Vectors

• Finish minus start, one point plus a multiple of one direction, and check the third equation: those three habits carry every question in this chapter.
• Finish minus start; solve two, check the third; directions only, and the angle between lines is acute. If those three come back instantly on day 30, the chapter has stuck.

• 3.7.1 I can use — use standard notations for vectors, i.e. the column vector (x, y), xi + yj, the column vector (x, y, z), xi + yj + zk, the displacement vector AB with an arrow over it, and a bold lower-case letter a
• 3.7.2 I can carry out — carry out addition and subtraction of vectors and multiplication of a vector by a scalar, and interpret these operations in geometrical terms
• 3.7.3 I can calculate — calculate the magnitude of a vector, and use unit vectors, displacement vectors and position vectors
• 3.7.4 I can understand — understand the significance of all the symbols used when the equation of a straight line is expressed in the form r = a + tb, and find the equation of a line, given sufficient information
• 3.7.5 I can determine — determine whether two lines are parallel, intersect or are skew, and find the point of intersection of two lines when it exists
• 3.7.6 I can use — use formulae to calculate the scalar product of two vectors, and use scalar products in problems involving lines and points

Accuracy for this chapter. Leave magnitudes and distances exact as surds (\(\sqrt{14}\), \(2\sqrt{5}\), \(3\sqrt{2}\)), giving a 3 significant figure decimal only as a second form or when asked. Keep \(\cos\theta\) unrounded (\(0.47541\), \(0.90453\)) until the angle is taken, then give degrees to 1 decimal place, or radians to 3 significant figures if radians are asked for. Write the scalar product with its components visible before the number; a bare angle from a calculator earns nothing.

