Vectors in Two Dimensions

• A vector is a pair of numbers with a direction attached. Keep the pair intact, operate on it row by row, and collapse it to a single number only at the end — and only if the question asked for one.
• Nothing in this chapter is a new calculation. Pythagoras gives magnitude, trigonometry gives components, simultaneous equations give the common time. What is new is only that two numbers travel together and must not be separated early.
• Direction is information, and taking a magnitude destroys it. Write the vector, keep both rows, and ask at the end whether the question wanted a pair of numbers or a single one.
• To get from \(A\) to \(B\), go back down \(\mathbf a\) and out along \(\mathbf b\): \(\overrightarrow{AB}=\mathbf b-\mathbf a\). Reverse the letters and you reverse the vector. Everything else in this chapter sits on top of that one line.
• Magnitude turns a vector into a number and cannot be undone. A unit vector divides by that number to keep the direction and throw the size away — and it is only correct once you have checked that it is \(1\) unit long.
• Add the rows, subtract the rows, scale both rows. One vector equation is two scalar equations — that single fact solves every unknown scalar and every collision in this chapter.
• A scalar multiple means parallel. A scalar multiple plus a shared point means collinear. The midpoint is the average of the two position vectors, and every vector proof in this topic is the displacement rule applied twice and then simplified.
• Velocities add like any other vectors. Cosine goes with the axis the angle is measured from, the quadrant supplies the signs, and an angle without a stated reference direction is not an answer.

• Read and write all four notations — \(\mathbf a\), \(\overrightarrow{AB}\), a column vector and \(x\mathbf i+y\mathbf j\) — and say what makes a quantity a vector rather than a scalar. Section A
• Derive \(\overrightarrow{AB}=\mathbf b-\mathbf a\) from the route \(O\to A\to B\), and explain why \(\overrightarrow{BA}=-\overrightarrow{AB}\) follows without any new work. Section B
• Distinguish a position vector from a displacement vector, and say what changes about a displacement if the origin is moved. Section B
• Find a magnitude with \(|\mathbf a|=\sqrt{x^2+y^2}\), and know that the answer is a scalar with no direction left in it. Section C
• Normalise a vector to \(\widehat{\mathbf a}=\mathbf a/|\mathbf a|\), verify that the result has magnitude \(1\), and state why \(\mathbf 0\) has no unit vector. Section C
• Add, subtract and scale component by component, and convert a single vector equation into two simultaneous scalar equations by equating corresponding components. Section D
• Use the triangle and parallelogram pictures of addition, and explain why \(|\mathbf a+\mathbf b|\) is almost never \(|\mathbf a|+|\mathbf b|\). Section D
• Prove the midpoint result \(\mathbf m=\tfrac12(\mathbf a+\mathbf b)\), and use scalar multiples to prove that vectors are parallel and that points are collinear. Section E
• Write a line as \(\overrightarrow{OP}=\mathbf a+\lambda\mathbf d\) and test whether a given point lies on it. Section E
• Compose velocities by adding them, and resolve a magnitude and direction into \(V\cos\theta\) and \(V\sin\theta\) with the signs checked against the quadrant. Section F
• Report a direction properly, with the reference direction attached — “\(36.9^\circ\) north of east”, never a bare \(36.9^\circ\). Section F
• Use \(\mathbf r=\mathbf r_0+t\mathbf v\) to write a position at any time, and decide whether two particles collide by demanding one common \(t\) from both component equations. Section G
• Recognise and repair the standard errors of this topic, above all the reversed displacement and the collision claimed from two different times. Mistake clinic

Three sentences that carry the whole topic. First: \(\overrightarrow{AB}=\mathbf b-\mathbf a\), endpoint minus start point — get this backwards once and every direction on the page is reversed. Second: a unit vector is a vector divided by a number, never by another vector, and you should always check that the result has magnitude \(1\). Third: two particles collide only if both component equations give the same time — crossing paths is a statement about geometry, and a collision is a statement about geometry and time.

Vector

A quantity that has both a magnitude and a direction, so that it is not fully described until both are given. In two dimensions a vector is written as a bold letter such as a, as a directed line segment such as AB with an arrow above it, as a column of two components, or in the component form x i + y j. A scalar, by contrast, has magnitude only. Displacement, velocity, acceleration and force are vectors; distance, speed, mass and temperature are scalars. Two vectors are equal when they have the same magnitude and the same direction, regardless of where each is drawn.

