A point tells you where something is. A displacement vector tells you how to move from one point to another. Confusing those two ideas can reverse a route or make a correct-looking vector calculation describe the wrong journey.
Begin each question by identifying the start point, end point and direction of the required vector. Then connect the geometry to components or the named vectors in the diagram.
This guide uses original O Level examples. Draw a small labelled sketch alongside the calculations and match your practice to the current syllabus. For wider planning, see Abeel’s O Level Mathematics guide.
Distinguish a point from a vector
A point A with coordinates (2,−1) is a location relative to the chosen coordinate system. A vector with components (2,−1) describes a movement of two units in the positive horizontal direction and one unit in the negative vertical direction.
The same numbers can therefore appear in related but different descriptions. The position vector OA connects the origin O to the point A.
A displacement AB connects A to B and need not start at the origin. Its components depend on the change from A’s coordinates to B’s coordinates.
Write the arrow direction or the ordered start and end labels clearly. AB and BA use the same two points but describe opposite movements.
Calculate displacement as end minus start
Let A = (2,−1) and B = (7,3). The vector AB has components (7−2, 3−(−1)) = (5,4).
The movement is five units horizontally in the positive direction and four units vertically in the positive direction. Starting from A and adding that movement reaches B.
The reverse vector BA is (2−7, −1−3) = (−5,−4). It is the negative of AB.
A reliable phrase is “end minus start,” applied to each coordinate. Do not change the subtraction order for the second component merely because a negative number appears.
Add vectors as linked movements
If you travel from A to B and then from B to C, the combined displacement is AC. In vector notation, AB + BC = AC.
Using A = (2,−1), B = (7,3) and C = (4,8), AB = (5,4) and BC = (−3,5). Their sum is (2,9).
Direct calculation confirms AC = (4−2, 8−(−1)) = (2,9). The route and component methods agree.
The end of the first route connects to the start of the next. Naming a chain of movements helps prevent adding a vector that points in the opposite direction from the path you intended.
Use subtraction to represent a reversed route
Subtracting a vector is equivalent to adding its negative. If a vector represents movement from A to B, its negative represents movement from B to A.
Suppose position vectors OA = a and OB = b. The route from A to B can be A to O, followed by O to B. Its vector is −a + b = b − a.
Writing a − b instead describes BA. This is one of the most common position-vector reversals.
When unsure, describe the route in words and write the vector for each leg. The signs then follow from direction rather than from a memorised ordering of letters.
Multiply by a scalar component by component
A scalar changes a vector’s length and may reverse its direction. If a = (3,−2), then 2a = (6,−4).
The vector −a = (−3,2) has the same magnitude as a but the opposite direction. Multiplying by −2 both reverses direction and doubles magnitude.
For b = (−1,4), the combination 2a + b is (6,−4) + (−1,4) = (5,0). Its vertical movement is zero because the two vertical components cancel.
Apply the multiplier to every component. Multiplying only the first component changes the vector into a different movement.
Keep magnitude distinct from components
The magnitude of a two-dimensional vector is its length. For AB = (5,4), the magnitude is √(5² + 4²) = √41.
The result is not nine, because adding horizontal and vertical components does not give the straight-line displacement length. Pythagoras connects the perpendicular components.
A vector with components (−5,−4) has the same magnitude √41. Negative components indicate direction; they do not make the length negative.
If the question asks for a vector, give its components or an appropriate vector expression. If it asks for a distance or magnitude, give the scalar length with any relevant units.
Distance travelled can exceed displacement magnitude
Imagine travelling three metres east and then four metres north along perpendicular straight segments. The total distance travelled is seven metres.
The displacement from the starting point to the ending point has components (3,4) in the stated east-north coordinate directions. Its magnitude is five metres.
Both answers describe the same journey but measure different things. Distance follows the route, while displacement connects the start directly to the end with a direction.
Do not replace one with the other because the numbers look more familiar. Read whether the question asks for path length, position change or the vector itself.
