The shape of the lesson
By the end of the lesson, learners can
- Two formulas, one number: The scalar (dot) product has two forms.
- Angles between lines: The angle between two lines is the angle between their direction vectors, and by convention it is the ACUTE angle.
- The foot of the perpendicular: The shortest route from a point \(P\) to a line meets the line at right angles, at the foot of the perpendicular \(N\).
- Calculate: \(\mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\).
- Angle: \(\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\), with both vectors pointing away from the vertex.
- Lines: use the direction vectors and give the acute angle.
- Perpendicular: set the scalar product equal to zero.
- Foot of perpendicular: general point \(N\), then \(\overrightarrow{PN} \cdot \mathbf{b} = 0\).
One picture every idea returns to
\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\). So \(\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\), and non-zero vectors are perpendicular exactly when \(\mathbf{a} \cdot \mathbf{b} = 0\).
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, Multiply two vectors, get an angle.. 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: Two formulas, one number | The scalar (dot) product has two forms. Model the worked example Angle between two vectors on the board, one line at a time. Your turn: Show that \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 5\mathbf{j} + \mathbf{k}\) are perpendicular. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Angles between lines | The angle between two lines is the angle between their direction vectors, and by convention it is the ACUTE angle. Model the worked example Acute angle between lines on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: The foot of the perpendicular | The shortest route from a point \(P\) to a line meets the line at right angles, at the foot of the perpendicular \(N\). Model the worked example Foot of the perpendicular and distance on the board, one line at a time. Your turn: Find the foot of the perpendicular from \(P(6, 6, 0)\) to the line \(\mathbf{r} = (1, 0, -1) + t(2, 1, 2)\), and the perpendicular distance. | Jigsaw |
| 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, 23 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 The foot of the perpendicular, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 |
|---|---|
| The angle in the formula is between vectors pointing away from the same point. | For the angle \(ABC\) in a triangle use \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\); using \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) gives \(180^\circ\) minus the angle you want. |
| Use the DIRECTION vectors of the lines, not the position vectors of the points on them. | The position vectors depend on where the origin is; the angle between the lines does not. |
| Dot \(\overrightarrow{PN}\) with the DIRECTION of the line, not with the position vector of \(N\) or of \(P\). | And remember to finish: the question usually wants \(N\) or \(PN\), not just the parameter. |
The scalar product: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
Show that the vectors \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) and \(4\mathbf{i} + \mathbf{j} - 5\mathbf{k}\) are perpendicular.
Find the value of \(p\) for which \(\mathbf{i} + p\mathbf{j} + 2p\mathbf{k}\) is perpendicular to \(4\mathbf{i} + 3\mathbf{j} - \mathbf{k}\).
The points \(A\) and \(B\) have position vectors \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) and \(3\mathbf{i} + 4\mathbf{k}\). Find angle \(AOB\).
The points \(A(1, 0, 2)\), \(B(3, 1, 0)\) and \(C(2, 4, 3)\) form a triangle. (a) Find angle \(BAC\). (b) Hence find the area of triangle \(ABC\).
The line \(l\) has equation \(\mathbf{r} = (1, 0, -1) + t(2, 1, 2)\) and \(P\) is the point \((6, 6, 0)\). Find the position vector of the foot of the perpendicular from \(P\) to \(l\), and hence the perpendicular distance from \(P\) to \(l\).
A pyramid \(OABCV\) has a square base \(OABC\) in the horizontal plane, with \(\overrightarrow{OA} = 6\mathbf{i}\), \(\overrightarrow{OC} = 6\mathbf{j}\), and apex \(V\) at \((3, 3, 4)\), vertically above the centre of the base.
(a) Find \(\overrightarrow{VA}\) and \(\overrightarrow{VB}\). (b) Find angle \(AVB\). (c) Show that the edge \(VA\) is perpendicular to the diagonal \(OB\).
Answer key and marking notes
- \(2(4) + (-3)(1) + 1(-5) = 8 - 3 - 5 = 0\).
- The scalar product of two non-zero vectors is zero, so they are perpendicular.
- \(4 + 3p - 2p = 0\).
- \(p = -4\).
- \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 3 + 0 + 8 = 11\).
- \(|\overrightarrow{OA}| = 3\), \(|\overrightarrow{OB}| = 5\), so \(\cos\theta = \tfrac{11}{15}\).
- Angle \(AOB = 42.8^\circ\) (1 d.p.).
- (a) \(\overrightarrow{AB} = (2, 1, -2)\), \(\overrightarrow{AC} = (1, 4, 1)\); product \(2 + 4 - 2 = 4\).