Position Vector

The vector from a fixed origin O to a point, written OA with an arrow above it or as the bold letter a. Its components are the coordinates of the point, so the point A with coordinates (6, 1) has position vector (6, 1) measured from O. A position vector fixes where a point is; a displacement vector fixes how to travel from one point to another and is found by subtracting the start point's position vector from the endpoint's, giving AB = b minus a. Moving the origin changes every position vector but leaves every displacement vector unchanged.

Unit Vector

A vector of magnitude exactly 1, used to carry a direction without carrying a size. The unit vector in the direction of a non-zero vector a is written a with a circumflex and equals a divided by the scalar magnitude of a, so each component is divided by the same single number. Its own magnitude should always be checked back to 1. The zero vector has no direction and its magnitude is 0, so it has no unit vector; dividing by zero is undefined. Multiplying a unit vector by any positive scalar k gives the vector of magnitude k in that same direction.

Collinear Points

Three or more points that all lie on a single straight line. Using vectors, points A, B and C are collinear when one of the displacement vectors between them is a scalar multiple of another and the two displacements share a common point: for example AB equal to k times BC, with B belonging to both. The scalar multiple alone establishes only that the two segments are parallel, since parallel segments can lie on different lines; the shared point is what forces them onto the same line. A drawing that appears straight is never sufficient evidence, because diagrams in questions are not drawn to scale.

Resultant Velocity

The single velocity that has the same effect as two or more velocities acting together, obtained by adding those velocities as vectors, component by component. A boat driven east at four metres per second through water that is itself moving north at three metres per second has resultant velocity four comma three metres per second relative to the ground. The magnitude of the resultant is the resulting speed, found by Pythagoras on its components, and its direction must always be reported against a stated reference such as north of east or anticlockwise from the positive x-axis. The resultant is generally shorter than the sum of the separate speeds, and is zero when the contributing velocities cancel exactly.

Vector Addition

The operation that combines two vectors into a single vector by adding their corresponding components, so that (a, b) plus (c, d) equals (a + c, b + d). Geometrically the same sum is obtained by placing the start of the second vector at the tip of the first and joining the original start to the final tip, which is the triangle law, or by drawing both vectors from a common point and taking the diagonal of the parallelogram they span. The result is called the resultant. Vector addition is commutative and its magnitude is generally less than the sum of the two separate magnitudes, being equal to it only when the two vectors point in the same direction.

Scalar Multiple

The vector obtained by multiplying every component of a vector by the same number k, written k a. Its magnitude is the modulus of k times the magnitude of the original vector, so the vector is stretched when the modulus of k exceeds 1 and compressed when it is less than 1. Its direction is unchanged when k is positive and exactly reversed when k is negative; multiplying by zero gives the zero vector. Two non-zero vectors are parallel precisely when one is a scalar multiple of the other, which is why the scalar multiple is the standard tool for proving that lines are parallel and that points are collinear.

Collision Condition

The requirement that two moving particles occupy the same position at the same instant. For particles with constant velocities, positions are written as r equals r nought plus t v, and the test is to set the two position vectors equal, giving one equation from the x components and one from the y components. Both equations must yield the same value of t for a collision to occur, and that value must be physically admissible, normally t greater than or equal to zero. If the two equations give different times, the paths may still intersect geometrically but the particles pass through the crossing point at different moments and do not meet. The collision position is found by substituting the common time into either particle's position vector, and confirming it with the other.

