The shape of the lesson
By the end of the lesson, learners can
- Notation and magnitude: In the lab, build a vector from its components and watch its length update: the length is Pythagoras on the components.
- Adding, subtracting and scaling: Vectors add component by component, and a scalar multiplies every component.
- Position vectors and midpoints: Every point is reached from the origin by its position vector, so to find a new point, start at a known point and add the right multiple of a direction.
- Displacement: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
- Magnitude: \(\sqrt{x^{2} + y^{2} + z^{2}}\); unit vector = vector ÷ magnitude.
- Parallel: one vector is a single multiple \(\lambda\) of the other.
- Collinear: \(\overrightarrow{AB}\) parallel to \(\overrightarrow{BC}\) with \(B\) shared.
- New points: known point + (fraction of a displacement) or + (distance × unit vector).
One picture every idea returns to
A vector in space is \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = \begin{pmatrix}x\\y\\z\end{pmatrix}\), with magnitude \(\sqrt{x^{2}+y^{2}+z^{2}}\). The displacement from \(A\) to \(B\) is \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\): end minus start.
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, Add a third direction.. 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: Notation and magnitude | In the lab, build a vector from its components and watch its length update: the length is Pythagoras on the components. Project tab 4 of the lab (Vectors lab, magnitude) and let a student drive it while the class predicts. Model the worked example Magnitude and unit vector on the board, one line at a time. Your turn: Find the possible values of \(k\) for which \(|2k\mathbf{i} + k\mathbf{j} + 2\mathbf{k}| = 7\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Adding, subtracting and scaling | Vectors add component by component, and a scalar multiplies every component. Model the worked example Completing a parallelogram on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Position vectors and midpoints | Every point is reached from the origin by its position vector, so to find a new point, start at a known point and add the right multiple of a direction. Model the worked example Midpoint on the board, one line at a time. Your turn: \(A\) is \((1, 2, 3)\) and \(B\) is \((7, 5, -3)\). Find the position vector of the point \(P\) on \(AB\) with \(AP: PB = 1: 2\). | Cooperative Concept Mapping |
| 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, 19 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 Position vectors and midpoints, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| \(\overrightarrow{AB}\) is \(\mathbf{b} - \mathbf{a}\), not \(\mathbf{a} - \mathbf{b}\). | Getting it backwards gives \(\overrightarrow{BA}\): the same length but the opposite direction, which ruins every later step that uses the direction. |
| Parallel vectors need ONE common factor for every component. \((2, 4, 6)\) and \((1, 2, 4)\) look alike, but the factors are \(2, 2, 1.5\), so they are not parallel. | Return to Adding, subtracting and scaling: ask a pair to state the correct version and show it on the lab before moving on. |
| The midpoint is half the SUM of the position vectors, \(\tfrac{1}{2}(\mathbf{a} + \mathbf{b})\). | Half the difference, \(\tfrac{1}{2}(\mathbf{b} - \mathbf{a})\), is the step from \(A\) to the midpoint, not the midpoint itself. |
Vectors in two and three dimensions: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
The points \(A\) and \(B\) have position vectors \(3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\) and \(-\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\). Find \(\overrightarrow{AB}\) and the exact length of \(AB\).
Find the unit vector in the direction of \(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\).
The vectors \(p\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) and \(2\mathbf{i} + q\mathbf{j} + 4\mathbf{k}\) are parallel. Find \(p\) and \(q\).
The points \(P\), \(Q\) and \(R\) have coordinates \((2, -1, 0)\), \((5, 1, -4)\) and \((11, 5, -12)\). Show that \(P\), \(Q\) and \(R\) are collinear and find the ratio \(PQ : QR\).
\(OABC\) is a parallelogram, where \(O\) is the origin, \(\overrightarrow{OA} = 4\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) and \(\overrightarrow{OB} = 7\mathbf{i} + 5\mathbf{j}\).
(a) Find \(\overrightarrow{OC}\). (b) Find the exact lengths of the diagonals \(OB\) and \(AC\).
Relative to a fixed origin \(O\), with distances in metres, a drone starts at the point \(A\) with position vector \(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\). It flies in a straight line in the direction of \(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) at a constant speed of \(12\ \text{m s}^{-1}\) for 4 seconds, arriving at \(P\).
(a) Find the position vector of \(P\). (b) Find the distance \(OP\), giving your answer to 3 significant figures.
Answer key and marking notes
- \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = -4\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\).
- \(AB = \sqrt{16 + 36 + 4} = \sqrt{56} = 2\sqrt{14}\).
- Magnitude \(\sqrt{36 + 4 + 9} = \sqrt{49} = 7\).
- Unit vector \(\tfrac{1}{7}(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})\).
