The shape of the lesson
By the end of the lesson, learners can
- What every symbol means: Read \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) as a set of directions: from the origin go to \(A\), then walk \(t\) lots of \(\mathbf{b}\).
- A line through two points: Two points fix a line.
- Distances along a line: Moving from \(t = 0\) to \(t = k\) covers a distance of \(|k|\,|\mathbf{b}|\).
- Form: \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\), with \(\mathbf{a}\) a point on the line and \(\mathbf{b}\) a direction.
- Two points: \(\mathbf{b} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}\); either point can be \(\mathbf{a}\).
- Point on line? Find \(t\) from one component, then check it in the other two.
- Same line? Parallel directions plus one shared point.
- Distance: a parameter change of \(t\) moves \(|t|\,|\mathbf{b}|\).
One picture every idea returns to
The line through the point with position vector \(\mathbf{a}\), in the direction of \(\mathbf{b}\), is \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\). Each value of the parameter \(t\) gives one point on the line, and every point on the line comes from exactly one \(t\).
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, One point, one direction, every point.. 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: What every symbol means | Read \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) as a set of directions: from the origin go to \(A\), then walk \(t\) lots of \(\mathbf{b}\). Model the worked example Points from parameters on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: A line through two points | Two points fix a line. Model the worked example Two points on the board, one line at a time. Your turn: Find a vector equation of the line through \(A(1, 4, -2)\) and \(B(3, 1, 2)\), and show that \(C(7, -5, 10)\) lies on it. | Pairs |
| 21 to 30 min | Teach and check: Distances along a line | Moving from \(t = 0\) to \(t = k\) covers a distance of \(|k|\,|\mathbf{b}|\). Model the worked example Points a given distance away on the board, one line at a time. | Numbered Heads Together |
| 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, 21 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 Distances along a line, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Finding \(t\) from ONE component proves nothing. | A point lies on the line only if the SAME \(t\) works in all three components; always check the remaining two. |
| The direction is \(\mathbf{b} - \mathbf{a}\), never the position vector of the second point. | Writing \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) with \(\mathbf{b} = \overrightarrow{OB}\) gives a line through \(A\) parallel to \(OB\), which usually misses \(B\) completely. |
| The parameter is not a distance unless \(\mathbf{b}\) is a unit vector. | With \(\mathbf{b} = (4, 2, -2)\), \(t = 5\) means \(5\sqrt{24}\) units, not 5 units. |
The vector equation of a line: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
The line \(l\) passes through \(A(3, 0, -2)\) and is parallel to \(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\). Write down a vector equation for \(l\) and find the point on \(l\) with parameter \(-3\).
Find a vector equation of the line through \(A(1, 4, -2)\) and \(B(3, 1, 2)\). Determine whether each of the points \(C(7, -5, 10)\) and \(D(-1, 7, -5)\) lies on the line.
Show that \(\mathbf{r} = (2, -1, 3) + s(1, 2, -1)\) and \(\mathbf{r} = (5, 5, 0) + t(-2, -4, 2)\) are equations of the same line.
The line \(l\) has equation \(\mathbf{r} = (1, 2, -1) + t(1, -2, 2)\). The point \((p, q, 7)\) lies on \(l\). Find \(p\) and \(q\).
The line \(l\) has equation \(\mathbf{r} = (3, 1, -2) + t(2, -2, 1)\) and \(A\) is the point \((3, 1, -2)\). Find the coordinates of the two points on \(l\) whose distance from \(A\) is 9.
Relative to an origin \(O\) on horizontal ground, with \(z\) measured vertically upwards in metres, a straight cable runs from the top of a mast at \(T(2, -1, 12)\) through a ring at \(R(4, 0, 9)\) and is anchored to the ground at \(G\).
(a) Find a vector equation for the line of the cable. (b) Find the coordinates of \(G\). (c) Find the length of cable from \(T\) to \(G\), correct to 3 significant figures.
Answer key and marking notes
- \(\mathbf{r} = 3\mathbf{i} - 2\mathbf{k} + t(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})\).
- \(t = -3\): \((3 - 3,\ 0 + 6,\ -2 - 12) = (0, 6, -14)\).
- \(\overrightarrow{AB} = (2, -3, 4)\), so \(\mathbf{r} = (1, 4, -2) + t(2, -3, 4)\).
- \(C\): \(1 + 2t = 7\) gives \(t = 3\); \(y = 4 - 9 = -5\) ✓, \(z = -2 + 12 = 10\) ✓. \(C\) lies on the line.
- \(D\): \(1 + 2t = -1\) gives \(t = -1\); \(y = 4 + 3 = 7\) ✓, but \(z = -2 - 4 = -6 \ne -5\). \(D\) does not lie on the line.
- \((-2, -4, 2) = -2(1, 2, -1)\), so the lines are parallel.
- \((5, 5, 0)\) on the first line: \(2 + s = 5\) gives \(s = 3\); \(y = -1 + 6 = 5\) ✓, \(z = 3 - 3 = 0\) ✓.
