The shape of the lesson
By the end of the lesson, learners can
- Pythagoras for vectors: In the lab, drag the head of the vector and watch the right-angled triangle beneath it.
- Magnitude of a combination: To find the magnitude of something like \(2\mathbf{a} - \mathbf{b}\), first work out the column vector, then use Pythagoras once on the result.
- Unknowns and speeds: If the magnitude is given, Pythagoras becomes an equation for a missing component.
- Column vector first: for \(\overrightarrow{AB}\) or \(2\mathbf{a} - \mathbf{b}\), work out the single column vector.
- Magnitude: \(\sqrt{x^2 + y^2}\), squaring negatives with brackets.
- Exact or rounded: a surd such as \(\sqrt{50} = 5\sqrt{2}\), or 3 significant figures.
- Unknowns: set \(x^2 + y^2\) equal to the magnitude squared; remember \(\pm\).
- In context: the magnitude of a velocity is a speed.
One picture every idea returns to
The magnitude of \(\begin{pmatrix}x\\y\end{pmatrix}\) is \(\sqrt{x^2 + y^2}\). It is written \(|\mathbf{a}|\) or \(|\overrightarrow{AB}|\), it is always positive, and \(|\mathbf{a} + \mathbf{b}|\) is usually NOT \(|\mathbf{a}| + |\mathbf{b}|\).
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, How long is the arrow?. 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: Pythagoras for vectors | In the lab, drag the head of the vector and watch the right-angled triangle beneath it. Project tab 4 of the lab (Vectors lab, magnitude) and let a student drive it while the class predicts. Model the worked example A whole-number magnitude on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Magnitude of a combination | To find the magnitude of something like \(2\mathbf{a} - \mathbf{b}\), first work out the column vector, then use Pythagoras once on the result. Model the worked example Sum first, then length on the board, one line at a time. Your turn: \(\mathbf{p} = \begin{pmatrix}-2\\5\end{pmatrix}\). Find \(|3\mathbf{p}|\), correct to 3 significant figures. | Pairs |
| 21 to 30 min | Teach and check: Unknowns and speeds | If the magnitude is given, Pythagoras becomes an equation for a missing component. Model the worked example A missing component 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, 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 Unknowns and speeds, 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 |
|---|---|
| Squaring a negative number on a calculator without brackets: typing \(-4^2\) gives \(-16\). | Type \((-4)^2\), or simply square the number without its sign, since the square is always positive. |
| Adding magnitudes. \(|\mathbf{a} + \mathbf{b}|\) is the length of ONE vector; find the column vector \(\mathbf{a} + \mathbf{b}\) first, then its length. \(|\mathbf{a}| + |\mathbf{b}|\) is a different number. | Return to Magnitude of a combination: ask a pair to state the correct version and show it on the lab before moving on. |
| Forgetting the negative root. \(k^2 = 25\) has two solutions, \(5\) and \(-5\): both \(\begin{pmatrix}5\\12\end{pmatrix}\) and \(\begin{pmatrix}-5\\12\end{pmatrix}\) have length 13. | Return to Unknowns and speeds: ask a pair to state the correct version and show it on the lab before moving on. |
Magnitude of a vector: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Find the magnitude of (a) \(\begin{pmatrix}-7\\24\end{pmatrix}\) (b) \(\begin{pmatrix}5\\-5\end{pmatrix}\), giving (b) correct to 3 significant figures.
\(A\) is the point \((1, -2)\) and \(B\) is the point \((7, 6)\). Find \(|\overrightarrow{AB}|\).
\(\mathbf{a} = \begin{pmatrix}2\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-5\\1\end{pmatrix}\). Find \(|2\mathbf{a} + \mathbf{b}|\), giving your answer as an exact surd in its simplest form.
\(\left|\begin{pmatrix}k\\2k\end{pmatrix}\right| = 10\) and \(k > 0\). Find \(k\), giving your answer as an exact surd.
\(\mathbf{a} = \begin{pmatrix}4\\-3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}0\\5\end{pmatrix}\).
(a) Show that \(|\mathbf{a}| = |\mathbf{b}|\) but \(\mathbf{a} \ne \mathbf{b}\).
(b) Find \(|\mathbf{a} - \mathbf{b}|\) as an exact surd.
A drone flies from its base by the vector \(\begin{pmatrix}30\\40\end{pmatrix}\) metres, then by \(\begin{pmatrix}-50\\80\end{pmatrix}\) metres.
(a) Find the total distance the drone flies.
(b) Find how far the drone finishes from its base.
Give your answers correct to 3 significant figures.
Answer key and marking notes
- (a) \(\sqrt{49 + 576} = \sqrt{625} = 25\)
- (b) \(\sqrt{25 + 25} = \sqrt{50} = 7.07\) (3 s.f.)
- \(\overrightarrow{AB} = \begin{pmatrix}6\\8\end{pmatrix}\).
- \(|\overrightarrow{AB}| = \sqrt{36 + 64} = 10\).
- \(2\mathbf{a} + \mathbf{b} = \begin{pmatrix}4 - 5\\6 + 1\end{pmatrix} = \begin{pmatrix}-1\\7\end{pmatrix}\).
- \(|2\mathbf{a} + \mathbf{b}| = \sqrt{1 + 49} = \sqrt{50}\).
- \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).
