A position vector locates a point from a chosen origin. A directed segment describes the movement from one point to another. Keeping those two meanings distinct makes ratio problems and collinearity proofs much easier: first find where each point is, then subtract in the direction that the proof needs.
Here O is the origin and bold letters represent vectors. Write OA = a to mean the vector from O to A, and AB to mean the vector from A to B. The same two-letter notation always keeps its direction. Our task is geometric internal division and proof; force resultants and motion are separate applications.
Subtraction means endpoint minus starting point
If OA = a and OB = b, travelling O to A and then A to B gives the same total displacement as O to B. Therefore a + AB = b, so AB = b − a. Reversing the trip gives BA = a − b = −AB.
For coordinates, subtract matching components. If A has coordinates (u, v) and B has coordinates (s, t), then AB = (s − u, t − v). A vector such as (4, −2) describes a directed change; the point with coordinates (4, −2) has that position vector only when measured from the chosen origin.
Internal division comes from a fraction of the whole segment
Suppose P lies on AB with AP:PB = m:n, where m and n are positive. The whole segment contains m + n ratio parts, and AP contains m of them. Thus AP = [m/(m + n)]AB. Add OA to reach P from O:
OP = a + [m/(m + n)](b − a) = [na + mb]/(m + n).
The coefficients on a and b look reversed compared with AP:PB. That is a consequence of starting at A and moving a fraction towards B. If m is small, P is near A, so the coefficient of a should be larger. This endpoint-near check is often more memorable than a formula alone.
Worked case 1: divide a triangle side in the ratio 2:5
Let O, A and B be non-collinear, with OA = a and OB = b. Point P divides AB internally so that AP:PB = 2:5. Find AP, OP and BP in terms of a and b.
Since AB has seven equal ratio parts, AP = (2/7)(b − a). Then OP = a + (2/7)(b − a) = (5/7)a + (2/7)b. Point P is closer to A, consistent with its larger coefficient of a.
For BP, start at B and finish at P: BP = OP − OB = (5/7)a − (5/7)b = (5/7)(a − b). It points opposite to AB. By contrast, PB = (5/7)(b − a). The lengths AP and PB have ratio 2:5, but their reversed directed versions need their signs preserved.
Diagram data: use O = (0, 0), A = (0, 7) and B = (14, 0) for an illustrative coordinate model. Then P = (4, 5). The arrows OA and OB locate the vertices; the arrow along AB points from A towards B. AP = (4, −2) and PB = (10, −5), which are in the stated 2:5 ratio. The coordinate model illustrates the algebra without restricting it to this particular triangle.
Worked case 2: prove that three constructed points are collinear
Keep the same triangle and point P. Choose C on OA with OC = (5/7)a, and D on OB with OD = (2/7)b. Let M be the midpoint of CD. Prove that O, M and P lie on one straight line, and locate M relative to OP.
The midpoint has position vector OM = (OC + OD)/2. Therefore OM = (5/14)a + (1/7)b = (1/2)[(5/7)a + (2/7)b] = (1/2)OP.
OM and OP share the starting point O, and one is a scalar multiple of the other. Since OP is nonzero in this non-degenerate triangle, both vectors lie on the same line through O. The scalar 1/2 is between zero and one, so M lies between O and P, halfway along OP. Hence OM:MP = 1:1.
For the coordinate model O = (0, 0), A = (0, 7), B = (14, 0), the construction gives C = (0, 5), D = (4, 0), M = (2, 2.5) and P = (4, 5). The midpoint of C and D is indeed (2, 2.5). Both components of OP are twice those of OM. This independent coordinate check also confirms the point order O, M, P.
Parallel vectors need a common point for a collinearity proof
A scalar multiple establishes parallel directions. To establish that three points lie on one line, use vectors with a common endpoint, such as PQ and PR, or show explicitly that the lines also share a point. Merely finding AB = 2CD does not put A, B, C and D on one line; AB and CD could lie on separate parallel lines.
A complete proof should name the vectors, give the scalar relation and state the common point. For example, “PR = (5/3)PQ, and both start at P, therefore P, Q and R are collinear.” The scalar can also locate the points: a factor above one places R beyond Q along the ray from P, while a negative factor places it on the opposite ray.
Worked case 3: repair reversed subtraction in coordinates
Let A = (−3, 2) and B = (11, 9). Point P divides AB in the ratio AP:PB = 3:4. A learner writes AB = (−14, −7), then calculates P = (−9, −1). Explain the error and find the correct point.
The learner subtracted B from A, obtaining BA. The correct directed segment is AB = (11 − (−3), 9 − 2) = (14, 7). Since AP is three sevenths of AB, AP = (6, 3). Add this change to A: P = (−3, 2) + (6, 3) = (3, 5).
Check PB = (11 − 3, 9 − 5) = (8, 4). Both components of AP:PB give 6:8 = 3:4 and 3:4 respectively. The correct point lies between the endpoints in both coordinates. The incorrect point (−9, −1) lies beyond A in the wrong direction, so the internal-division condition already rules it out.
Three practice tasks with full solutions
Practice 1: find a point near the first endpoint
A = (2, −1), B = (10, 7), and AP:PB = 1:3. Find P and check its location. AB = (8, 8), so AP = (1/4)(8, 8) = (2, 2). Therefore P = (4, 1). PB = (6, 6) = 3AP. The point is one quarter of the way from A to B and is closer to A.
Practice 2: prove collinearity and find a ratio
P = (2, 3), Q = (5, 9) and R = (7, 13). Show they are collinear and find PQ:QR. We obtain PQ = (3, 6) and PR = (5, 10) = (5/3)PQ. They share P, so the three points are collinear. Also QR = (2, 4) = (2/3)PQ. Hence PQ:QR = 3:2, with Q between P and R.
Practice 3: reject an incomplete proof
A = (0, 0), B = (2, 1), C = (0, 3) and D = (4, 5). Does CD = 2AB prove all four points collinear? AB = (2, 1) and CD = (4, 2), so the scalar relation is true. However AC = (0, 3) is not a scalar multiple of AB: its first component would force the scalar to be zero, while its second component is nonzero. Thus C is not on line AB, and the proposed conclusion is false.
Frequently asked questions
Which position vector do I subtract first?
Write endpoint minus starting point. AB is OB − OA, while BA is OA − OB.
Why are the weights reversed in the section formula?
Moving m/(m + n) of the way from A leaves coefficient n/(m + n) on A and m/(m + n) on B.
What does internally mean?
The dividing point lies between the two endpoints. Positive ratio parts make its fraction of the whole segment lie between zero and one.
Do equal vector directions imply equal vectors?
No. Equal vectors need equal magnitudes as well as the same direction. A positive scalar multiple may have a different magnitude.
Can a negative scalar prove collinearity?
Yes, for nonzero vectors sharing a point. It puts the endpoints on opposite rays from that common point.
Why mention a common point?
Parallel directions alone can belong to different lines. A shared point identifies the same line for the collinearity argument.
Must I use coordinates in every proof?
No. Algebra with position vectors can prove the general result. Coordinates are useful for checking a construction or solving a numerical problem.
Choose the relevant next resource
The Mathematics topical revision collection offers further navigation and may include member-only resources. The existing Physics explanation of vectors is optional background on different applications; this lesson’s task is geometric position-vector proof.
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