• Writing \(\overrightarrow{AB}=\mathbf a-\mathbf b\). Why it is silent The answer has the right size and the right shape. Only the sign of both components is wrong, and nothing in the arithmetic looks unusual. Every direction, angle and unit vector derived from it is then reversed. Fix Say it out loud as a journey: to get from \(A\) to \(B\) you go backwards along \(\mathbf a\) and forwards along \(\mathbf b\). Endpoint minus start point.
• Collapsing to a magnitude too early, then continuing to work with it. Why it is silent \(|\mathbf a|\) is a perfectly good number, so the arithmetic continues without complaint. But the direction has been destroyed, and \(|\mathbf a|+|\mathbf b|\) is not \(|\mathbf a+\mathbf b|\) except in the one case where \(\mathbf a\) and \(\mathbf b\) already point the same way. Fix Take the square root last. Add the columns, then take the magnitude of the result.
• Reading \(\tan^{-1}\) off the calculator and reporting it as the direction. Why it is silent \(\tan^{-1}\) always returns an angle between \(-90^\circ\) and \(90^\circ\), so for a vector in the second or third quadrant it returns an angle for the opposite vector, and reports it without any sign of trouble. Fix Sketch the components first. Decide the quadrant from the two signs, then use \(\tan^{-1}\) only on the acute angle to the axis and place it in the quadrant yourself.
• Giving a direction as a bare angle. Why it is silent A number appears on the answer line and looks complete. But \(36.9^\circ\) describes four different directions depending on where it is measured from, so it identifies none of them. Fix Every angle in this chapter is written with its reference attached: “north of east”, “below the positive \(x\)-axis”, “anticlockwise from \(Ox\)”. If the question chose a reference, use that one.
• Finding where two paths cross and calling it a collision. Why it is silent The crossing point exists and is computable, so a full page of correct-looking algebra ends in a coordinate. Two cars can cross the same junction all day without ever meeting. Fix Use one shared \(t\) for both particles from the start. Equate the \(x\)-components, equate the \(y\)-components, and require the same \(t\) from both. If the two values differ, write “they do not collide” as your answer.
• Dividing one vector by another to normalise. Why it is silent Written as \(\mathbf a/|\mathbf a|\) it looks like a division of two similar objects, and dividing the components by each other produces numbers. But vector division is not defined; only the denominator here is a scalar. Fix Read the bars. \(|\mathbf a|\) is a number. Divide each component by that one number, then check that the result has magnitude \(1\) — a check that takes five seconds and catches this instantly.
• Treating a vector as its magnitude — writing \(\mathbf a=5\) when \(\mathbf a=\binom{3}{-4}\). Why it fails \(5\) is one of infinitely many vectors’ magnitudes. Once the components are gone, the direction is gone, and no later step can recover it — every angle, unit vector and component operation afterwards is working with an object that no longer contains the information it needs. Fix Keep the column intact until the question asks for a length, a distance or a speed. Write \(|\mathbf a|=5\), with the bars, so the conversion is visible and deliberate.
• Calling \(15\ \mathrm{km\,h^{-1}}\) a velocity, or calling \(\binom{4}{3}\ \mathrm{m\,s^{-1}}\) a speed. Why it fails They are different kinds of quantity, so the words are not interchangeable even when the number is right. A question that asks for a velocity and receives a single number has not been answered, and vice versa. Fix Speed is the magnitude of the velocity. If your answer has two components it is a velocity; if it has one non-negative number it is a speed. Check that against the command word before writing it down.
• Adding magnitudes: “\(|\mathbf a|=4\) and \(|\mathbf b|=3\), so \(|\mathbf a+\mathbf b|=7\).” Why it fails Magnitudes are non-negative, so nothing can cancel. Vectors that partly oppose each other produce a resultant shorter than either estimate suggests. For \(\binom{4}{-1}\) and \(\binom{-3}{5}\), the magnitudes total \(9.95\) while the resultant has magnitude \(4.12\) — an error of a factor of \(2.4\). Fix Add the vectors first, take one magnitude last. The two operations do not commute, and \(|\mathbf a+\mathbf b|=|\mathbf a|+|\mathbf b|\) only in the special case where the vectors point the same way.
• Omitting the unit from a velocity, force or displacement answer. Why it fails \(\binom{4}{3}\) is a pair of numbers; \(\binom{4}{3}\ \mathrm{m\,s^{-1}}\) is a velocity. In a contextual question the unit is part of the answer, and a component that is stated without one cannot be checked dimensionally by you or by anyone marking it. Fix Attach the unit once, outside the bracket, to the whole vector — not separately to each row. Check that it matches the unit given in the question.
• Writing \(\overrightarrow{AB}=\mathbf a-\mathbf b\). Why it fails The route from \(A\) to \(B\) is \(\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b\). Reversing the subtraction gives the journey from \(B\) to \(A\) instead, so every component has the wrong sign. The magnitude is unaffected, which is exactly what makes the error survive undetected into the next part of the question. Fix Say it as a sentence before substituting: endpoint minus start point. Then sketch the two points roughly and confirm that the signs of your answer agree with the direction of travel you can see.
