01 · End minus start
A(2,−3), B(−4,5). Find AB and BA.
Hint
Subtract coordinates in the requested order.
Worked solution
AB = −6i + 8j
BA = 6i − 8j
Understand · explore · practise
Position vectors
Use end minus start to find displacements and distances. Recover endpoints, find possible parallelogram vertices and solve position constraints involving lines and circles.
Before you startVector arithmetic, coordinates and quadratic equations
01 / Positions need an origin
A position vector points from the chosen origin to a point.
If A has coordinates (−2,1) relative to O, its position vector is OA = −2i + j. The point A is a location; its position vector describes how to reach it from O.
Move the origin while keeping the axes parallel and the same scale, and the position vector changes. The displacement between two fixed points does not. The model keeps A and B fixed while you choose an origin.
Move the origin; keep the displacementMove at your pace
The original reference coordinates of the chosen origin are (0,0). The fixed points now have positions OA = (-2,1) and OB = (4,5). Their difference AB is still (6,4). Axes remain parallel and use equal scales.
Watch two position vectors reveal the displacement between their endpoints
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / End minus start
The route through the origin gives the difference.
To travel from A to B via O, first reverse OA and then follow OB:
AB = AO + OB
= OB − OA
For A(−2,1) and B(4,5):
AB = − =
The reverse displacement BA is −6i − 4j. Changing the origin subtracts the same origin-shift vector c from both position vectors: (b − c) − (a − c) = b − a.
03 / Distance between points
Take the magnitude after subtracting the positions.
For the same A and B, the straight-line distance is the magnitude of AB:
|AB| = √(6² + 4²) = √52 = 2√13
This is different from subtracting their distances from the origin. |OB| − |OA| would compare two separate lengths and would depend on where O is placed.
Distances are symmetric: |AB| = |BA|. Displacements are directed: AB = −BA. Keep those two statements distinct when interpreting a question.
04 / Recover an endpoint
Add the displacement to its starting position.
If OA = −2i + j and AB = 6i + 4j, then:
OB = OA + AB
= 4i + 5j
If the supplied vector were BA = 6i + 4j instead, use b − a = −BA, giving OB = −8i − 3j. Read the order of the endpoint labels before choosing a sign.
Recover the starting point
If OB = 7i − 2j and AB = 3i + 5j, then OA = OB − AB = 4i − 7j. Adding AB back to this result returns OB, which is a useful check.
05 / A missing parallelogram vertex
Without a vertex order, three answers can fit.
Three distinct non-collinear points A, B and C do not say which one lies opposite the missing fourth vertex. With position vectors a,b,c, the three possibilities are:
d = b + c − a
d = a + c − b
d = a + b − c
These come from the three choices of diagonal pairing: opposite-vertex position vectors have the same sum because their diagonal midpoints agree.
For A(−1,0), B(3,2), C(0,5), the possible fourth points are (4,7), (−4,3), (2,−3). If the question specifies the consecutive order A,B,C,D, only d = a + c − b applies, giving D(−4,3).
06 / Positions on lines and circles
Translate each geometric condition into an equation.
A point P lies on x + 2y = 5 and is √10 from O. Its position vector is xi + yj, and its magnitude condition is x² + y² = 10.
x = 5 − 2y
(5 − 2y)² + y² = 10
y² − 4y + 3 = 0
y = 1 or 3
So OP = 3i + j or −i + 3j. Both points satisfy the line and the magnitude condition.
How many positions are possible?
The point on x + 2y = 5 nearest O is (1,2), distance √5. You can see this by writing x² + y² = 5(y − 2)² + 5 after substituting x = 5 − 2y. A required distance less than √5 gives no point; exactly √5 gives the single tangent point; a larger distance gives two points.
This connects position-vector questions to circle and line intersections. Keep every allowed root, then check the original conditions.
07 / Your turn
Start by labelling what each vector begins and ends at.
Give exact magnitudes where asked.
02 · A distance
Find the distance between the points in question 1.
Hint
Take the magnitude of either displacement.
Worked solution
√(36 + 64) = 10
03 · Find B
OA = 3i − 2j and AB = −5i + 7j. Find OB.
Hint
Follow OA then AB.
Worked solution
OB = −2i + 5j
04 · Reversed data
OB = 2i + 6j and BA = −4i + j. Find OA.
Hint
Follow OB then BA.
Worked solution
OA = −2i + 7j
05 · Translate the origin
Relative to O, A = (2,1), B = (5,−3). Move the origin to C = (1,2), keeping parallel axes. Find the new positions and show AB is unchanged.
Hint
Subtract C from both positions.
Worked solution
New A = (1,−1), new B = (4,−5)
AB = (4 − 1, −5 − (−1)) = (3,−4)
This agrees with the original coordinate difference (5 − 2, −3 − 1).
06 · Three fourth vertices
Find all possible fourth vertices of a parallelogram with three vertices (0,0), (2,0), (0,3), with no order specified.
Hint
Use each of the three diagonal pairings.
Worked solution
(2,3), (−2,3), (2,−3).
07 · Two constrained positions
P lies on x − y = 1 and is √5 from O. Find all possible OP.
Hint
Use x = y + 1 and x² + y² = 5.
Worked solution
y² + y − 2 = 0 ⇒ y = 1 or −2
OP = 2i + j or −i − 2j
08 · A claim about distances
A = (3,0), B = (0,4). A student says AB has length 4 − 3 = 1. Correct the method.
Hint
Find the vector from A to B before its magnitude.
Worked solution
AB = −3i + 4j
|AB| = 5
Subtracting distances from O does not give the distance between A and B.
08 / Recap
A displacement is a difference of positions.
- Position vectors depend on the chosen origin.
- AB = b − a; BA = a − b.
- Distance is the magnitude of the displacement.
- Recover an endpoint by following a connected vector route.
- Three unordered parallelogram vertices can give three fourth points.
- Turn line and magnitude conditions into simultaneous equations, then check every root.
Section 1 of 8 · Positions need an origin