- Parallel lines with a common point coincide, so they are the same line.
- \(z\): \(-1 + 2t = 7\), so \(t = 4\).
- \(p = 1 + 4 = 5\) and \(q = 2 - 8 = -6\).
- \(|(2, -2, 1)| = \sqrt{4 + 4 + 1} = 3\), so the distance from \(A\) is \(3|t|\).
- \(3|t| = 9\) gives \(t = 3\) or \(t = -3\).
- \(t = 3\): \((9, -5, 1)\). \(t = -3\): \((-3, 7, -5)\).
- (a) \(\overrightarrow{TR} = (2, 1, -3)\), so \(\mathbf{r} = (2, -1, 12) + t(2, 1, -3)\).
- (b) Ground: \(12 - 3t = 0\), so \(t = 4\) and \(G = (2 + 8,\ -1 + 4,\ 0) = (10, 3, 0)\).
- (c) \(TG = 4 \times |(2, 1, -3)| = 4\sqrt{14} = 15.0\) 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.
Find a vector equation of the line through \(A(1, 3, -2)\) and \(B(4, 1, 0)\).
- A\(\mathbf{r} = \begin{pmatrix}1\\3\\-2\end{pmatrix} + t\begin{pmatrix}3\\-2\\2\end{pmatrix}\)
- B\(\mathbf{r} = \begin{pmatrix}1\\3\\-2\end{pmatrix} + t\begin{pmatrix}4\\1\\0\end{pmatrix}\)
- C\(\mathbf{r} = \begin{pmatrix}1\\3\\-2\end{pmatrix} + t\begin{pmatrix}5\\4\\-2\end{pmatrix}\)
- D\(\mathbf{r} = \begin{pmatrix}3\\-2\\2\end{pmatrix} + t\begin{pmatrix}1\\3\\-2\end{pmatrix}\)
The line \(l\) has equation \(\mathbf{r} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k} + t(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})\). Find the point on \(l\) where \(t = -2\).
- A\((0, 5, -11)\)
- B\((4, -3, 5)\)
- C\((-2, 4, -8)\)
- D\((0, -1, -5)\)
The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix}1\\-2\\3\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}\). Which of these points lies on \(l\)?
- A\((-5, -5, 6)\)
- B\((3, -1, -2)\)
- C\((5, 0, -1)\)
- D\((2, 1, -1)\)
Which line is parallel to \(\mathbf{r} = \begin{pmatrix}5\\0\\-1\end{pmatrix} + t\begin{pmatrix}2\\-1\\3\end{pmatrix}\)?
- A\(\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + s\begin{pmatrix}-4\\2\\-6\end{pmatrix}\)
- B\(\mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + s\begin{pmatrix}5\\0\\-1\end{pmatrix}\)
- C\(\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + s\begin{pmatrix}2\\1\\3\end{pmatrix}\)
- D\(\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + s\begin{pmatrix}4\\-2\\3\end{pmatrix}\)
The line \(l\) passes through \(A(1, 0, 3)\) and has direction \(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\). Find the two points on \(l\) that are at a distance 6 from \(A\).
- A\((5, -2, 7)\) and \((-3, 2, -1)\)
- B\((13, -6, 15)\) and \((-11, 6, -9)\)
- C\((5, -2, 7)\) only
- D\((3, -1, 5)\) and \((-1, 1, 1)\)
The points \(P(2, k, 1)\), \(Q(5, 1, 7)\) and \(R(8, 7, 13)\) all lie on one straight line. Find \(k\).
- A\(k = -5\)
- B\(k = 5\)
- C\(k = 7\)
- D\(k = -6\)
The line \(\mathbf{r} = \begin{pmatrix}3\\a\\-1\end{pmatrix} + t\begin{pmatrix}1\\2\\b\end{pmatrix}\) passes through the points \((5, 1, 5)\) and \((7, 5, 11)\). Find \(a\) and \(b\).
- A\(a = -3,\ b = 3\)
- B\(a = 5,\ b = 3\)
- C\(a = -3,\ b = 6\)
- D\(a = -1,\ b = 3\)
Which equation describes the SAME line as \(\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + t\begin{pmatrix}2\\-1\\4\end{pmatrix}\)?
