A diagram can suggest a straight line, but a proof needs a vector relationship. Parallel vectors give you direction only, and collinear points also need a shared point.
This lesson follows finding a ratio from two expressions for one vector. It builds on the basic test in using parallel vectors and collinearity.
What can a diagram tell me?
Only what the given vector relationships say. Questions use three kinds of statement, and each needs different working.
| Statement | Means | Working needed |
|---|---|---|
| AB is parallel to CD | Same direction, may be different lines | AB = k CD |
| A, B, C are collinear | All three on one line | AB = k BC, with B shared |
| AB = 2 CD | Parallel, and AB is twice as long | Stated, or shown by components |
Worked example: a trapezium
In trapezium OABC, OA = 4a, CB = 2a and OC = b, where a and b are not parallel.
Is CB parallel to OA? CB = 2a and OA = 4a, so CB = (1/2)OA. Yes, they are parallel, and CB is half the length.
Are O, A and B collinear? Find AB by going A to O to C to B.
AB = AO + OC + CB = −4a + b + 2a = b − 2a
For collinearity, AB would need to be a multiple of OA = 4a. But AB contains b, which is not a multiple of a. So O, A and B are not collinear.
Both statements used the same diagram, yet one is true and one is false. The working decides, not the look of the sketch.
A case where collinear is true
Let D be a point with OD = 6a. Then OD = (3/2)OA, so the vectors OD and OA are parallel. Both start at O, which is the shared point, so O, A and D are collinear.
That is the difference: the trapezium sides were parallel with no shared point, and here there is one.
The mistake that loses the mark
Writing “CB is parallel to OA, so the points are collinear” is the usual slip. Parallel gives direction only.
The fix is a three-part sentence: the relation, the shared point, the conclusion. For the trapezium, the correct statement is “CB = (1/2)OA, so CB is parallel to OA”, and nothing more.
Check yourself
PQ = 3u, QR = 6u and ST = 6u. Which of these can you conclude: (a) P, Q and R are collinear; (b) PQ is parallel to ST; (c) P, Q and S are collinear?
Answer
(a) Yes. QR = 2PQ, so the vectors are parallel, and Q is a common point. So P, Q and R are collinear.
(b) Yes. ST = 2PQ, so the two vectors are parallel.
(c) No. The information gives no link between S and the line PQ. ST is parallel to PQ, but S and T could sit on a different line. You need a shared point before you can claim collinear.
In each part, give the reason and name the shared point.
What to study next
Continue with finding an internal intersection using two vector paths, which needs two routes to one point. The integrated practice set then mixes all four lessons.
A teacher in online one-to-one Additional Mathematics tuition can go through your written conclusions with you. The mistake log and paper-error review tool helps you spot whether the lost marks are in the working or the wording.