The shape of the lesson
By the end of the lesson, learners can
- Three possible relationships: Picture the edges of a box (Fig. 1).
- Finding the intersection: Write each line as a general point, using a DIFFERENT parameter for each line, and set them equal.
- Lines with an unknown: If you are told the lines intersect, the test equation becomes an extra condition.
- Parallel? Is one direction a multiple of the other? If so, test one point to decide parallel or the same line.
- Otherwise equate general points, with different parameters for the two lines.
- Solve two of the three component equations for the two parameters.
- Test the third equation: true means intersecting, false means skew.
- Point: substitute each parameter into its OWN line, and check both agree.
One picture every idea returns to
First compare the directions: if one is a multiple of the other, the lines are parallel (or the same line). If not, set the two general points equal, solve two component equations for the two parameters, and test the third: it holds when the lines intersect and fails when they are skew.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Lines that never meet and are not parallel.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Three possible relationships | Picture the edges of a box (Fig. 1). Model the worked example Parallel or the same line? on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Finding the intersection | Write each line as a general point, using a DIFFERENT parameter for each line, and set them equal. Model the worked example Lines that intersect on the board, one line at a time. Your turn: Show that the lines \(\mathbf{r} = (1, 0, 2) + s(1, 1, -1)\) and \(\mathbf{r} = (1, 3, -1) + t(2, -1, 1)\) intersect, and find the point of intersection. | Pairs |
| 21 to 30 min | Teach and check: Lines with an unknown | If you are told the lines intersect, the test equation becomes an extra condition. Model the worked example Find the constant on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Lines with an unknown, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| In two dimensions, non-parallel lines must meet, so students skip the check. | In three dimensions non-parallel is not enough: you must show the lines share a point before calling them intersecting. |
| Using the same letter for both parameters (say \(t\) and \(t\)) forces the two points to have equal parameters, which they almost never do. | You then wrongly conclude 'skew'. Always use \(s\) and \(t\), or \(\lambda\) and \(\mu\). |
| Intersecting PATHS do not mean a collision. | The parameters are separate; a collision needs both objects at the point at the same time, which means equal times, not just equal positions. |
Parallel, intersecting and skew lines: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
State, with a reason, whether the lines with direction vectors (a) \(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(-4\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}\), (b) \(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) and \(2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) are parallel.
Show that the lines \(\mathbf{r} = (1, 0, 2) + s(1, 1, -1)\) and \(\mathbf{r} = (1, 3, -1) + t(2, -1, 1)\) intersect, and find the position vector of the point of intersection.
Show that the lines \(\mathbf{r} = (1, 0, 2) + s(1, 1, -1)\) and \(\mathbf{r} = (2, 1, 4) + t(3, 0, 1)\) are skew.
The lines \(\mathbf{r} = (3, -1, 2) + s(2, 1, -1)\) and \(\mathbf{r} = (5, 1, c) + t(-1, 1, 3)\) intersect. Find \(c\) and the point of intersection.
The line \(l\) has equation \(\mathbf{r} = (2, 1, 0) + s(1, -1, 2)\). Determine whether \(l\) is parallel to, intersects, or is skew to each of the lines
\(m\): \(\mathbf{r} = (1, 2, 1) + t(-2, 2, -4)\) and \(n\): \(\mathbf{r} = (6, -3, 8) + u(0, 1, -1)\).
A tunnel is being drilled from both ends along straight lines. Relative to a fixed origin, with distances in metres, the first drill moves along \(\mathbf{r} = (0, 0, -10) + s(4, 2, -1)\) and the second along \(\mathbf{r} = (60, 40, -30) + t(-2, -2, 1)\), where \(s, t \ge 0\).
(a) Show that the two bores meet, and find the meeting point. (b) Find the total length of tunnel drilled when they meet, to 3 significant figures.
Answer key and marking notes
- (a) \(-4\mathbf{i} + 2\mathbf{j} - 6\mathbf{k} = -2(2\mathbf{i} - \mathbf{j} + 3\mathbf{k})\): parallel.
- (b) Factors \(2, 2, \tfrac{5}{3}\) are not all equal: not parallel.
- Not parallel. \(x\): \(1 + s = 1 + 2t\), so \(s = 2t\).
- \(y\): \(s = 3 - t\), so \(2t = 3 - t\), \(t = 1\), \(s = 2\).
- \(z\): \(2 - 2 = 0\) and \(-1 + 1 = 0\) ✓.
- Point: \(3\mathbf{i} + 2\mathbf{j}\).
- \((3, 0, 1)\) is not a multiple of \((1, 1, -1)\) (zero \(\mathbf{j}\)-component), so the lines are not parallel.