• “From the \(x\) and \(y\) equations \(s = t = 1\), so the lines meet at \((3, 1, 4)\).” Correct Solve two equations, check the third. Two equations in two unknowns almost always have a solution; only the third equation says whether the lines meet. If it fails, the lines are skew and there is no point of intersection (section E).
• “\(\cos\theta = -\tfrac{1}{6}\), so the angle between the lines is \(99.6^\circ\).” Correct The angle between two lines uses the direction vectors and is acute. Use \(\mathbf{b}_1\) and \(\mathbf{b}_2\), never the position vectors \(\mathbf{a}_1\), \(\mathbf{a}_2\); when \(\cos\theta < 0\), the lines make \(180^\circ - \theta = 80.4^\circ\). Between two vectors the obtuse value stands (section F).
• “\(\overrightarrow{AB} = \overrightarrow{OA} - \overrightarrow{OB}\).” Correct Finish minus start: \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}\). The other order is \(\overrightarrow{BA}\), and the sign error then runs through every line built on it.
• “\(1 + t = 4\) gives \(t = 3\), so \(D(4, -4, 5)\) is on the line.” Correct One value of \(t\) must satisfy all three components. Here \(t = 3\) gives \(z = 4\), not 5: \(D\) is not on the line (section D).
• “The lines do not meet, so they are parallel.” Correct Parallel means the direction vectors are multiples of each other. In three dimensions two lines can fail to meet without being parallel: they are then skew.
• “The shortest distance from \(P\) to the line is \(|\overrightarrow{PA}|\), where \(A\) is the point given in the equation.” Correct \(A\) is just some point of the line. Find the foot \(N\) from \(\overrightarrow{PN}\cdot\mathbf{b} = 0\) first; the shortest distance is \(|\overrightarrow{PN}|\).
• Wrong: “\(\overrightarrow{AB} = \overrightarrow{OA} - \overrightarrow{OB}\).” Repair \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\), finish minus start; the other order is \(\overrightarrow{BA}\).
• Wrong: “The midpoint of \(AB\) is \(\tfrac{1}{2}(\mathbf{b} - \mathbf{a})\).” Repair That is half of the displacement. The midpoint is \(\mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})\).
• Wrong: “\(|3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}| = \sqrt{9 + 16} = 5\).” Repair All three components: \(\sqrt{9 + 16 + 144} = 13\).
• Wrong: “The unit vector is \(\mathbf{a}/|\mathbf{a}|^2\).” Repair Divide by \(|\mathbf{a}|\) once, then check the result has magnitude 1.
• Wrong: “The line through \(A(1, 2, 3)\) and \(B(3, 1, 4)\) is \(\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}3\\1\\4\end{pmatrix}\).” Repair The direction is \(\overrightarrow{AB} = \begin{pmatrix}2\\-1\\1\end{pmatrix}\), a difference of position vectors, not the position vector of a point.
• Wrong: “\(1 + t = 4\) gives \(t = 3\), so \(D(4, -4, 5)\) is on the line.” Repair The other two components must agree with the same \(t\); \(D\) fails in \(z\) (4, not 5).
• Wrong: using the same letter \(t\) for both lines when looking for an intersection. Repair The lines reach a common point at their own parameter values, which need not be equal; use \(s\) and \(t\).
• Wrong: “\(s = t = 1\) from \(x\) and \(y\), so the lines meet at \((3, 1, 4)\)”, with no \(z\) check. Repair Substitute into the third equation. If it fails the lines are skew and there is no point.
• Wrong: “They do not intersect, so they are parallel.” Repair Parallel means the directions are multiples. Not parallel and not meeting is skew.
• Wrong: finding the angle between two lines from their position vectors \(\mathbf{a}_1\) and \(\mathbf{a}_2\). Repair Only the directions \(\mathbf{b}_1\) and \(\mathbf{b}_2\) matter; the fixed points say where a line is, not which way it points.
• Wrong: “\(\cos\theta = -\tfrac{1}{6}\), so the angle between the lines is \(99.6^\circ\).” Repair Between two lines give the acute angle, \(180^\circ - 99.6^\circ = 80.4^\circ\).
• Wrong: “Perpendicular from \(P\): \(\overrightarrow{PN}\cdot\overrightarrow{OP} = 0\).” Repair The condition is \(\overrightarrow{PN}\cdot\mathbf{b} = 0\): perpendicular to the line's direction.
• Wrong: “The shortest distance from \(P\) to the line is \(|\overrightarrow{PA}|\)” for the point \(A\) in the line's equation. Repair Find the foot \(N\) first; the distance is \(|\overrightarrow{PN}|\), which is never more than \(|\overrightarrow{PA}|\).
• Wrong: “\(\mathbf{a}\cdot\mathbf{b} = \begin{pmatrix}a_1b_1\\a_2b_2\\a_3b_3\end{pmatrix}\).” Repair The scalar product is one number, \(a_1b_1 + a_2b_2 + a_3b_3\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. When you reach chapter 17, re-answer retrieval questions 2 and 20 in two dimensions and compare them with adding complex numbers and finding a modulus on an Argand diagram; when you revise chapter 4, where the cosine rule and \(\tfrac{1}{2}ab\sin C\) reappear, redo worked example 8 and compare the scalar-product angle with the cosine rule. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3. Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. This chapter's content is in Paper 3 only; none of it is in Paper 2. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A structured vectors question can be built on one set of points or lines and work through them in parts: a line from two points, a point test or a “show that the lines are skew”, an angle, then a foot of a perpendicular and a distance. A solid (a cuboid, a pyramid) starts with setting up coordinates. When each part stands on the one before, a wrong direction vector in part (a) carries into every later part.
• MF19 prints only the scalar product, a.b = a1b1 + a2b2 + a3b3 = |a||b| cos θ. The line r = a + tb, the magnitude, the unit vector, the midpoint, finish minus start and the foot-of-perpendicular condition must all be known (see the MF19 card).
• A missing third-equation check, an obtuse angle given between two lines, an angle found from position vectors instead of directions, and a distance rounded too early. Keep surds exact (2√5) with a 3 significant figure decimal as a second form, keep cos θ unrounded, and give angles in degrees to 1 decimal place.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.