- \(|\overrightarrow{AB}| = 3\), \(|\overrightarrow{AC}| = \sqrt{18}\), \(\cos A = \dfrac{4}{3\sqrt{18}}\), so angle \(BAC = 71.7^\circ\).
- (b) Area \(= \tfrac{1}{2} \times 3 \times \sqrt{18} \times \sin 71.7^\circ = 6.04\) (3 s.f.).
- General point \(N = (1 + 2t, t, -1 + 2t)\), so \(\overrightarrow{PN} = (2t - 5,\ t - 6,\ 2t - 1)\).
- \(\overrightarrow{PN} \cdot (2, 1, 2) = 4t - 10 + t - 6 + 4t - 2 = 9t - 18 = 0\), so \(t = 2\).
- Foot: \(5\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\).
- \(\overrightarrow{PN} = (-1, -4, 3)\), distance \(\sqrt{26} = 5.10\) (3 s.f.).
- (a) \(B = (6, 6, 0)\). \(\overrightarrow{VA} = (6, 0, 0) - (3, 3, 4) = (3, -3, -4)\), \(\overrightarrow{VB} = (3, 3, -4)\).
- (b) \(\overrightarrow{VA} \cdot \overrightarrow{VB} = 9 - 9 + 16 = 16\); both magnitudes are \(\sqrt{34}\).
- \(\cos AVB = \tfrac{16}{34}\), so angle \(AVB = 61.9^\circ\) (1 d.p.).
- (c) \(\overrightarrow{OB} = (6, 6, 0)\); \(\overrightarrow{VA} \cdot \overrightarrow{OB} = 18 - 18 + 0 = 0\), so \(VA\) is perpendicular to \(OB\).
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 \(\mathbf{a} \cdot \mathbf{b}\), where \(\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = 2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\).
- A\(-7\)
- B\(6\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}\)
- C\(19\)
- D\(7\)
Find the angle between the vectors \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) and \(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).
- A\(63.6^{\circ}\)
- B\(116.4^{\circ}\)
- C\(26.4^{\circ}\)
- D\(81.5^{\circ}\)
The vectors \(p\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} - p\mathbf{j} + 4\mathbf{k}\) are perpendicular. Find \(p\).
- A\(p = 4\)
- B\(p = -4\)
- C\(p = \dfrac{4}{5}\)
- D\(p = 1\)
For vectors \(\mathbf{a}\) and \(\mathbf{b}\), \(|\mathbf{a}| = 3\), \(|\mathbf{b}| = 4\) and \(\mathbf{a} \cdot \mathbf{b} = -6\). Find the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
- A\(120^{\circ}\)
- B\(60^{\circ}\)
- C\(150^{\circ}\)
- D\(30^{\circ}\)
Find the acute angle between the lines \(\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + s\begin{pmatrix}1\\-1\\0\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}3\\1\\4\end{pmatrix} + t\begin{pmatrix}-2\\1\\2\end{pmatrix}\).
- A\(45^{\circ}\)
- B\(135^{\circ}\)
- C\(15.3^{\circ}\)
- D\(80.4^{\circ}\)
The points \(A(1, 2, 0)\), \(B(3, 0, 1)\) and \(C(2, 4, 5)\) are given. Find angle \(ABC\).
- A\(69.6^{\circ}\)
- B\(110.4^{\circ}\)
- C\(48.2^{\circ}\)
- D\(20.4^{\circ}\)
The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-2\end{pmatrix}\) and \(P\) is the point \((6, 1, 3)\). Find the foot of the perpendicular from \(P\) to \(l\).
- A\((3, 1, 0)\)
- B\((7, 3, -4)\)
- C\((19, 9, -16)\)
- D\((-1, -1, 4)\)
Find the perpendicular distance from \(P(6, 3, 3)\) to the line \(\mathbf{r} = \begin{pmatrix}0\\1\\-1\end{pmatrix} + t\begin{pmatrix}1\\2\\2\end{pmatrix}\).
- A\(2\sqrt{5}\)
- B\(2\sqrt{14}\)
- C\(3\sqrt{6}\)
- D\(2\sqrt{41}\)
Quiz key
\(\mathbf{a} \cdot \mathbf{b} = 3(2) + (-2)(4) + 1(-5) = 6 - 8 - 5 = -7\). The tempting \(6\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}\) multiplies the components but never adds them: a scalar product is a scalar, not a vector. \(19\) ignores the negative signs, and \(7\) drops the overall sign.
