The shape of the lesson
By the end of the lesson, learners can
- Position vectors: In the lab, drag the points \(A\) and \(B\).
- Routes through a shape: Most questions give two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), on a diagram and ask for other vectors in terms of them.
- Proving with vectors: In the lab, drag the heads of \(\mathbf{a}\) and \(\mathbf{b}\) and watch the proof hold for every shape.
- Label: mark \(\mathbf{a}\) and \(\mathbf{b}\) with arrows on the diagram.
- Route: go along known lines; backwards along a vector means minus it.
- Ratio: \(m : n\) along a line means \(\dfrac{m}{m + n}\) of that line's vector.
- Simplify: collect the \(\mathbf{a}\) terms and the \(\mathbf{b}\) terms.
- Parallel: show one vector is a multiple of the other.
- Collinear: parallel AND a common point; say so in words.
One picture every idea returns to
To find any vector, follow a route of known vectors: \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\). Two vectors are parallel when one is a number times the other; three points are collinear when two such vectors are parallel AND share a point.
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, Prove it with vectors.. Extended. 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: Position vectors | In the lab, drag the points \(A\) and \(B\). Project tab 5 of the lab (Vectors lab, position vectors) and let a student drive it while the class predicts. Model the worked example From coordinates on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Routes through a shape | Most questions give two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), on a diagram and ask for other vectors in terms of them. Model the worked example Midpoint of a side on the board, one line at a time. Your turn: With \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(Q\) lies on \(AB\) with \(AQ: QB = 2: 1\). Find \(\overrightarrow{OQ}\). | Pairs |
| 21 to 30 min | Teach and check: Proving with vectors | In the lab, drag the heads of \(\mathbf{a}\) and \(\mathbf{b}\) and watch the proof hold for every shape. Project tab 6 of the lab (Vectors lab, vector proofs) and let a student drive it while the class predicts. Model the worked example Showing lines are parallel on the board, one line at a time. Your turn: Use vectors to show that \(A(1, 2)\), \(B(3, 5)\) and \(C(7, 11)\) are collinear. | Rally Coach |
| 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, 17 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 Proving with vectors, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Writing \(\overrightarrow{AB} = \mathbf{a} - \mathbf{b}\). | The route from \(A\) to \(B\) goes BACK along \(\mathbf{a}\) (that is \(-\mathbf{a}\)) and then out along \(\mathbf{b}\): \(-\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). |
| Turning a ratio straight into a fraction. \(AP: PB = 1: 3\) means \(AP\) is \(\dfrac{1}{4}\) of \(AB\), not \(\dfrac{1}{3}\): add the parts of the ratio to find the denominator. | Return to Routes through a shape: ask a pair to state the correct version and show it on the lab before moving on. |
| Stopping at 'parallel'. | Two parallel vectors could lie on two different lines. For collinear you must also say that the vectors share a common point. |
Vector geometry: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
\(A\) is the point \((-1, 4)\) and \(B\) is the point \((5, 2)\).
(a) Write down the position vector of \(A\). (b) Find \(\overrightarrow{AB}\).
In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(OB\).
Find \(\overrightarrow{AM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) lies on \(AB\) with \(AP : PB = 2 : 1\).
Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), in its simplest form.
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), \(\overrightarrow{OC} = 2\mathbf{a}\) and \(\overrightarrow{OD} = 2\mathbf{b}\).
Show that \(CD\) is parallel to \(AB\) and twice as long.
\(\overrightarrow{OP} = \mathbf{a} + \mathbf{b}\), \(\overrightarrow{OQ} = 2\mathbf{a} + 3\mathbf{b}\) and \(\overrightarrow{OR} = 4\mathbf{a} + 7\mathbf{b}\).
Show that \(P\), \(Q\) and \(R\) are collinear, and find \(PQ : QR\).
The frame of a sign is a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). A crossbar joins \(M\), the midpoint of \(OA\), to \(N\), the midpoint of \(OB\).
