda) + (4 + 4\lambda) = 9\lambda + 3 = 0\), so \(\lambda = -\tfrac{1}{3}\) and \(F = \big(-\tfrac{1}{3}, \tfrac{5}{3}, \tfrac{7}{3}\big)\). ✓
\(\overrightarrow{AF} = \begin{pmatrix}-10/3\\-1/3\\4/3\end{pmatrix}\), so \(|AF| = \dfrac{\sqrt{100 + 1 + 16}}{3} = \dfrac{\sqrt{117}}{3} = \sqrt{13}\). ✓
Every point of \(l_{2}\) is the same distance from \(l_{1}\), so the distance between the lines is \(\sqrt{13} \approx 3.61\). ✓

Analysis

[6 marks]

The lines \[l_{1}: \mathbf{r} = \begin{pmatrix}5\\2\\-1\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\0\end{pmatrix} \quad\text{and}\quad l_{2}: \mathbf{r} = \begin{pmatrix}5\\-6\\0\end{pmatrix} + \mu\begin{pmatrix}0\\1\\1\end{pmatrix}\] are skew.

(a) The point \(P\) on \(l_{1}\) and the point \(Q\) on \(l_{2}\) are such that \(PQ\) is perpendicular to both lines. Find \(P\) and \(Q\).

(b) Find the shortest distance between the lines, and an equation of their common perpendicular.

(a) \(P = (5 + 2\lambda,\; 2 + \lambda,\; -1)\), \(Q = (5,\; -6 + \mu,\; \mu)\), so \(\overrightarrow{PQ} = \begin{pmatrix}-2\lambda\\-8 + \mu - \lambda\\\mu + 1\end{pmatrix}\). ✓
\(\overrightarrow{PQ} \cdot \mathbf{b}_{1} = -4\lambda + (-8 + \mu - \lambda) = \mu - 5\lambda - 8 = 0\). ✓
\(\overrightarrow{PQ} \cdot \mathbf{b}_{2} = (-8 + \mu - \lambda) + (\mu + 1) = 2\mu - \lambda - 7 = 0\), so \(\lambda = 2\mu - 7\). ✓
Then \(\mu - 10\mu + 35 - 8 = 0\), so \(\mu = 3\) and \(\lambda = -1\): \(P(3, 1, -1)\) and \(Q(5, -3, 3)\). ✓
(b) \(\overrightarrow{PQ} = \begin{pmatrix}2\\-4\\4\end{pmatrix}\), so \(|PQ| = \sqrt{4 + 16 + 16} = 6\). ✓
Check: \(\mathbf{b}_{1} \times \mathbf{b}_{2} = \begin{pmatrix}1\\-2\\2\end{pmatrix}\) and \((\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n} = 0 + 16 + 2 = 18\), \(18 \div 3 = 6\). Common perpendicular: \(\mathbf{r} = \begin{pmatrix}3\\1\\-1\end{pmatrix} + t\begin{pmatrix}1\\-2\\2\end{pmatrix}\). ✓

Analysis

[5 marks]

The lines \[l_{1}: \mathbf{r} = \begin{pmatrix}1\\-2\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\2\end{pmatrix} \quad\text{and}\quad l_{2}: \mathbf{r} = \begin{pmatrix}k\\0\\3\end{pmatrix} + \mu\begin{pmatrix}2\\-1\\1\end{pmatrix},\] where \(k\) is a constant.

(a) Show that the shortest distance between \(l_{1}\) and \(l_{2}\) is \(\dfrac{|k - 2|}{\sqrt{3}}\).

(b) Find the value of \(k\) for which the lines intersect, and the point of intersection.