- \((p, 3, -6) = \lambda(2, q, 4)\).
- \(\mathbf{k}\): \(-6 = 4\lambda\), so \(\lambda = -\tfrac{3}{2}\).
- \(\mathbf{i}\): \(p = 2\lambda = -3\). \(\mathbf{j}\): \(3 = -\tfrac{3}{2}q\), so \(q = -2\).
- \(\overrightarrow{PQ} = (3, 2, -4)\).
- \(\overrightarrow{QR} = (6, 4, -8) = 2\overrightarrow{PQ}\), so \(\overrightarrow{PQ}\) and \(\overrightarrow{QR}\) are parallel.
- They share the point \(Q\), so \(P\), \(Q\), \(R\) are collinear, and \(PQ : QR = 1 : 2\).
- (a) \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\).
- (b) \(OB = \sqrt{49 + 25 + 0} = \sqrt{74}\).
- \(\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (3-4,\ 4-1,\ 2-(-2)) = (-1, 3, 4)\), so \(AC = \sqrt{1 + 9 + 16} = \sqrt{26}\).
- Distance flown \(= 12 \times 4 = 48\) m.
- \(|2\mathbf{i} + 2\mathbf{j} + \mathbf{k}| = 3\), so the unit direction is \(\tfrac{1}{3}(2, 2, 1)\).
- Displacement \(= 48 \times \tfrac{1}{3}(2, 2, 1) = (32, 32, 16)\).
- (a) \(\overrightarrow{OP} = (2 + 32,\ -1 + 32,\ 3 + 16) = 34\mathbf{i} + 31\mathbf{j} + 19\mathbf{k}\).
- (b) \(OP = \sqrt{34^{2} + 31^{2} + 19^{2}} = \sqrt{2478} = 49.8\) 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.
Find the magnitude of the vector \(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\).
- A\(7\)
- B\(5\)
- C\(49\)
- D\(\sqrt{31}\)
The points \(A\) and \(B\) have position vectors \(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\) and \(3\mathbf{i} + \mathbf{j} - 2\mathbf{k}\). Find \(\overrightarrow{AB}\).
- A\(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\)
- B\(-2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\)
- C\(4\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)
- D\(2\mathbf{i} - \mathbf{j} - 6\mathbf{k}\)
Find the unit vector in the direction of \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\).
- A\(\dfrac{1}{3}\begin{pmatrix}1\\-2\\2\end{pmatrix}\)
- B\(\dfrac{1}{9}\begin{pmatrix}1\\-2\\2\end{pmatrix}\)
- C\(\dfrac{1}{5}\begin{pmatrix}1\\-2\\2\end{pmatrix}\)
- D\(3\begin{pmatrix}1\\-2\\2\end{pmatrix}\)
The points \(P\) and \(Q\) have coordinates \((2, -1, 5)\) and \((6, 3, -1)\). Find the position vector of the midpoint \(M\) of \(PQ\).
- A\(4\mathbf{i} + \mathbf{j} + 2\mathbf{k}\)
- B\(2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\)
- C\(4\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\)
- D\(8\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\)
The vectors \(2\mathbf{i} + p\mathbf{j} - 3\mathbf{k}\) and \(q\mathbf{i} + 6\mathbf{j} + 9\mathbf{k}\) are parallel. Find \(p\) and \(q\).
- A\(p = -2,\ q = -6\)
- B\(p = 2,\ q = 6\)
- C\(p = -2,\ q = 6\)
- D\(p = -18,\ q = -6\)
The points \(A(2, -1, 3)\), \(B(5, 1, -1)\) and \(C(p, 5, q)\) are collinear. Find \(p\) and \(q\).
- A\(p = 11,\ q = -9\)
- B\(p = 9,\ q = -12\)
- C\(p = 9.5,\ q = -7\)
- D\(p = 11,\ q = 15\)
The vector \(\mathbf{v} = t\mathbf{i} + (t + 1)\mathbf{j} + 2t\mathbf{k}\) has magnitude 3. Find the possible values of \(t\).
- A\(t = 1\) or \(t = -\dfrac{4}{3}\)
- B\(t = 1\) only
- C\(t = -1\) or \(t = \dfrac{4}{3}\)
- D\(t = \dfrac{-1 \pm \sqrt{13}}{6}\)
\(OABC\) is a parallelogram, with \(O\) the origin, \(\overrightarrow{OA} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\overrightarrow{OB} = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\). The vertices are in the order \(O, A, B, C\). Find \(\overrightarrow{OC}\).