• Treating \(\overrightarrow{BA}\) as though it were \(\overrightarrow{AB}\), on the grounds that “it is the same line”. Why it fails It is the same line but not the same vector: \(\overrightarrow{BA}=-\overrightarrow{AB}\). The two have equal magnitudes and opposite directions, so any answer that depends on direction — a unit vector, an angle, a resultant — comes out reversed. Fix Check with the closed-journey test: \(\overrightarrow{AB}+\overrightarrow{BA}\) must be \(\mathbf 0\). If it is not, one of them has been written the wrong way round.
• Scaling only one row: \(3\binom{4}{-1}=\binom{12}{-1}\). Why it fails A scalar multiplies the whole vector, which means every component. Scaling one row changes the direction as well as the length, so the result is not parallel to the original at all — and parallelism is the one property scalar multiplication is supposed to preserve. Fix Write the scaled vector on its own line before combining it with anything else, and check the ratio: both rows must have been multiplied by the same number.
• Dividing one vector by another to normalise, for instance dividing the components of \(\mathbf a\) by each other. Why it fails Division by a vector is not a defined operation at all. In \(\mathbf a/|\mathbf a|\) the denominator is a number, and the expression means “multiply \(\mathbf a\) by the scalar \(1/|\mathbf a|\)”. What looks like a fraction of two similar objects is a scalar multiplication in disguise. Fix Read the vertical bars every time you see them: they turn the thing inside into a number. Then divide each component by that one number.
• Producing a “unit vector” for \(\mathbf 0\), usually written as \(\binom{0}{0}\). Why it fails Two reasons that are really the same reason. Arithmetically, \(|\mathbf 0|=0\) and division by zero is undefined. Geometrically, \(\mathbf 0\) has no direction, so there is nothing for a unit vector to record. And \(\binom{0}{0}\) has magnitude \(0\), not \(1\), so it is not a unit vector under any reading. Fix State plainly that the zero vector has no unit vector. If a normalisation in a longer question produces a division by zero, that is the answer, not an obstacle to work around.
• Stopping as soon as the division is done, without checking the magnitude. Why it fails Every error available in this calculation — a mis-computed magnitude, one row divided and not the other, dividing twice — shows up instantly in the check and is invisible without it. The check costs one line. Fix Square both components of your answer and add. The total must be exactly \(1\). If it comes to \(25\) you forgot to divide; if it comes to \(1/25\) you divided twice.
• Reading positive components off a sketch without checking the quadrant. Why it fails A component’s sign comes from the axes you chose, not from how the arrow was drawn. A velocity described as “south” is negative in the \(\mathbf j\) direction when \(\mathbf j\) points north, and a sketch that puts it on the page pointing “down and right” can easily be transcribed as two positives. Fix Write down what \(\mathbf i\) and \(\mathbf j\) mean before the first component appears, then read every direction against that statement rather than against the drawing.
• Using \(V\cos\theta\) horizontally when \(\theta\) was measured from the vertical. Why it fails Cosine belongs to the axis the angle is measured from. Measuring from the \(y\)-axis makes the vertical component \(V\cos\theta\) and the horizontal component \(V\sin\theta\) — the roles swap. Using the wrong pairing gives an answer of the right size that is completely wrong except at \(\theta=45^\circ\), which is exactly the value at which the error hides. Fix Identify the reference axis first, in writing. Then test your choice on \(\theta=0\): the component along the reference axis must come out as the full \(V\), and the other must come out as \(0\).
• Taking \(\tan^{-1}\) straight off the calculator and reporting the result as the direction. Why it fails \(\tan^{-1}\) returns only values from \(-90^\circ\) to \(90^\circ\), so it can never name a direction in the second or third quadrant. For \(\binom{-7}{24}\) it returns \(-73.7^\circ\), which points south-east — the exact opposite of the truth — and it reports this with no warning at all. Fix Get the quadrant from the two signs first. Then apply \(\tan^{-1}\) to the ratio of the component sizes to obtain an acute angle, and place that angle in the quadrant yourself.
• Giving a direction as a bare angle, with no reference stated. Why it fails “\(36.9^\circ\)” could be measured from east, from north, clockwise or anticlockwise, and each reading names a different direction. The number by itself identifies nothing. Fix Always write the reference: “\(36.9^\circ\) north of east”, “\(106.3^\circ\) anticlockwise from \(Ox\)”, “on a bearing of \(053^\circ\)”. If the question chose a convention, answer in that one.
• Converting a bearing with \(V\cos\beta\) horizontally, as though bearings were measured from the \(x\)-axis. Why it fails Bearings run clockwise from north, so with \(\mathbf i\) east and \(\mathbf j\) north the components are \(\binom{V\sin\beta}{V\cos\beta}\) — sine and cosine the other way round. Using the anticlockwise-from-\(Ox\) formula on a bearing produces a direction reflected in the line \(y=x\). Fix Test any convention on a case you already know: a bearing of \(090^\circ\) must give \(\binom{V}{0}\). Better still, convert the bearing to an angle from the positive \(x\)-axis and use the single rule you trust.
