Cambridge O Level Additional Mathematics · Syllabus 4037 · Vectors in Two Dimensions
Unit Vector
What is 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.
This definition is part of the Vectors in Two Dimensions chapter in Cambridge O Level Additional Mathematics.
Unit Vector in context
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 Unit 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.
- 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.
- 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.
- 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.
Examiner tips on Unit Vector
- 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 Unit Vector
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.