\(\mathbf{a} \cdot \mathbf{b} = 2 - 2 + 4 = 4\) and \(|\mathbf{a}| = |\mathbf{b}| = 3\), so \(\cos\theta = \dfrac{4}{9}\) and \(\theta = 63.6^{\circ}\). The tempting \(116.4^{\circ}\) comes from a sign slip that makes the scalar product \(-4\), and \(26.4^{\circ}\) uses \(\sin^{-1}\) instead of \(\cos^{-1}\). \(81.5^{\circ}\) divides by \(3 \times 9\), using \(|\mathbf{b}|^{2}\) instead of \(|\mathbf{b}|\).
\(3p + 2(-p) + (-1)(4) = 0\) gives \(p - 4 = 0\), so \(p = 4\). The tempting \(p = -4\) slips the sign of the constant term. \(\dfrac{4}{5}\) writes \(2(-p)\) as \(+2p\), giving \(5p = 4\), and \(p = 1\) ignores the \(\mathbf{k}\) components.
\(\cos\theta = \dfrac{-6}{3 \times 4} = -\dfrac{1}{2}\), so \(\theta = 120^{\circ}\). The tempting \(60^{\circ}\) drops the minus sign; the angle between two vectors (as opposed to two lines) can be obtuse. \(150^{\circ}\) and \(30^{\circ}\) mix up the exact values: \(\cos 150^{\circ} = -\dfrac{\sqrt{3}}{2}\), not \(-\dfrac{1}{2}\).
\(\begin{pmatrix}1\\-1\\0\end{pmatrix} \cdot \begin{pmatrix}-2\\1\\2\end{pmatrix} = -2 - 1 + 0 = -3\), with magnitudes \(\sqrt{2}\) and 3, so \(\cos\theta = -\dfrac{1}{\sqrt{2}}\) and \(\theta = 135^{\circ}\). Lines make two angles, \(135^{\circ}\) and \(45^{\circ}\); the acute one is \(45^{\circ}\). The tempting \(135^{\circ}\) is the obtuse angle. \(15.3^{\circ}\) uses the position vectors \(\begin{pmatrix}1\\0\\2\end{pmatrix}\) and \(\begin{pmatrix}3\\1\\4\end{pmatrix}\), and \(80.4^{\circ}\) uses \(|\mathbf{b}|^{2}\) values (2 and 9) in the denominator.
\(\overrightarrow{BA} = \begin{pmatrix}-2\\2\\-1\end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix}-1\\4\\4\end{pmatrix}\). Scalar product \(2 + 8 - 4 = 6\); magnitudes 3 and \(\sqrt{33}\). \(\cos ABC = \dfrac{6}{3\sqrt{33}} = \dfrac{2}{\sqrt{33}}\), so \(ABC = 69.6^{\circ}\). The tempting \(110.4^{\circ}\) uses \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\): one vector points into \(B\) and one out, which gives the supplement. \(48.2^{\circ}\) uses the position vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) (the angle \(AOC\)), and \(20.4^{\circ}\) is \(90^{\circ} - 69.6^{\circ}\), from using \(\sin^{-1}\).
\(\overrightarrow{PN} = \begin{pmatrix}-5 + 2t\\-1 + t\\-1 - 2t\end{pmatrix}\). Then \(2(-5 + 2t) + (-1 + t) - 2(-1 - 2t) = 9t - 9 = 0\), so \(t = 1\) and \(N = (3, 1, 0)\). Check: \(\overrightarrow{PN} = \begin{pmatrix}-3\\0\\-3\end{pmatrix}\) is perpendicular to the direction. The tempting \((7, 3, -4)\) divides the scalar product 9 by \(|\mathbf{b}| = 3\) instead of by \(|\mathbf{b}|^{2} = 9\), giving \(t = 3\). \((19, 9, -16)\) uses \(t = 9\), forgetting to divide, and \((-1, -1, 4)\) has a sign error giving \(t = -1\).
\(\overrightarrow{PN} = \begin{pmatrix}t - 6\\2t - 2\\2t - 4\end{pmatrix}\). Setting \((t - 6) + 2(2t - 2) + 2(2t - 4) = 9t - 18 = 0\) gives \(t = 2\), \(N = (2, 5, 3)\), and \(PN = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}\). The tempting \(2\sqrt{14}\) is the distance from \(P\) to the point \((0, 1, -1)\), which is on the line but is not the nearest point. \(3\sqrt{6}\) is \(OP\), and \(2\sqrt{41}\) comes from dividing by \(|\mathbf{b}|\) instead of \(|\mathbf{b}|^{2}\), giving \(t = 6\).
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.