• Concluding that three points are collinear because they look collinear on the diagram. Why it fails Diagrams in questions are not necessarily drawn to scale, so their proportions carry no guarantee. Three points can look collinear and be a long way from it; the human eye cannot resolve a gradient difference of a few per cent. Fix Compute two displacement vectors and show one is a scalar multiple of the other. That is the proof, and no amount of drawing substitutes for it.
• Calling two vectors parallel because they “go roughly the same way”, or because one component is a multiple of the other. Why it fails Parallelism requires the same scalar on both rows. \(\binom{2}{5}\) and \(\binom{4}{9}\) have a top-row ratio of \(2\) and a bottom-row ratio of \(1.8\), so they are not parallel — even though they point in visibly similar directions. Fix Test both rows and state the common \(k\) explicitly. If the two ratios differ, say so: that is a complete answer to “are they parallel?”.
• Reading \(\mathbf u=-3\mathbf v\) as “\(\mathbf u\) is three times \(\mathbf v\) in the same direction”. Why it fails The negative sign reverses the direction. \(\mathbf u\) is three times as long as \(\mathbf v\) and points the opposite way. In a collinearity question this also changes the geometry: a negative scalar puts the shared point outside the segment joining the other two rather than between them. Fix Report the sign and the size separately: “parallel, opposite in direction, three times the length”. \(|k|\) gives the length ratio and the sign of \(k\) gives the orientation.
• Proving \(\overrightarrow{AB}=k\,\overrightarrow{CD}\) for four distinct points and concluding that \(A\), \(B\), \(C\), \(D\) are collinear. Why it fails Nothing in that statement forces the two segments onto the same line; parallel segments sitting on different lines satisfy it perfectly well. The shared point is the missing ingredient, and it cannot be inferred from the scalar multiple. Fix Choose displacements that already share a letter — \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\), or \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) — and then say in words which point is common to both.
• Finding where the two paths cross and reporting it as the collision point. Why it fails Crossing is a statement about the two routes; colliding is a statement about the routes and the clock. Two vehicles can use the same junction all day without ever meeting. The crossing point is computable in almost every case, which is why this error produces a confident-looking wrong answer. Fix Ask which question was set. “Do they collide?” requires one shared \(t\); “where do the paths cross?” requires two independent parameters. Never answer one with the method for the other.
• Getting \(t=2\) from the \(x\)-row and \(t=4\) from the \(y\)-row, then averaging, choosing one, or declaring a collision anyway. Why it fails The two equations must hold simultaneously. Two different values means no value works for both, so there is no instant at which the two particles share a position. There is nothing to average: \(t=3\) satisfies neither equation. Fix Solve each row independently and compare. If they disagree, write the conclusion as a sentence: “the times differ, so the particles do not collide.” Two numbers left side by side are not an answer to a question that asked you to determine something; the sentence is.
• Writing \(\mathbf r_A=\mathbf r_{A0}+t\mathbf v_A\) and \(\mathbf r_B=\mathbf r_{B0}+s\mathbf v_B\) when testing for a collision. Why it fails Two independent letters permit the particles to arrive at different moments, so the equations can be satisfied by a near miss. The set-up has quietly answered “do the paths cross?” instead, and it will usually return a solution — which is precisely the trap. Fix One clock for both particles. Use the same \(t\) in both position vectors from the first line, and the requirement of a common time is built into the algebra rather than remembered at the end.
• Answering “yes, at \(t=2\)” when the question asked when and where they collide. Why it fails The position is a separate piece of information, and a question that asked for both has only been half answered. It also serves as the check on the time: substituting \(t\) into both position vectors and getting the same point confirms the whole calculation. Fix Finish every collision answer by substituting the common \(t\) into both position vectors and stating the point. Also check that \(t\) is admissible — a negative \(t\) means the meeting would have happened before the motion started, so the particles do not collide.
• Reaching for a dot product, a cross product or a third component. Why it fails None of these is in Additional Mathematics 4037, so a method that depends on one cannot be what a question is testing. More practically, a solution built on an unavailable tool tends to bypass exactly the reasoning the question was set to draw out — the component equations, the scalar multiple, the shared time. Fix If your method needs \(\mathbf a\cdot\mathbf b\) or a \(\mathbf k\) component, stop and re-read the question: there is a two-dimensional route through it using only the operations in this chapter. Perpendicularity, when it appears at this level, is set up geometrically rather than by a dot product.

