For a point P on the line AB but outside the segment, the ratio AP:PB still compares lengths. The vectors AP and PB point in opposite directions, so one gets a minus sign.
This lesson is the last in vector dependence and ratio reasoning. It extends the inside-the-segment method from solving geometric vector ratios.
How is an external point different?
Take AP:PB = 3:1 in two ways. If P is between A and B, then AP and PB point the same way. If P is beyond B, then AP points towards B and continues past it, while PB points back.
So for the external case, AP = −3PB, because the lengths are in ratio 3:1 and the directions are opposite.
Worked example: P beyond B
OA = a and OB = b. P lies on AB produced beyond B, such that AP:PB = 3:1. Find OP.
Let OP = p. Use AP = −3PB in terms of position vectors.
- AP = p − a and PB = b − p.
- So p − a = −3(b − p) = −3b + 3p.
- Collect: −a + 3b = 2p.
- So OP = −(1/2)a + (3/2)b.
Sign check: the coefficients add up to −1/2 + 3/2 = 1, so P is on the line AB. The coefficient of a is negative, which says P is beyond B, on the side away from A. Both checks agree with the wording.
The same ratio inside the segment
If P were between A and B with AP:PB = 3:1, then AP = (3/4)AB and OP = a + (3/4)(b − a) = (1/4)a + (3/4)b. Both coefficients are positive and less than 1.
Test both with real coordinates. Let a = (2, 0) and b = (6, 4).
| Case | OP | AP | PB | Lengths AP:PB |
|---|---|---|---|---|
| Internal | (1/4)(2, 0) + (3/4)(6, 4) = (5, 3) | (3, 3) | (1, 1) | 3:1 |
| External | −(1/2)(2, 0) + (3/2)(6, 4) = (8, 6) | (6, 6) | (−2, −2) | 3:1 |
In the external case, AP = −3PB, since (6, 6) = −3 × (−2, −2). The lengths are in the required ratio, and the directions are opposite.
The mistake that loses the marks
The usual slip is to use the internal formula without checking the wording. The answer then looks tidy but puts P in the wrong place.
The repair is a two-second sign check. Sketch the line, mark P, and ask whether the answer should have a negative coefficient.
Check yourself
Q lies on AB produced beyond A, so that AQ:QB = 1:3. Express OQ in terms of a and b.
Answer
Let AQ have length t, so QB has length 3t. Since Q is beyond A, QB = QA + AB in length, so AB has length 3t − t = 2t. Then AQ is half of AB, pointing from A away from B.
OQ = OA − (1/2)AB = a − (1/2)(b − a) = (3/2)a − (1/2)b.
Check: QA has length (1/2)AB and QB has length (3/2)AB, so the ratio is 1:3. The negative coefficient is on b, which says Q is on the far side from B, beyond A.
What to study next
Try the integrated practice set, which mixes internal, external, parallel and intersection questions. The timed original practice session builder can set up a session.
For a teacher to check your sketches and signs, see online one-to-one Additional Mathematics tuition. The mistake log and paper-error review tool helps you find repeat slips.