Two vectors are parallel when one is a multiple of the other, a = kb. Three points are collinear when two vectors between them are parallel and share a point.
This lesson belongs to SPM Additional Mathematics vectors. It assumes you can already write a vector in component form, as covered in expressing vectors in component and unit-vector form.
What is the test for parallel vectors?
Write one vector as k times the other and match the components. Both components must give the same k.
Here AB means the vector from A to B. In your exam it is written with an arrow.
Take a = (m + 1)i + 6j and b = 4i + 8j, and find m so that a is parallel to b.
- Write a = kb, so (m + 1)i + 6j = k(4i + 8j).
- Match the j parts: 6 = 8k, so k = 3/4.
- Match the i parts: m + 1 = 4k = 3, so m = 2.
Check: a = 3i + 6j, and (3/4)(4i + 8j) = 3i + 6j. The two sides agree.
How is collinear different?
Collinear is a statement about points, so the working needs a point that both vectors share. Parallel is only about direction.
For points A, B and C, the usual test is AB = k BC. B is in both vectors, so A, B and C sit on one line.
The contrast matters in diagrams. If AB = 2 CD, then AB is parallel to CD, and nothing more can be said until a point is shared.
Worked example: finding a missing coordinate
The points are A(1, 2), B(3, 7) and C(5, h). Find h if A, B and C are collinear.
First find the two vectors, each as end point minus start point.
- AB = (3 − 1, 7 − 2) = (2, 5)
- BC = (5 − 3, h − 7) = (2, h − 7)
Write AB = k BC. The first components give 2 = 2k, so k = 1. The second components give 5 = k(h − 7) = h − 7, so h = 12.
Check by a different route. The vector from A to C(5, 12) is (4, 10), and 2 × (2, 5) = (4, 10). The points lie on one line, and B is the shared point.
The mistake that loses the mark
A common slip is to stop at “AB = 2 BC, so the vectors are parallel”. The question asked for collinear points.
| Step | Incomplete | Complete |
|---|---|---|
| Relation | AB = 2BC | AB = 2BC |
| Reason | “so parallel” | “AB is parallel to BC and B is a common point” |
| Conclusion | none | “therefore A, B and C are collinear” |
Another slip is to match only one component, then assume the other works. Always test both components, because the unknown may sit in the second.
Check yourself
The points are P(−1, 3), Q(2, 4) and R(8, t). Find t so that P, Q and R are collinear.
Answer
PQ = (3, 1) and QR = (6, t − 4).
Write QR = k PQ. The first components give 6 = 3k, so k = 2. The second components give t − 4 = 2 × 1 = 2, so t = 6.
Check: PR = (9, 3) = 3 × (3, 1), so the point R(8, 6) lies on the line through P and Q. The shared point is Q, so P, Q and R are collinear.
What to study next
The next lesson, solving geometric vector ratios, uses the same idea to find where a point divides a line. For harder diagram questions, vector dependence and ratio reasoning goes further, and the vectors practice set tests the whole chapter.
If the working is right but you lose marks on the written conclusion, a teacher in online one-to-one Additional Mathematics tuition can read each line with you. The mistake log and paper-error review tool helps you track which step you skip.