Cambridge O Level Additional Mathematics · Syllabus 4037 · Vectors in Two Dimensions
Vector
What is 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.
This definition is part of the Vectors in Two Dimensions chapter in Cambridge O Level Additional Mathematics.
Vector in context
A vector is a quantity with both a magnitude (how much) and a direction (which way), such as displacement, velocity or force; a scalar has magnitude only, such as distance, speed or mass. A vector can be written as \(\mathbf a\), as \(\overrightarrow{AB}\), as a column \(\binom{x}{y}\), or as \(x\mathbf i+y\mathbf j\), and all four notations mean the same thing. The displacement from point \(A\) to point \(B\) is \(\overrightarrow{AB}=\mathbf b-\mathbf a\), endpoint minus start point, and reversing the letters reverses the sign of every component. Its magnitude, \(|\mathbf a|=\sqrt{x^2+y^2}\), is a single non-negative number found by Pythagoras and carries no direction of its own.
A unit vector has magnitude exactly \(1\), found by dividing a vector by its own magnitude, \(\widehat{\mathbf a}=\mathbf a/|\mathbf a|\); it is undefined for the zero vector, since division by zero has no meaning. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, and points \(A\), \(B\), \(C\) are collinear only when that scalar multiple also shares a common point, since parallel segments can still lie on different lines. Two particles moving with constant velocity collide only when their \(x\)- and \(y\)-component equations give the same time \(t\), not merely when their paths cross.
Common mistakes with Vector
- 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.
- 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.
- 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.
- 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.
- 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.
- 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?”.
- 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.
Examiner tips on Vector
- 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.
Questions students ask about Vector
What is the difference between a vector and a scalar?
A vector carries both a magnitude and a direction, such as displacement, velocity, acceleration or force; a scalar carries magnitude only, such as distance, speed, mass or temperature. Two vectors are equal when they have the same magnitude and the same direction, regardless of where each is drawn, while two scalars are equal when their numbers match. Mixing the two up in a contextual answer — giving a single number where a velocity was asked for — means the question has not really been answered.
How do you find the displacement vector \(\overrightarrow{AB}\) from two position vectors?
Subtract the start point's position vector from the endpoint's: \(\overrightarrow{AB}=\mathbf b-\mathbf a\), which comes from the route \(\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b\). Get the order backwards and every component's sign flips, reversing every direction, angle and unit vector calculated from it afterwards. \(\overrightarrow{BA}=-\overrightarrow{AB}\) follows automatically: it is the same line but not the same vector, since the two point in opposite directions.
How do you find a unit vector, and why is there no unit vector for \(\mathbf 0\)?
Divide the vector by its own magnitude: \(\widehat{\mathbf a}=\mathbf a/|\mathbf a|\), which keeps the direction and scales the length to exactly \(1\); check the result's magnitude equals \(1\) as a final line. The zero vector has magnitude \(0\) and no direction of its own, so dividing by \(0\) is undefined and there is nothing for a unit vector to point towards. A unit vector is a vector divided by a scalar, never by another vector.
Why is \(|\mathbf a+\mathbf b|\) almost never equal to \(|\mathbf a|+|\mathbf b|\)?
Because magnitudes add like lengths only when the two vectors point in exactly the same direction; in every other case the vectors partly work against each other, and the resultant is shorter than the sum of the two lengths. For \(\binom{4}{-1}\) and \(\binom{-3}{5}\), the individual magnitudes do not simply add. Always add the two vectors component by component first, then take the magnitude of the result, rather than adding the two magnitudes directly.
How do you prove that two vectors are parallel, or that three points are collinear?
Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, using the same scalar on both rows. Three points \(A\), \(B\), \(C\) are collinear when a displacement such as \(\overrightarrow{AB}\) is a scalar multiple of \(\overrightarrow{BC}\) and the two share the point \(B\); the scalar multiple alone only proves the segments are parallel, since parallel segments can lie on different lines. A diagram that looks straight is never sufficient evidence, since exam diagrams are not drawn to scale.
How do you decide whether two moving particles actually collide?
Write each particle's position as \(\mathbf r=\mathbf r_0+t\mathbf v\), using the same variable \(t\) for both, then set the two position vectors equal to get one equation from the \(x\)-components and one from the \(y\)-components. A collision needs both equations to give the same value of \(t\); if they give different values, the paths cross geometrically but the particles pass through that point at different moments, which is a miss, not a collision.