Parallel and collinear vector arguments
Use scalar multiples plus a shared point to justify parallelism or collinearity.
Extended only.
Before you begin
- Simplify expressions in two coplanar vectors.
- Find segment vectors from position vectors.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What must be true for two nonzero vectors to be parallel?
What you will learn
- Recognise parallel nonzero vectors as scalar multiples.
- Prove collinearity using directed vectors from a shared point.
A scalar multiple gives a parallel direction
Nonzero vectors u and v are parallel when u = kv for a nonzero scalar k. For (6,−4) and (3,−2), the factor is two. For (−6,4) it is minus two, indicating the opposite direction along parallel lines. Both components must have the same factor.
The zero vector has no direction, so do not use it as evidence that a line is parallel. Sharing equal horizontal components alone is insufficient: (2,3) and (2,5) are not scalar multiples. Explain the common factor rather than merely saying that the diagram looks parallel.
A common point makes the line claim precise
If AB = 2AC with both nonzero, A, B and C are collinear because both vectors start at A and have the same line direction. Parallel segments with different starting points do not by themselves establish that all endpoints lie on one line.
For OA = a, OB = b, and OC = 3b − 2a, AC = OC − OA = 3(b − a) = 3AB. Thus A, B and C are collinear and C lies beyond B on the ray from A. Ratios and midpoint statements can similarly be converted into expressions in a and b before comparing coefficients.
Pause and explain
AB = 3AC, with nonzero vectors. What follows?
Worked example
OA = a, OB = b, OC = 3b − 2a. Prove A, B, C are collinear.
Show the worked solution
- AB = b − a.
- AC = (3b−2a)−a = 3(b−a) = 3AB.
- The vectors share starting point A and are parallel, so the three points lie on one line.
Answer AC = 3AB; A, B and C are collinear.
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.
Are (6,−4) and (3,−2) parallel?
Give me a hint
Compare both component ratios.
Compare my reasoning
- Six is twice three.
- Minus four is twice minus two.
- The first vector is twice the second, so they are parallel.
Yes; factor 2.
Look for these in your work
- Checked both components.
- Named the scalar factor.
Are (2,3) and (4,5) parallel?
Give me a hint
A factor must work on both components.
Compare my reasoning
- The horizontal components suggest factor two.
- Twice three is six, not five.
- No common scalar exists, so the vectors are not parallel.
No.
Look for these in your work
- Tested a candidate factor.
- Rejected inconsistent components.
OA = a, OB = b and OD = (a+b)/2. Prove D lies on line AB and identify its position.
Give me a hint
Subtract OA to form AD.
Compare my reasoning
- AD = (a+b)/2 − a = (b−a)/2.
- AB = b−a, so AD = ½AB with common starting point A.
- D is collinear with A and B and lies at their midpoint.
D is the midpoint of AB.
Look for these in your work
- Used directed segments from A.
- Connected the scalar to the point's position.
Common mistakes
- Checking only one component.
- Confusing parallel with equal magnitude.
- Claiming collinearity without a shared point.
What can you explain now?
Is parallelism alone enough to place four endpoints on one line?
Compare with the explanation
No.
Different parallel lines can contain distinct endpoints; a common point is needed for the collinearity argument here.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, turn a ratio statement into a scalar-multiple argument and state the common point.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.
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