To show that A, B and C are collinear, show that →AC is a scalar multiple of →AB. Because both displacements start at A, they lie on one line, and so A, B and C do too.
This lesson uses the dividing-point skills from find a dividing point on a segment.
What makes a collinearity proof complete?
A full proof has three parts:
- Write two displacements from a common point, for example →AB and →AC.
- Show one is a scalar multiple of the other, such as →AC = 3→AB, by matching both the a and b coefficients.
- Write the conclusion: they are parallel and share the point A, so A, B and C are collinear.
Worked example
→OA = a + 2b, →OB = 3a + b and →OC = 7a − b. Show that A, B and C are collinear, and find AB : BC.
→AB: →OB − →OA = (3a + b) − (a + 2b) = 2a − b.
→AC: →OC − →OA = (7a − b) − (a + 2b) = 6a − 3b.
Compare: 6a − 3b = 3(2a − b), so →AC = 3→AB.
Conclusion: →AC and →AB are parallel and share the point A, so A, B and C are collinear.
Ratio: →BC = →AC − →AB = 3→AB − →AB = 2→AB. Therefore AB : BC = 1 : 2.
The mistake to watch for
A common slip is to check only one coefficient.
Mistaken working: →AB = 2a − b and →AC = 6a − 2b. “6 ÷ 2 = 3, so →AC = 3→AB.”
The student compared the a terms and ignored the b terms.
Multiplying through: 3(2a − b) = 6a − 3b, not 6a − 2b. The b coefficients are not in the same ratio, so these vectors are not parallel. Always multiply the whole vector and compare both coefficients.
Check yourself
1. →AB = 2a + 3b and →BC = 6a + 9b. Are A, B and C collinear?
Show answer
→BC = 3(2a + 3b) = 3→AB. They are parallel and share B.
Yes, collinear.
2. →PQ = 3a − 2b and →QR = 6a + kb. Find k so that P, Q, R are collinear.
Show answer
The a coefficient doubles (3 to 6), so →QR = 2→PQ = 6a − 4b. Check: 2 × (−2) = −4.
k = −4
3. →OA = a, →OB = b and →OC = 3b − 2a. Show A, B, C are collinear.
Show answer
→AB = b − a. →AC = →OC − →OA = 3b − 2a − a = 3b − 3a = 3(b − a).
→AC = 3→AB with A shared, so A, B, C are collinear.
Where this leads next
Collinearity gives you a condition for a point being on a line. The next step is finding where two such lines meet in find an intersection of vector-defined lines. The module overview shows how the pieces fit.
If you understand each step but find conclusions hard to word, see how we work in online one-to-one Additional Mathematics tuition.