One multiplier for both components
Two vectors are collinear when one is a scalar multiple of the other: b = ka for a single number k. Drawn from the same point, the two arrows then lie along one straight line through that point. Collinear vectors are also called parallel vectors.
Take a = (2, 1) and the vector (6, 3). Divide each component of the second by the matching component of the first: 6 ÷ 2 = 3 and 3 ÷ 1 = 3. The same number, 3, works for both components, so (6, 3) = 3a, and the two vectors are collinear.
a = (2, 1) and 3a = (6, 3), both drawn from the origin, lie along the one line .
When no single multiplier fits
Now compare b = (3, 4) with a = (2, 1). The first components give 3 ÷ 2 = 1.5, and the second components give 4 ÷ 1 = 4.
To be a multiple of a, b would need one number k with 3 = k × 2 and 4 = k × 1 at the same time. The first equation needs k = 1.5 and the second needs k = 4. No single multiplier fits both components, so a and b are not collinear: they point in different directions.
The line through a is . The head of b = (3, 4) is far off that line, because b is not a multiple of a.
The test without dividing
For two vectors (p, q) and (r, s), the test can be written as one equation: they are collinear exactly when ps = qr. If (r, s) = k(p, q), then r = kp and s = kq, so ps and qr are both kpq. And if ps = qr with p not 0, then fits both components.
For a = (2, 1) and 3a = (6, 3): ps = 2 × 3 = 6 and qr = 1 × 6 = 6, which are equal. For a = (2, 1) and b = (3, 4): ps = 2 × 4 = 8 and qr = 1 × 3 = 3, which are not.
This form also settles a component of 0, where dividing has no value. For (0, 3) and (0, −5), the first division would be 0 ÷ 0. Cross-multiplying gives 0 × (−5) = 0 and 3 × 0 = 0, which are equal, and indeed : both vectors point straight along the y-axis.
A negative multiplier
The multiplier can be negative. For (2, 1) and (−4, −2), the divisions give −4 ÷ 2 = −2 and −2 ÷ 1 = −2, the same number for both, so (−4, −2) = −2a. The two vectors are collinear and point in opposite directions along the same line.
The signs never decide the question on their own. Only whether one multiplier fits every component does.
λ = −1.5 < 0: r = a + λd walks the other way along d, behind A, and it is still the same line
Slide P to λ = 2 and read r = a + 2d
From the point A(1, 1), the gold arrow AP is , a multiple of d = (2, 1). At it is (−3, −1.5), pointing back, and P is at (1 − 3, 1 − 1.5) = (−2, −0.5). Drag P: AP is always a multiple of d, so P never leaves the dashed line through A.
Collinear points
Three points A, B and C lie on one straight line when the vectors AB and BC are collinear and the two share the point B. Take A(0, 1), B(2, 2) and C(6, 4). Then AB = (2 − 0, 2 − 1) = (2, 1) and BC = (6 − 2, 4 − 2) = (4, 2). Both components give 4 ÷ 2 = 2 and 2 ÷ 1 = 2, so BC = 2AB.
A multiple shows only that the two vectors have the same direction. The shared point B is what puts them on the same line: a line through B in that direction is only one line. The multiplier also gives the ratio of the lengths, so AB : BC = 1 : 2.
Without a shared point, a multiple shows parallel lines and nothing more. With D(0, 3) and E(2, 4), DE = (2, 1) equals AB exactly, but D is not on the line through A and B: the vector AD = (0, 2) is not a multiple of (2, 1). AB and DE are opposite sides of a parallelogram, parallel and apart.
AB = (2, 1) and BC = (4, 2) share B, so A, B and C lie on the line . DE = (2, 1) is parallel to AB but starts above that line.
The usual mistakes
Checking one component only. For (2, 3) and (7, 9), the second components give 9 = 3 × 3, but 3 × 2 = 6, not 7. One component that scales is not enough; every component must scale by the same number.
Adding the same number instead of multiplying. (5, 4) is (2, 1) with 3 added to each component, but 5 ÷ 2 = 2.5 and 4 ÷ 1 = 4, so it is not a multiple of (2, 1) and not collinear with it.
Rejecting a negative multiplier. (−4, −2) = −2 × (2, 1) is collinear with (2, 1); it points the other way along the same line.
Stopping at parallel when the question is about points. A multiple gives the same direction; a shared point is needed to put the points on one line.
Three posts on one fence line
In the application below, three survey posts are given by their coordinates. Find the vector from the first post to the second and from the second to the third, show that one is a multiple of the other, and use the post they share to put all three on one line. The multiplier then gives the ratio of the two lengths of fence.
Worked example: Three Survey Posts on One Straight Fence Line
Question A surveyor marks three posts on flat ground. Measured in meters east and north of a corner post O, they are at P(1, 1), Q(5, 4) and R(13, 10). (a) Show that P, Q and R lie on one straight line. (b) Find the ratio PQ : QR and the length of fence from P to R.
1.PQ = 5 − 14 − 1 = 43 and QR = 13 − 510 − 4 = 86.
PQ = 5 − 14 − 1 = 43 and QR = 13 − 510 − 4 = 86. 2.86 = 243, so QR = 2PQ. One vector is a scalar multiple of the other, so the two are parallel.
86 = 243, so QR = 2PQ: one is a scalar multiple of the other, so they are parallel. 3.(a) PQ and QR are parallel and share the point Q. Two parallel lines through one point are the same line, so P, Q and R lie on one straight line.
(a) The two vectors are parallel and share the point Q, so P, Q and R lie on one straight line. 4.PQ = √42 + 32 = √25 = 5 m, and QR is twice as long, so QR = 10 m.
PQ = √42 + 32 = 5 m, and QR is twice as long, 10 m. 5.(b) PQ : QR = 5 : 10 = 1 : 2, and the fence from P to R is 5 + 10 = 15 m. Check: PR = 129 and √122 + 92 = √225 = 15 m.
(b) PQ : QR = 1 : 2, and the fence from P to R is 15 m.
Answer: (a) QR = 2PQ and Q is shared, so the posts are in one line; (b) 1 : 2, and 15 m
Common mistakes
- Stopping at parallel. Two parallel vectors can lie on two different lines, like the opposite sides of a rectangle; it is the shared point Q that makes them one line.
- Writing the ratio as 2 : 1 because QR = 2PQ. The ratio PQ : QR puts PQ first, and PQ is the shorter part, so the ratio is 1 : 2.