- \(y\): \(s = 1\). \(x\): \(1 + 1 = 2 + 3t\), so \(t = 0\).
- \(z\): \(2 - 1 = 1\) but \(4 + 0 = 4\); \(1 \ne 4\).
- No common point and not parallel, so the lines are skew.
- \(x\): \(3 + 2s = 5 - t\); \(y\): \(-1 + s = 1 + t\), so \(t = s - 2\).
- Then \(3 + 2s = 7 - s\), so \(s = \tfrac{4}{3}\) and \(t = -\tfrac{2}{3}\).
- \(z\): \(2 - \tfrac{4}{3} = c + 3\left(-\tfrac{2}{3}\right)\), so \(\tfrac{2}{3} = c - 2\) and \(c = \tfrac{8}{3}\).
- Point (\(s = \tfrac{4}{3}\) on the first line): \(\left(\tfrac{17}{3}, \tfrac{1}{3}, \tfrac{2}{3}\right)\).
- \(m\): \((-2, 2, -4) = -2(1, -1, 2)\), parallel. Is \((1, 2, 1)\) on \(l\)? \(2 + s = 1\) gives \(s = -1\), then \(y = 2\) ✓ and \(z = -2 \ne 1\). So \(l\) and \(m\) are parallel and distinct.
- \(n\): not parallel to \(l\). \(x\): \(2 + s = 6\), so \(s = 4\).
- \(y\): \(1 - 4 = -3 + u\), so \(u = 0\). \(z\): \(2(4) = 8\) and \(8 - 0 = 8\) ✓.
- \(l\) and \(n\) intersect at \((6, -3, 8)\).
- (a) \(x\): \(4s = 60 - 2t\); \(y\): \(2s = 40 - 2t\). Subtract: \(2s = 20\), so \(s = 10\), \(t = 10\).
- \(z\): \(-10 - 10 = -20\) and \(-30 + 10 = -20\) ✓. They meet at \((40, 20, -20)\).
- (b) First bore: \(10 \times |(4, 2, -1)| = 10\sqrt{21}\). Second: \(10 \times |(-2, -2, 1)| = 10 \times 3 = 30\).
- Total \(= 10\sqrt{21} + 30 = 75.8\) m (3 s.f.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The lines \(\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + s\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}4\\1\\5\end{pmatrix} + t\begin{pmatrix}1\\1\\-2\end{pmatrix}\) intersect. Find the point of intersection.
- A\((3, 0, 7)\)
- B\((5, 2, 3)\)
- C\((6, 3, 1)\)
- D\((2, 1, 5)\)
Two lines have direction vectors \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\begin{pmatrix}-4\\2\\-6\end{pmatrix}\), and they have no point in common. How are they related?
- AThey are parallel.
- BThey are skew.
- CThey intersect.
- DThey are perpendicular.
Determine the relationship between the lines \(\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + s\begin{pmatrix}1\\1\\0\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + t\begin{pmatrix}0\\1\\1\end{pmatrix}\).
- AThey are skew.
- BThey intersect at \((0, -1, 2)\).
- CThey are parallel.
- DThey intersect at \((0, 1, 0)\).
Lines with direction vectors \(3\mathbf{i} + p\mathbf{j} - 6\mathbf{k}\) and \(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) are parallel. Find \(p\).
- A\(p = -6\)
- B\(p = 6\)
- C\(p = -\dfrac{2}{3}\)
- D\(p = \dfrac{2}{3}\)
The lines \(\mathbf{r} = \begin{pmatrix}1\\1\\0\end{pmatrix} + s\begin{pmatrix}2\\1\\a\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}4\\3\\5\end{pmatrix} + t\begin{pmatrix}1\\0\\-1\end{pmatrix}\) intersect. Find \(a\).
- A\(a = 2\)
- B\(a = 3\)
- C\(a = -2\)
- D\(a = \dfrac{1}{2}\)
The lines \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + s(2\mathbf{i} - \mathbf{j} + \mathbf{k})\) and \(\mathbf{r} = -3\mathbf{i} + 4\mathbf{j} + 6\mathbf{k} + t(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\) are not parallel. Equating the \(x\)- and \(y\)-components gives the same equation, \(s - t = -2\). What is the correct conclusion?
- AThey intersect at \((3, 1, 0)\).
- BThey are skew.
- CThey are the same line.
- DThey intersect at \((-1, 3, -2)\).
The lines \(\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + s\begin{pmatrix}1\\0\\1\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}2\\1\\k\end{pmatrix} + t\begin{pmatrix}0\\1\\1\end{pmatrix}\) are given. For which values of \(k\) are the lines skew?