- A\(2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\)
- B\(8\mathbf{i} + 5\mathbf{j} + \mathbf{k}\)
- C\(-2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\)
- D\(4\mathbf{i} + 2.5\mathbf{j} + 0.5\mathbf{k}\)
Quiz key
\(\sqrt{2^{2} + (-3)^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\). The tempting \(\sqrt{31}\) comes from writing \(-3^{2} = -9\), so the negative component is subtracted instead of added. \(49\) forgets the square root, and \(5\) just adds the components \(2 - 3 + 6\), which is not a length.
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (3 - 1)\mathbf{i} + (1 - (-2))\mathbf{j} + (-2 - 4)\mathbf{k} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\). The tempting \(-2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\) is \(\mathbf{a} - \mathbf{b} = \overrightarrow{BA}\), start minus end. \(4\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) adds the position vectors, and \(2\mathbf{i} - \mathbf{j} - 6\mathbf{k}\) mishandles the double negative in \(1 - (-2)\).
The magnitude is \(\sqrt{1 + 4 + 4} = 3\), so the unit vector is \(\dfrac{1}{3}\begin{pmatrix}1\\-2\\2\end{pmatrix}\), which has length 1. The tempting \(\dfrac{1}{9}\) divides by the magnitude squared (forgetting the square root), giving a vector of length \(\dfrac{1}{3}\). \(\dfrac{1}{5}\) divides by \(1 + 2 + 2\), the sum of the sizes of the components, and multiplying by 3 makes the vector longer, not length 1.
\(\overrightarrow{OM} = \dfrac{1}{2}\left[(2 + 6)\mathbf{i} + (-1 + 3)\mathbf{j} + (5 - 1)\mathbf{k}\right] = 4\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The tempting \(2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\) is \(\dfrac{1}{2}\overrightarrow{PQ}\), the half-way step from \(P\), not a position vector from \(O\); you would still need to add \(\mathbf{p}\). \(4\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\) is \(\overrightarrow{PQ}\) itself, and \(8\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\) forgets to halve the sum.
Write \(2\mathbf{i} + p\mathbf{j} - 3\mathbf{k} = \lambda(q\mathbf{i} + 6\mathbf{j} + 9\mathbf{k})\). The \(\mathbf{k}\) components give \(-3 = 9\lambda\), so \(\lambda = -\dfrac{1}{3}\). Then \(p = 6\lambda = -2\) and \(2 = \lambda q\) gives \(q = -6\). The tempting \(p = 2,\ q = 6\) uses \(\lambda = \dfrac{1}{3}\), dropping the sign from \(-3\). \(p = -2,\ q = 6\) gets the sign right only once, and \(p = -18\) divides 6 by \(\lambda\) instead of multiplying.
\(\overrightarrow{AB} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\) and \(\overrightarrow{AC} = (p - 2)\mathbf{i} + 6\mathbf{j} + (q - 3)\mathbf{k}\). The \(\mathbf{j}\) components give \(6 = 2t\), so \(t = 3\). Then \(p - 2 = 9\), \(p = 11\), and \(q - 3 = -12\), \(q = -9\). The tempting \(p = 9,\ q = -12\) is the vector \(3\overrightarrow{AB}\) itself: you must add \(\mathbf{a}\) to get the position of \(C\). \(p = 9.5,\ q = -7\) takes \(t = \dfrac{5}{2}\) from the \(y\)-coordinate 5 instead of the change \(5 - (-1) = 6\), and \(q = 15\) adds 12 instead of subtracting.
\(t^{2} + (t + 1)^{2} + 4t^{2} = 9\) gives \(6t^{2} + 2t + 1 = 9\), so \(3t^{2} + t - 4 = 0\), \((3t + 4)(t - 1) = 0\) and \(t = 1\) or \(t = -\dfrac{4}{3}\). Both are valid: a magnitude condition does not restrict the sign of \(t\). The tempting \(\dfrac{-1 \pm \sqrt{13}}{6}\) sets the sum of squares equal to 3 instead of \(3^{2} = 9\). \(t = 1\) only discards a valid root, and \(t = -1\) or \(\dfrac{4}{3}\) comes from factorising with the signs reversed.
\(\overrightarrow{OC} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\). Check: \(\mathbf{a} + \mathbf{c} = \mathbf{b}\), as the diagonal rule for a parallelogram requires. The tempting \(8\mathbf{i} + 5\mathbf{j} + \mathbf{k}\) is \(\mathbf{a} + \mathbf{b}\), which would be the fourth vertex if \(OA\) and \(OB\) were the sides, but \(OB\) is a diagonal. \(-2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\) is \(\overrightarrow{BA}\), the wrong direction, and \(4\mathbf{i} + 2.5\mathbf{j} + 0.5\mathbf{k}\) is the midpoint of \(AB\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.