- \(k^2 + (2k)^2 = 100\), so \(5k^2 = 100\).
- \(k^2 = 20\), so \(k = \sqrt{20} = 2\sqrt{5}\) (taking \(k > 0\)).
- (a) \(|\mathbf{a}| = \sqrt{16 + 9} = 5\) and \(|\mathbf{b}| = \sqrt{0 + 25} = 5\). Equal lengths, but the components differ, so the directions differ and \(\mathbf{a} \ne \mathbf{b}\).
- (b) \(\mathbf{a} - \mathbf{b} = \begin{pmatrix}4\\-8\end{pmatrix}\), so \(|\mathbf{a} - \mathbf{b}| = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\).
- (a) First leg: \(\sqrt{900 + 1600} = 50\) m. Second leg: \(\sqrt{2500 + 6400} = \sqrt{8900} = 94.33\ldots\) m.
- Total \(= 50 + 94.33\ldots = 144\) m (3 s.f.).
- (b) Overall vector \(\begin{pmatrix}-20\\120\end{pmatrix}\), so the distance is \(\sqrt{400 + 14\,400} = \sqrt{14\,800} = 122\) m (3 s.f.).
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.
Find \(|\mathbf{p}|\) when \(\mathbf{p} = \begin{pmatrix}-5\\12\end{pmatrix}\).
- A\(13\)
- B\(17\)
- C\(7\)
- D\(10.9\)
Find the magnitude of \(\begin{pmatrix}3\\-7\end{pmatrix}\), correct to 3 significant figures.
- A\(7.62\)
- B\(6.32\)
- C\(10\)
- D\(4\)
\(A\) is the point \((-2, 5)\) and \(B\) is the point \((4, -3)\). Find \(|\overrightarrow{AB}|\).
- A\(10\)
- B\(14\)
- C\(2.83\)
- D\(6.32\)
\(\mathbf{a} = \begin{pmatrix}2\\-3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\7\end{pmatrix}\). Find \(|\mathbf{a} + \mathbf{b}|\).
- A\(5\)
- B\(10.7\)
- C\(10.0\)
- D\(10.4\)
\(\left|\begin{pmatrix}k\\8\end{pmatrix}\right| = 17\). Find the possible values of \(k\).
- A\(15\) or \(-15\)
- B\(15\) only
- C\(9\) or \(-9\)
- D\(18.8\) or \(-18.8\)
A boat moves by \(\begin{pmatrix}8\\-6\end{pmatrix}\) km every 30 minutes. Find the speed of the boat in km/h.
- A20 km/h
- B10 km/h
- C28 km/h
- D5 km/h
Quiz key
\(|\mathbf{p}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). The tempting 17 adds the sizes of the components, \(5 + 12\), which is the route along two sides, not the length of the vector. 7 is \(12 - 5\), and 10.9 is \(\sqrt{144 - 25}\), from squaring \(-5\) as \(-25\).
\(\sqrt{3^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58} = 7.62\) (3 s.f.). The tempting 6.32 is \(\sqrt{49 - 9} = \sqrt{40}\): typing \(-7^2\) without brackets gives \(-49\), and the signs get muddled. 10 adds \(3 + 7\) without Pythagoras, and 4 adds the components with their signs, \(3 + (-7)\), and drops the minus.
\(\overrightarrow{AB} = \begin{pmatrix}4 - (-2)\\-3 - 5\end{pmatrix} = \begin{pmatrix}6\\-8\end{pmatrix}\), so \(|\overrightarrow{AB}| = \sqrt{36 + 64} = 10\). The tempting 14 adds \(6 + 8\) instead of using Pythagoras. 2.83 adds the coordinates instead of subtracting them, \(\sqrt{2^2 + 2^2}\), and 6.32 works out \(-3 - 5\) as 2, giving \(\sqrt{6^2 + 2^2}\).
\(\mathbf{a} + \mathbf{b} = \begin{pmatrix}3\\4\end{pmatrix}\), so \(|\mathbf{a} + \mathbf{b}| = \sqrt{9 + 16} = 5\). The tempting 10.7 is \(|\mathbf{a}| + |\mathbf{b}| = \sqrt{13} + \sqrt{50}\): adding magnitudes is not the same as the magnitude of the sum. 10.0 is \(|\mathbf{a} - \mathbf{b}| = \sqrt{1 + 100}\), and 10.4 gets \(-3 + 7\) wrong as 10, giving \(\sqrt{9 + 100}\).
\(k^2 + 64 = 289\), so \(k^2 = 225\) and \(k = 15\) or \(k = -15\). Both \(\begin{pmatrix}15\\8\end{pmatrix}\) and \(\begin{pmatrix}-15\\8\end{pmatrix}\) have length 17. The tempting "15 only" forgets the negative square root. \(\pm 9\) subtracts \(17 - 8\) without squaring, and \(\pm 18.8\) adds the squares, \(\sqrt{289 + 64}\), instead of subtracting.
Distance in 30 minutes \(= \sqrt{8^2 + (-6)^2} = \sqrt{100} = 10\) km, so in one hour the boat travels 20 km: 20 km/h. The tempting 10 km/h is the distance in half an hour, not per hour. 28 km/h doubles \(8 + 6\) (no Pythagoras), and 5 km/h halves the distance instead of doubling it.
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.