- AAll real \(k\) except \(k = 3\)
- B\(k = 3\) only
- CAll real values of \(k\)
- DNo value of \(k\)
The lines \(\mathbf{r} = \begin{pmatrix}3\\-1\\2\end{pmatrix} + s\begin{pmatrix}1\\1\\-2\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}-1\\4\\-5\end{pmatrix} + t\begin{pmatrix}2\\-1\\1\end{pmatrix}\) intersect at \(P\). Find the distance \(OP\), where \(O\) is the origin.
- A\(\sqrt{30}\)
- B\(\sqrt{14}\)
- C\(30\)
- D\(2\)
Quiz key
\(x\): \(1 + s = 4 + t\); \(y\): \(2 - s = 1 + t\). Adding gives \(3 = 5 + 2t\), so \(t = -1\) and \(s = 2\). Check \(z\): \(3 + 4 = 7\) and \(5 + 2 = 7\). The point is \((1 + 2, 2 - 2, 3 + 4) = (3, 0, 7)\). The tempting \((5, 2, 3)\) substitutes \(t = +1\) into the second line, dropping the sign. \((6, 3, 1)\) puts the value \(s = 2\) into the second line's equation, and \((2, 1, 5)\) uses \(s = 1\).
\(\begin{pmatrix}-4\\2\\-6\end{pmatrix} = -2\begin{pmatrix}2\\-1\\3\end{pmatrix}\), so the directions are parallel; with no common point they are distinct parallel lines. The tempting "skew" also describes lines that never meet, but skew lines must have NON-parallel directions. They cannot intersect (no common point), and perpendicular directions would need a scalar product of 0, whereas here it is \(-28\).
\(x\): \(1 + s = 0\), so \(s = -1\). \(y\): \(s = 1 + t\), so \(t = -2\). \(z\): \(2 = t\) would need \(t = 2\), a contradiction. Not parallel and no common point, so the lines are skew. The tempting "intersect at \((0, -1, 2)\)" is the point on the first line at \(s = -1\), found by skipping the check in the third component. They are not parallel because \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is not a multiple of \(\begin{pmatrix}0\\1\\1\end{pmatrix}\), and \((0, 1, 0)\) is only the starting point of the second line.
\(3 = -\lambda\) gives \(\lambda = -3\) (and \(-6 = 2\lambda\) agrees), so \(p = 2\lambda = -6\). The tempting \(p = 6\) uses \(\lambda = 3\), losing the sign of \(-\mathbf{i}\). \(-\dfrac{2}{3}\) divides 2 by \(\lambda\) instead of multiplying, and \(\dfrac{2}{3}\) makes both errors.
\(y\): \(1 + s = 3\), so \(s = 2\). \(x\): \(1 + 4 = 4 + t\), so \(t = 1\). \(z\): \(2a = 5 - 1 = 4\), so \(a = 2\), and the lines meet at \((5, 3, 4)\). The tempting \(a = 3\) uses \(5 + t\) in the \(z\)-component (ignoring the minus in the direction), giving \(2a = 6\). \(a = -2\) is a sign slip, and \(a = \dfrac{1}{2}\) solves \(2a = 4\) as \(a = \dfrac{2}{4}\).
Use \(t = s + 2\) in the \(z\)-components: \(-1 + s = 6 - 2(s + 2)\), so \(3s = 3\), \(s = 1\) and \(t = 3\). All three components now agree, and the point is \((1 + 2, 2 - 1, -1 + 1) = (3, 1, 0)\). The tempting "skew" (or "same line") comes from thinking the method has failed when the first two equations coincide; you simply choose a different pair. \((-1, 3, -2)\) uses \(s = -1\).
\(x\): \(1 + s = 2\) gives \(s = 1\); \(y\): \(2 = 1 + t\) gives \(t = 1\); \(z\): \(3 + 1 = k + 1\) gives \(k = 3\). So the lines intersect only when \(k = 3\) (at \((2, 2, 4)\)) and, as they are never parallel, they are skew for every other \(k\). The tempting "\(k = 3\) only" is the value where they intersect, the opposite of skew. "All \(k\)" and "no \(k\)" ignore that one special value.
\(x\): \(3 + s = -1 + 2t\); \(y\): \(-1 + s = 4 - t\). Subtracting: \(4 = -5 + 3t\), so \(t = 3\) and \(s = 2\). Check \(z\): \(2 - 4 = -2\) and \(-5 + 3 = -2\). So \(P = (5, 1, -2)\) and \(OP = \sqrt{25 + 1 + 4} = \sqrt{30}\). The tempting \(\sqrt{14}\) is the distance of the first line's starting point \((3, -1, 2)\) from \(O\), not of \(P\). \(30\) forgets the square root, and \(2\) adds the coordinates \(5 + 1 - 2\).
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.