Find a midpoint using positions
For points A and B with position vectors a and b, the midpoint M has position vector (a + b)/2.
Using A = (2,−1) and B = (7,3), the midpoint coordinates are ((2+7)/2, (−1+3)/2) = (4.5,1).
The vector AM is half of AB, so it is (2.5,2). Adding it to A gives (4.5,1), which confirms the midpoint position.
Notice the difference between OM and AM. The position vector OM locates the midpoint from the origin; AM describes a displacement starting at A. They are related but not interchangeable.
Divide a segment using the required fraction
Suppose A = (1,2), B = (11,7), and P lies on AB with AP:PB = 2:3. The whole segment consists of five equal parts, so AP is two fifths of AB.
Since AB = (10,5), AP = (4,2). Adding that displacement to A gives P = (5,4).
Using three fifths instead would locate the point corresponding to the opposite portion of the segment. Read which part begins at the stated starting point.
For a point between A and B, a position expression is OP = a + (2/5)(b−a). This formula combines the starting position with the required fraction of the displacement.
Compare non-zero vectors for parallel directions
If one non-zero vector is a scalar multiple of another, their directions are parallel. A negative multiple points in the opposite sense along a parallel direction.
For example, (6,−4) is twice (3,−2), while (−3,2) is its negative. Both comparisons establish the relationship between the vector directions.
Check both components. Two vectors with matching first components or similar-looking numbers are not necessarily scalar multiples.
The zero vector has no unique direction. Be careful when using a direction argument in a case where a vector might be zero, and follow any non-degeneracy conditions in the problem.
Prove collinearity using a common starting point
To show that A, P and B lie on one straight line, one useful approach is to show AP = kAB for an appropriate scalar k, with the relevant non-zero displacement.
In the ratio example, AP = (2/5)AB. Because the vectors start at A and have proportional directions, P lies on the line through A and B. The factor between zero and one places P between them.
Comparing unrelated position vectors without tracking their starting point can prove a different relationship. For example, OA being a multiple of OB concerns O, A and B.
State the points involved in the conclusion. “The vectors are parallel” alone may not fully explain the requested collinearity claim.
Build a short midpoint proof
Let O, A and B form a non-degenerate triangle, with OA = a and OB = b. Let P be the midpoint of OA and Q the midpoint of OB.
Their position vectors are OP = a/2 and OQ = b/2. Therefore PQ = OQ − OP = (b−a)/2.
But AB = b−a, so PQ = AB/2. Thus the midpoint segment has a direction parallel to AB and half its magnitude.
The proof depends on keeping position and displacement clear. Subtracting in the opposite order would give QP instead of PQ, changing the sign but not the underlying parallel-direction relationship.
Check a vector answer through its route
After calculating AB, add it to A’s position and see whether you reach B. For a multi-stage route, compare the combined movement with direct end-minus-start calculation.
Check scalar factors against the diagram’s stated ratios. A midpoint displacement should be half the whole displacement, not twice it.
For a magnitude, confirm that the result is non-negative and consistent with the components. Do not attach a direction label to a scalar length unless you separately describe the vector.
Use Abeel’s Cambridge past-paper guide and O Level revision plan to practise these checks after independent attempts.
Explore Abeel Eschool and the course catalogue for current subject options. Ask through admissions and FAQs, sharing your labelled route and the expression you attempted.
Frequently asked questions
Is AB the same as BA?
No. They have opposite directions, so BA = −AB. Their magnitudes are equal.
How do I calculate a displacement from coordinates?
Subtract the starting coordinates from the ending coordinates, component by component.
Is magnitude the sum of the components?
Not generally. In two perpendicular coordinate directions, use the square root of the sum of the squared components.
How do position vectors give AB?
If OA = a and OB = b, then AB = b−a. The route goes from A back to O and then from O to B.
How can I show three points are collinear?
One method is to compare non-zero displacement vectors with a common starting point and show that one is a scalar multiple of the other. Explain which points the relation concerns.
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