(a) \(\begin{pmatrix}1\\1\\2\end{pmatrix} \times \begin{pmatrix}2\\-1\\1\end{pmatrix} = \begin{pmatrix}3\\3\\-3\end{pmatrix}\); take \(\mathbf{n} = \begin{pmatrix}1\\1\\-1\end{pmatrix}\), \(|\mathbf{n}| = \sqrt{3}\). ✓
\(\mathbf{a}_{2} - \mathbf{a}_{1} = \begin{pmatrix}k - 1\\2\\3\end{pmatrix}\), scalar product \((k - 1) + 2 - 3 = k - 2\), so \(d = \dfrac{|k - 2|}{\sqrt{3}}\). ✓
(b) The directions are not parallel, so the lines meet when \(d = 0\): \(k = 2\). Then \(y\): \(-2 + \lambda = -\mu\) and \(z\): \(2\lambda = 3 + \mu\); adding, \(3\lambda - 2 = 3\), so \(\lambda = \tfrac{5}{3}\), \(\mu = \tfrac{1}{3}\). ✓
Point: \(\big(1 + \tfrac{5}{3},\; -2 + \tfrac{5}{3},\; \tfrac{10}{3}\big) = \big(\tfrac{8}{3}, -\tfrac{1}{3}, \tfrac{10}{3}\big)\); on \(l_{2}\): \(\big(2 + \tfrac{2}{3}, -\tfrac{1}{3}, 3 + \tfrac{1}{3}\big)\), the same point. ✓
(c) \(\dfrac{|k - 2|}{\sqrt{3}} = 2\sqrt{3}\) gives \(|k - 2| = 6\), so \(k = 8\) or \(k = -4\). ✓

Proof

[5 marks]

The skew lines \(l_{1}: \mathbf{r} = \mathbf{a}_{1} + \lambda\mathbf{b}_{1}\) and \(l_{2}: \mathbf{r} = \mathbf{a}_{2} + \mu\mathbf{b}_{2}\) have common perpendicular \(PQ\), with \(P\) on \(l_{1}\) and \(Q\) on \(l_{2}\). Let \(X\) be any point on \(l_{1}\) and \(Y\) any point on \(l_{2}\).

(a) Show that \(\overrightarrow{XY} = \overrightarrow{PQ} + \mathbf{w}\), where \(\mathbf{w}\) is perpendicular to \(\overrightarrow{PQ}\), and deduce that \(|XY| \ge |PQ|\).

(b) By taking \(X = A_{1}\) and \(Y = A_{2}\), the points with position vectors \(\mathbf{a}_{1}\) and \(\mathbf{a}_{2}\), deduce that \(|PQ| = \dfrac{|(\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n}|}{|\mathbf{n}|}\), where \(\mathbf{n} = \mathbf{b}_{1} \times \mathbf{b}_{2}\).

(a) \(X\) and \(P\) are both on \(l_{1}\), so \(\overrightarrow{XP} = s\mathbf{b}_{1}\) for some \(s\); similarly \(\overrightarrow{QY} = t\mathbf{b}_{2}\). ✓
\(\overrightarrow{XY} = \overrightarrow{XP} + \overrightarrow{PQ} + \overrightarrow{QY} = \overrightarrow{PQ} + \mathbf{w}\) with \(\mathbf{w} = s\mathbf{b}_{1} + t\mathbf{b}_{2}\). Since \(\overrightarrow{PQ} \cdot \mathbf{b}_{1} = \overrightarrow{PQ} \cdot \mathbf{b}_{2} = 0\), \(\overrightarrow{PQ} \cdot \mathbf{w} = 0\). ✓
So \(|XY|^{2} = (\overrightarrow{PQ} + \mathbf{w}) \cdot (\overrightarrow{PQ} + \mathbf{w}) = |PQ|^{2} + 2\,\overrightarrow{PQ} \cdot \mathbf{w} + |\mathbf{w}|^{2} = |PQ|^{2} + |\mathbf{w}|^{2} \ge |PQ|^{2}\), hence \(|XY| \ge |PQ|\). ✓
(b) \(\overrightarrow{PQ}\) is perpendicular to both \(\mathbf{b}_{1}\) and \(\mathbf{b}_{2}\), so it is parallel to \(\mathbf{n}\), and \(\mathbf{w} \cdot \mathbf{n} = 0\). Dotting \(\mathbf{a}_{2} - \mathbf{a}_{1} = \overrightarrow{PQ} + \mathbf{w}\) with \(\mathbf{n}\) gives \((\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n} = \overrightarrow{PQ} \cdot \mathbf{n}\). ✓
For parallel vectors \(|\overrightarrow{PQ} \cdot \mathbf{n}| = |PQ|\,|\mathbf{n}|\), so \(|PQ| = \dfrac{|(\mathbf{a}_{2} - \mathbf{a}_{1}) \cdot \mathbf{n}|}{|\mathbf{n}|}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.