Add vectors and use midpoint and parallelogram geometry
Interpret component arithmetic as directed paths, use scalar multipliers, and find midpoints relative to a common origin.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.7. Paper 1 coordinate geometry and algebra are assumed. Work in two or three dimensions as stated. The general section theorem is not included; midpoint geometry is required. The cross product, shortest distance between skew lines and their common perpendicular are not required. Scalar products include angles between lines and the foot of a perpendicular from a point to a line, including 3D solids. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Distinguish positions and finish-minus-start displacements.
- Add signed components and interpret a scalar multiplier.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is −2(3, −1)ᵀ?
What you will learn
- Add, subtract and multiply vectors by scalars with a geometric interpretation.
- Use the parallelogram rule and the midpoint position (a + b)/2.
Vector addition joins directed paths
Place the tail of a second displacement at the head of the first. Their sum is the displacement from the original start to the final endpoint. Component arithmetic records the same operation: (3, 1)ᵀ + (1, 3)ᵀ = (4, 4)ᵀ. A scalar multiplies every component. Positive scalars keep direction, negative scalars reverse it, and zero gives the zero vector; the magnitude is multiplied by the scalar's absolute value.
For parallelogram OABC, with vertices in that order and O the origin, →OB = →OA + →OC. Taking A = (3, 1) and C = (1, 3) gives B = (4, 4). The two routes O→A→B and O→C→B produce the same diagonal displacement. More generally, for parallelogram PQRS in that order, r = q + s − p; subtracting the non-origin starting position is essential.
A midpoint averages positions, not magnitudes
If A and B have positions a and b, the midpoint M has position m = (a + b)/2. Then →AM = m − a = (b − a)/2 and →MB = b − m = (b − a)/2, verifying both equal directed halves. The average applies component by component in two or three dimensions. It is not the average of the position-vector lengths, and it does not depend on either point being the origin.
In the plotted parallelogram, the midpoint of diagonal OB is (2, 2). The midpoint of the other diagonal AC is also (2, 2), so the diagonals bisect one another. The midpoint of side AB is instead (3.5, 2.5); check which pair of endpoints the question names. The syllabus requires this midpoint result, while the general section theorem for arbitrary division ratios is not included.
Pause and explain
For parallelogram OABC with O the origin, which vector reaches the opposite vertex B?
Worked example
OABC is a parallelogram with O = (0, 0), A = (3, 1) and C = (1, 3). Find B and the common midpoint of its diagonals.
Show the worked solution
- Use →OB = →OA + →OC to obtain B = (4, 4).
- The midpoint of OB is (2, 2).
- The midpoint of AC is [(3, 1) + (1, 3)]/2 = (2, 2), confirming that both diagonals share it.
Answer B = (4, 4); common diagonal midpoint (2, 2).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the midpoint of A = (2, −1, 4) and B = (−2, 5, 2), then verify the two half-displacements.
Give me a hint
Average each coordinate and use finish minus start for both halves.
Compare my reasoning
- The midpoint is M = (0, 2, 3).
- →AM = M − A = (−2, 3, −1)ᵀ.
- →MB = B − M = (−2, 3, −1)ᵀ, equal to the first half-displacement.
M = (0, 2, 3); →AM = →MB = (−2, 3, −1)ᵀ.
Look for these in your work
- Average the signed endpoint coordinates.
- Verify the two directed halves rather than their lengths alone.
PQRS is a parallelogram in that order. P = (1, 2, −1), Q = (4, 1, 1) and S = (−1, 5, 2). Find R and check the opposite-edge displacements.
Give me a hint
Use r = q + s − p because P is not the origin.
Compare my reasoning
- R = (4, 1, 1) + (−1, 5, 2) − (1, 2, −1) = (2, 4, 4).
- →PQ = (3, −1, 2)ᵀ equals →SR = R − S.
- →PS = (−2, 3, 3)ᵀ equals →QR = R − Q, verifying the vertex order.
R = (2, 4, 4); →PQ = →SR = (3, −1, 2)ᵀ; →PS = →QR = (−2, 3, 3)ᵀ.
Look for these in your work
- Subtract the non-origin starting position.
- Check both pairs of opposite directed edges.
A supplied 2D motion model adds velocity vectors u = (3, 1) m/s and v = (−1, 2) m/s. Starting at (5, −2) m, find its constant resultant velocity, displacement after 4 s and final position.
Give me a hint
Add velocities before multiplying by the elapsed time.
Compare my reasoning
- The resultant velocity is (2, 3) m/s.
- In 4 s the displacement is 4(2, 3) = (8, 12) metres.
- The final position is (5, −2) + (8, 12) = (13, 10) metres.
Velocity (2, 3) m/s; displacement (8, 12) m; final position (13, 10) m.
Look for these in your work
- Add quantities with matching vector units before scaling.
- Distinguish velocity, displacement and final position.
Common mistakes
- A parallelogram sum assumes the stated vertex order; identify the opposite vertex correctly.
- Average the endpoint positions rather than the vector magnitudes.
- A negative scalar reverses direction; it does not give a negative magnitude.
What can you explain now?
What is the midpoint position between positions a and b?
Compare with the explanation
(a + b)/2.
Its displacement from a and the displacement from it to b both equal (b − a)/2.
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Make it stick
Explain vector addition as a path, find a non-origin parallelogram vertex, and verify a midpoint by the two equal directed half-displacements.
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