• Write the vector line before the numbers. A student who writes \(\overrightarrow{AB}=\mathbf b-\mathbf a\) and then subtracts in the wrong order has one arithmetic slip sitting under a correct rule, and can find it by re-reading two lines. A student who writes only \(\binom{-6}{6}\) has an unexplained wrong answer and nothing to check it against. In a topic where every step is a single line, the temptation to do it all in your head is strong — and this is the topic where doing it in your head reverses signs most easily.
• The two-word audit. Before you write any final answer in this topic, ask: vector or scalar? Table 6 exists to make that question answerable from the wording alone. A vector answer with one number in it, or a distance answer with two, is wrong regardless of how correct the working was.
• What links almost all twenty-four. Every one of them is an answer that arrives without a check attached. The three checks in this chapter are cheap and catch nearly everything: \(\overrightarrow{AB}+\overrightarrow{BA}=\mathbf 0\); a unit vector squares to \(1\); a claimed collision point must come out the same from both particles. Build them in as habits, not as afterthoughts.
• What these three have in common. None of them can be started by choosing a formula. Each begins by asking what kind of object the answer must be — a vector, a length, a time, a conclusion in words — and works backwards from that. In W1 the words “\(AD=2.5\) cm” force a unit vector into the middle of a position-vector question; in W2 the phrase “relative to the water” forces an addition; in W3 the phrase “\(6\sqrt2\) m apart” forces a magnitude and therefore a modulus. Reading the question for the kind of answer, rather than for the topic, is what turns a five-mark problem into a two-line one.
• The one-line reminder to carry forward. Write it on the inside cover of your notes: endpoint minus start point; magnitude last; one clock. Those three are the conventions the rest of the topic is built on, and none of them requires remembering a formula.

• Additional Mathematics 4037 is assessed by two written papers of two hours each, worth \(80\) marks apiece and carrying \(50\%\) of the qualification each. Vectors can appear in either. The difference between them decides how a vectors question can be asked:
• Both halves of that table are already in this chapter. The exact-value route is Worked example F3, retrieval question 4 and Worked example W2(a); the decimal route is Worked examples F1, F2 and F4. Neither is optional, because you do not choose which paper the topic appears in.
• A vectors question tends to arrive in one of three costumes, and recognising which one you are looking at settles the method before any arithmetic happens.
• Costume 1 · Points and displacements Two or three named points with position vectors. Wanted: a displacement, a magnitude, a unit vector, a midpoint, or a proof that three points are collinear. Everything is \(\mathbf b-\mathbf a\) and scalar multiples.
• Costume 2 · Magnitude and direction A speed or a force with an angle. Wanted: components, or a resultant, or the speed and direction of a resultant. Everything is \(V\cos\theta\), \(V\sin\theta\), and a quadrant check.
• Costume 3 · Motion in time One or two particles with initial positions and constant velocities. Wanted: a position at a given time, a distance apart, or whether they collide. Everything is \(\mathbf r=\mathbf r_0+t\mathbf v\) and one common \(t\).