(a) Find \(\overrightarrow{MN}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b) Show that the crossbar is parallel to \(AB\).
(c) The side \(AB\) is 1.8 m long. How long is the crossbar?
Answer key and marking notes
- (a) \(\begin{pmatrix}-1\\4\end{pmatrix}\)
- (b) \(\mathbf{b} - \mathbf{a} = \begin{pmatrix}5 - (-1)\\2 - 4\end{pmatrix} = \begin{pmatrix}6\\-2\end{pmatrix}\)
- \(\overrightarrow{OM} = \dfrac{1}{2}\mathbf{b}\).
- \(\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OM} = -\mathbf{a} + \dfrac{1}{2}\mathbf{b}\).
- \(\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\).
- \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}\mathbf{b} - \dfrac{2}{3}\mathbf{a}\).
- \(\overrightarrow{OP} = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\).
- \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
- \(\overrightarrow{CD} = -2\mathbf{a} + 2\mathbf{b} = 2(\mathbf{b} - \mathbf{a}) = 2\overrightarrow{AB}\).
- \(\overrightarrow{CD}\) is a multiple of \(\overrightarrow{AB}\), so \(CD\) is parallel to \(AB\), and the multiplier 2 means it is twice as long.
- \(\overrightarrow{PQ} = (2\mathbf{a} + 3\mathbf{b}) - (\mathbf{a} + \mathbf{b}) = \mathbf{a} + 2\mathbf{b}\).
- \(\overrightarrow{QR} = (4\mathbf{a} + 7\mathbf{b}) - (2\mathbf{a} + 3\mathbf{b}) = 2\mathbf{a} + 4\mathbf{b} = 2\overrightarrow{PQ}\).
- \(PQ\) and \(QR\) are parallel and share the point \(Q\), so \(P\), \(Q\), \(R\) are collinear. \(PQ : QR = 1 : 2\).
- (a) \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\).
- (b) \(\overrightarrow{MN} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\overrightarrow{AB}\), a multiple of \(\overrightarrow{AB}\), so \(MN\) is parallel to \(AB\).
- (c) \(MN\) is half as long as \(AB\): \(1.8 \div 2 = 0.9\) m.
6 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.
In the diagram, \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
- A\(\mathbf{b} - \mathbf{a}\)
- B\(\mathbf{a} - \mathbf{b}\)
- C\(\mathbf{a} + \mathbf{b}\)
- D\(-\mathbf{a} - \mathbf{b}\)
\(A\) is the point \((-1, 4)\) and \(B\) is the point \((5, 0)\). Find the position vector of the midpoint of \(AB\).
- A\(\begin{pmatrix}2\\2\end{pmatrix}\)
- B\(\begin{pmatrix}3\\-2\end{pmatrix}\)
- C\(\begin{pmatrix}6\\-4\end{pmatrix}\)
- D\(\begin{pmatrix}4\\4\end{pmatrix}\)
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) lies on \(AB\) with \(AP : PB = 2 : 1\). Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
- A\(\dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
- B\(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
- C\(\mathbf{a} + \dfrac{2}{3}\mathbf{b}\)
- D\(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\)
\(OACB\) is a parallelogram with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AC\). Find \(\overrightarrow{BM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
- A\(\mathbf{a} - \dfrac{1}{2}\mathbf{b}\)
- B\(\mathbf{a} + \dfrac{1}{2}\mathbf{b}\)
- C\(\dfrac{1}{2}\mathbf{a} - \mathbf{b}\)
- D\(\dfrac{1}{2}\mathbf{b} - \mathbf{a}\)
\(\overrightarrow{PQ} = 2\mathbf{a} - 5\mathbf{b}\) and \(\overrightarrow{RS} = k\mathbf{a} - 15\mathbf{b}\). \(PQ\) is parallel to \(RS\). Find \(k\).
