Vector geometry
Section: Transformations and vectors | Syllabus: Cambridge IGCSE Mathematics (0580)
Position Vectors
Extended Only
- A position vector describes where a point is relative to a fixed origin O
- The position vector of a point A is written OA=a; its components are always equal to the coordinates of A
- The displacement vector between two points with known position vectors follows a simple rule: AB=b-a (finish's position vector minus start's position vector)
- The position vector of the midpoint M of a line joining points with position vectors a and b is OM=(a+b/2)
Worked Example: Finding a Displacement Vector from Position Vectors
- Question: Point C has position vector c= 3. -1 and point D has position vector d= -2. 5 . Find CD.
- Step 1: Apply the rule CD=d-c
- Step 2: -2. 5 - 3. -1 = -5. 6
- Answer: CD= -5. 6
Common Mistakes
MistakeComputing a - b instead of b - a when finding the vector from A to B
Fixthe displacement vector always uses finish minus start: AB = b - a, not a - b
MistakeForgetting to divide by 2 when finding the position vector of a midpoint
Fixa midpoint's position vector is the average of the two endpoints' position vectors - always halve the sum
Finding Vector Routes in Geometric Figures
Extended Only
- An unknown vector can often be found by travelling along a route of known vectors from a start point to an end point, adding each stage tip-to-tail
- In shapes with parallel or equal sides (parallelograms, trapezia), a side that is parallel to a given vector is a scalar multiple of it - the scalar depends on the relative length
- Always simplify a final vector expression fully, and give the answer in terms of the letters given in the question
Worked Example: Finding a Vector Using a Route
- Question: In trapezium OABC, OA=p and OC=q. AB is parallel to OC, and AB=2OC. M is the midpoint of BC. Find OM in terms of p and q, in its simplest form.
- Step 1: Since AB is parallel to OC and twice its length: AB=2q
- Step 2: Find the position vector of B by routing through A: OB=OA+AB=p+2q
- Step 3: C already has position vector q; use the midpoint rule on B and C: OM=((p+2q)+q/2)
- Answer: OM=(1/2)p+(3/2)q
Common Mistakes
MistakeAdding vectors in the wrong order along the route, e.g. going from B to O instead of O to B
Fixsketch the route with arrows before writing any expression - each vector in the sum must point the same way you are travelling
MistakeLeaving the final answer as an unsimplified sum of vectors instead of collecting like terms
Fixalways collect the p terms and q terms together at the end and cancel any common factors, exactly as with ordinary algebra
Using Vectors to Prove Lines Are Parallel
Extended Only
- Two lines are parallel if their vectors are scalar multiples of each other, e.g. if PQ=k×AB for some number k
- If, in addition, the two lines share a common point, then the points involved are collinear (they all lie on the same straight line)
- To prove a geometric result with vectors: find both vectors in terms of the same letters, then show algebraically that one is a scalar multiple of the other
Worked Example: Proving Two Lines Are Parallel
- Question: In triangle OAB, OA=a and OB=b. Point P lies on OA such that OP:PA=1:2, and point Q lies on OB such that OQ:QB=1:2. Show that PQ is parallel to AB.
- Step 1: Since OP:PA=1:2, P is (1/3) of the way from O to A: OP=(1/3)a. Similarly OQ=(1/3)b
- Step 2: Find PQ=OQ-OP=(1/3)b-(1/3)a=(1/3)(b-a)
- Step 3: Find AB=b-a
- Answer: Since PQ=(1/3)AB, PQ is a scalar multiple of AB, so PQ is parallel to AB (and exactly (1/3) of its length)
Common Mistakes
MistakeConcluding two lines are parallel just because their vector expressions "look similar", without showing one is an exact scalar multiple of the other
Fixa parallel proof must end with an explicit statement such as "PQ = (1/3)AB, so PQ is parallel to AB" - the scalar multiple is the proof
MistakeMisreading a ratio such as OP:PA = 1:2 as meaning P is halfway along OA
FixOP:PA = 1:2 splits OA into 3 equal parts in total, so P is 1/3 of the way from O to A, not 1/2
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