- A\(\mathbf{r} = \begin{pmatrix}5\\0\\8\end{pmatrix} + s\begin{pmatrix}-4\\2\\-8\end{pmatrix}\)
- B\(\mathbf{r} = \begin{pmatrix}5\\0\\8\end{pmatrix} + s\begin{pmatrix}2\\1\\4\end{pmatrix}\)
- C\(\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + s\begin{pmatrix}-2\\-1\\-4\end{pmatrix}\)
- D\(\mathbf{r} = \begin{pmatrix}3\\3\\4\end{pmatrix} + s\begin{pmatrix}4\\-2\\8\end{pmatrix}\)
Quiz key
Direction \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}3\\-2\\2\end{pmatrix}\); starting at \(A\) gives \(\mathbf{r} = \begin{pmatrix}1\\3\\-2\end{pmatrix} + t\begin{pmatrix}3\\-2\\2\end{pmatrix}\). The tempting direction \(\begin{pmatrix}4\\1\\0\end{pmatrix}\) is the position vector of \(B\), so that line runs from \(A\) parallel to \(OB\) and misses \(B\). \(\begin{pmatrix}5\\4\\-2\end{pmatrix}\) is \(\mathbf{a} + \mathbf{b}\), and the last option swaps the roles of point and direction: \((3, -2, 2)\) is not on the line.
\(\mathbf{r} = (2 - 2)\mathbf{i} + (1 + 4)\mathbf{j} + (-3 - 8)\mathbf{k} = 5\mathbf{j} - 11\mathbf{k}\), the point \((0, 5, -11)\). The tempting \((4, -3, 5)\) uses \(t = +2\). \((-2, 4, -8)\) is \(-2\) times the direction only, forgetting the starting point, and \((0, -1, -5)\) subtracts 2 from every coordinate instead of multiplying the direction by \(-2\).
For \((-5, -5, 6)\): \(1 + 2t = -5\) gives \(t = -3\); then \(y = -2 - 3 = -5\) and \(z = 3 + 3 = 6\), both match. For \((3, -1, -2)\), \(t = 1\) from \(x\) and \(y\), but \(z = 3 - 1 = 2\), not \(-2\). For \((5, 0, -1)\), \(t = 2\) gives \(z = 1\), not \(-1\). The point \((2, 1, -1)\) is the tempting choice because its numbers appear in the equation, but it is the direction vector, not a point: \(x = 2\) needs \(t = \dfrac{1}{2}\), which gives \(y = -1.5\).
\(\begin{pmatrix}-4\\2\\-6\end{pmatrix} = -2\begin{pmatrix}2\\-1\\3\end{pmatrix}\), so the first line is parallel (a negative multiple still gives a parallel direction). The tempting second option has \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\) as its position vector, which says nothing about direction. \(\begin{pmatrix}2\\1\\3\end{pmatrix}\) differs in the sign of one component, and \(\begin{pmatrix}4\\-2\\3\end{pmatrix}\) doubles only two of the three components.
\(|2\mathbf{i} - \mathbf{j} + 2\mathbf{k}| = 3\), so the distance from \(A\) is \(3|t|\). Then \(3|t| = 6\) gives \(t = \pm 2\): the points \((1 + 4, 0 - 2, 3 + 4) = (5, -2, 7)\) and \((1 - 4, 0 + 2, 3 - 4) = (-3, 2, -1)\). The tempting \(t = \pm 6\) assumes the direction vector has length 1, putting the points 18 from \(A\). Taking only \(t = 2\) misses the point on the other side of \(A\), and \(t = \pm 1\) gives points only 3 away.
\(\overrightarrow{QR} = \begin{pmatrix}3\\6\\6\end{pmatrix}\) and \(\overrightarrow{PQ} = \begin{pmatrix}3\\1 - k\\6\end{pmatrix}\). The \(x\)- and \(z\)-components show \(\overrightarrow{PQ} = \overrightarrow{QR}\), so \(1 - k = 6\) and \(k = -5\). The tempting \(k = 5\) loses the sign when solving \(1 - k = 6\), and \(k = 7\) writes the \(y\)-component as \(k - 1\) (that is \(\overrightarrow{QP}\)) while keeping \(+6\). \(k = -6\) ignores the 1 and sets \(-k = 6\).
The displacement is \(\begin{pmatrix}2\\4\\6\end{pmatrix} = 2\begin{pmatrix}1\\2\\3\end{pmatrix}\), so \(b = 3\). For \((5, 1, 5)\): \(3 + t = 5\) gives \(t = 2\); then \(a + 4 = 1\), so \(a = -3\), and \(z = -1 + 6 = 5\) checks. The tempting \(b = 6\) takes the displacement as the direction without dividing by 2 (the \(\mathbf{i}\) component must be 1). \(a = 5\) adds 4 instead of subtracting, and \(a = -1\) uses \(t = 1\).
\(\begin{pmatrix}-4\\2\\-8\end{pmatrix} = -2\begin{pmatrix}2\\-1\\4\end{pmatrix}\) is parallel, and \((5, 0, 8)\) is on the line (\(t = 2\)), so it is the same line. The second option starts at the right point but \(\begin{pmatrix}2\\1\\4\end{pmatrix}\) is not parallel. The third negates only some components, so \(\begin{pmatrix}-2\\-1\\-4\end{pmatrix}\) is not parallel either. The tempting last option has a parallel direction, but \((3, 3, 4)\) is not on the line (\(x = 3\) needs \(t = 1\), giving \(y = 1\)), so it is a different, parallel line.
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.