- A\(6\)
- B\(-6\)
- C\(-8\)
- D\(\dfrac{2}{3}\)
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). The points \(A\), \(B\) and \(C\) lie on a straight line. Find the ratio \(AB : BC\).
- A\(1 : 2\)
- B\(1 : 3\)
- C\(2 : 1\)
- D\(1 : 1\)
Quiz key
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}\). Going from \(A\) to \(O\) is backwards along \(\mathbf{a}\), so \(\overrightarrow{AO} = -\mathbf{a}\), and \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). The tempting \(\mathbf{a} - \mathbf{b}\) is \(\overrightarrow{BA}\), the journey the wrong way. \(\mathbf{a} + \mathbf{b}\) goes forwards along both vectors, and \(-\mathbf{a} - \mathbf{b}\) goes backwards along both.
\(\overrightarrow{AB} = \begin{pmatrix}6\\-4\end{pmatrix}\), so the midpoint is \(\overrightarrow{OA} + \dfrac{1}{2}\overrightarrow{AB} = \begin{pmatrix}-1\\4\end{pmatrix} + \begin{pmatrix}3\\-2\end{pmatrix} = \begin{pmatrix}2\\2\end{pmatrix}\). The tempting \(\begin{pmatrix}3\\-2\end{pmatrix}\) is half of \(\overrightarrow{AB}\), a movement, not a position: it forgets to start at \(A\). \(\begin{pmatrix}6\\-4\end{pmatrix}\) is \(\overrightarrow{AB}\) itself, and \(\begin{pmatrix}4\\4\end{pmatrix}\) adds the position vectors but forgets to halve.
The ratio has \(2 + 1 = 3\) parts, so \(AP = \dfrac{2}{3}AB\). \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\). The tempting \(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\) measures the 2 parts from \(B\) instead of from \(A\), putting \(P\) nearer \(A\). \(\mathbf{a} + \dfrac{2}{3}\mathbf{b}\) uses \(\mathbf{b}\) for \(\overrightarrow{AB}\), and \(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\) is the midpoint.
\(\overrightarrow{BM} = \overrightarrow{BC} + \overrightarrow{CM} = \mathbf{a} + \left(-\dfrac{1}{2}\mathbf{b}\right) = \mathbf{a} - \dfrac{1}{2}\mathbf{b}\), since \(C\) to \(M\) goes halfway BACK along \(\overrightarrow{AC} = \mathbf{b}\). The tempting \(\mathbf{a} + \dfrac{1}{2}\mathbf{b}\) is \(\overrightarrow{OM}\), the route from \(O\), not from \(B\). \(\dfrac{1}{2}\mathbf{a} - \mathbf{b}\) halves the wrong side, and \(\dfrac{1}{2}\mathbf{b} - \mathbf{a}\) is \(\overrightarrow{MB}\), the reverse journey.
\(-15 = 3 \times (-5)\), so \(\overrightarrow{RS} = 3\overrightarrow{PQ} = 6\mathbf{a} - 15\mathbf{b}\) and \(k = 6\). The tempting \(-6\) uses the multiplier \(-3\) because of the minus signs, but \(-5 \times (-3) = +15\), not \(-15\). \(-8\) subtracts 10 from 2 because \(-5\) fell by 10 to \(-15\) (a difference, not a multiple), and \(\dfrac{2}{3}\) divides by 3 instead of multiplying.
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BC} = (3\mathbf{b} - 2\mathbf{a}) - \mathbf{b} = 2\mathbf{b} - 2\mathbf{a} = 2(\mathbf{b} - \mathbf{a})\). So \(\overrightarrow{BC} = 2\overrightarrow{AB}\): they are parallel and share the point \(B\), and \(AB : BC = 1 : 2\). The tempting \(1 : 3\) compares \(AB\) with \(\overrightarrow{AC} = 3(\mathbf{b} - \mathbf{a})\), the whole line, instead of with \(BC\). \(2 : 1\) is the ratio the wrong way round, and \(1 : 1\) would make \(B\) the midpoint of \(AC\).
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.