Collinearity
Master collinearity for Year 12 Specialist Mathematics in Queensland (QCAA). In Unit 3 you use vectors to decide when points lie on one straight line: \(A\), \(B\) and \(C\) are collinear exactly when \(\overrightarrow{AC}\) is a scalar multiple of \(\overrightarrow{AB}\).
You will learn to test collinearity in two and three dimensions, use the cross-product test, solve for an unknown coordinate that makes points collinear, and find the ratio in which a point divides a segment — skills that underpin vector lines and geometric proofs later in the course.
Theory
Collinearity tests whether points lie on one straight line, a core skill in Unit 3 of Year 12 Specialist Mathematics (QCAA, Queensland). Points \(A\), \(B\) and \(C\) are collinear exactly when \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\) — the vectors are parallel and share the point \(A\). This page shows the test in 2D and 3D, how to find an unknown coordinate, and the ratio in which a point divides a segment.
Three or more points are collinear when they all lie on one straight line. Vectors turn this into a quick algebraic test.
First form the vector between two points as end minus start: for points with position vectors \(\mathbf{a}\) and \(\mathbf{b}\), \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\). This works the same way in two and three dimensions.
The collinearity test: \(A\), \(B\) and \(C\) are collinear if and only if \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\) for some scalar \(\lambda\). The two vectors are then parallel, and because they both start at \(A\) they share a common point, so all three points sit on the one line.
In three dimensions an equivalent statement uses the cross product: the points are collinear exactly when \(\overrightarrow{AB}\times\overrightarrow{AC}=\mathbf{0}\), since parallel vectors have zero cross product.
When the points are collinear, the scalar \(\lambda\) also fixes the ratio in which \(B\) divides \(AC\): comparing \(\overrightarrow{AB}\) with \(\overrightarrow{BC}\) gives \(AB:BC\).
Vector between two points (end minus start):
Collinearity test — \(A\), \(B\), \(C\) are collinear iff:
Equivalent 3D test using the cross product:
Dividing ratio — when \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\):
How to test three points for collinearity
- Form the vectors: compute \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) and \(\overrightarrow{AC}=\mathbf{c}-\mathbf{a}\), each as end minus start.
- Test for a scalar multiple: ask whether \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\). In 2D check the components give one common \(\lambda\) (or the cross \((AB_x)(AC_y)-(AB_y)(AC_x)=0\)); in 3D check \(\overrightarrow{AB}\times\overrightarrow{AC}=\mathbf{0}\).
- State the conclusion: if one scalar \(\lambda\) works, the points are collinear (they share \(A\)); if the components need different scalars, they are not.
- Find any ratio or unknown: solve the matching equation for an unknown coordinate, or compare \(\overrightarrow{AB}\) with \(\overrightarrow{BC}\) to read off the ratio \(AB:BC\).
Form both vectors as end minus start, then check for a common scalar:
| \(\overrightarrow{AB}\) | \(=\) | \(\mathbf{b}-\mathbf{a}=(4-2,\ 3-0)\) |
| \(=\) | \((2,3)\) | |
| \(\overrightarrow{AC}\) | \(=\) | \(\mathbf{c}-\mathbf{a}=(8-2,\ 9-0)\) |
| \(=\) | \((6,9)\) | |
| \(\overrightarrow{AC}\) | \(=\) | \(3(2,3)\) |
| \(=\) | \(3\,\overrightarrow{AB}\) |
\(\overrightarrow{AC}=3\,\overrightarrow{AB}\), so \(A\), \(B\), \(C\) are collinear.
Form \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), fix \(\lambda\) from a known component, then match the rest:
| \(\overrightarrow{AB}\) | \(=\) | \((2,3,3)\) |
| \(\overrightarrow{AC}\) | \(=\) | \((8,\ k,\ 12)\) |
| \(8\) | \(=\) | \(2\lambda \Rightarrow \lambda=4\) |
| \(12\) | \(=\) | \(4\times 3\ \checkmark\) |
| \(k\) | \(=\) | \(4\times 3\) |
| \(=\) | \(12\) |
\(k=12\).
Form the vectors and test the cross \((AB_x)(AC_y)-(AB_y)(AC_x)\):
| \(\overrightarrow{AB}\) | \(=\) | \((4-1,\ 2-1)=(3,1)\) |
| \(\overrightarrow{AC}\) | \(=\) | \((9-1,\ 5-1)=(8,4)\) |
| \(\text{cross}\) | \(=\) | \((3)(4)-(1)(8)\) |
| \(=\) | \(12-8\) | |
| \(=\) | \(4\ne 0\) |
The cross is \(4\ne0\), so the points are not collinear.
Compare \(\overrightarrow{AB}\) with \(\overrightarrow{BC}\); the ratio is \(AB:BC\):
| \(\overrightarrow{AB}\) | \(=\) | \((2-0,\ 4-1)=(2,3)\) |
| \(\overrightarrow{BC}\) | \(=\) | \((6-2,\ 10-4)=(4,6)\) |
| \(\overrightarrow{BC}\) | \(=\) | \(2(2,3)\) |
| \(=\) | \(2\,\overrightarrow{AB}\) | |
| \(AB:BC\) | \(=\) | \(1:2\) |
\(B\) divides \(AC\) in the ratio \(1:2\).
Common pitfalls
Frequently asked questions
How do you show three points are collinear using vectors?
Form \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) and \(\overrightarrow{AC}=\mathbf{c}-\mathbf{a}\). If \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\) for one scalar \(\lambda\), the vectors are parallel and share \(A\), so the points are collinear.
What does collinear mean?
Collinear points all lie on one straight line. Three points are collinear when the vector from the first to the third is a scalar multiple of the vector from the first to the second.
How do you test collinearity in 3D?
Either check that \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\) with the same \(\lambda\) for all three components, or use the cross product: the points are collinear when \(\overrightarrow{AB}\times\overrightarrow{AC}=\mathbf{0}\).
How do you find the ratio in which a point divides a segment?
When \(A\), \(B\), \(C\) are collinear, compare \(\overrightarrow{AB}\) with \(\overrightarrow{BC}\). If \(\overrightarrow{BC}=2\,\overrightarrow{AB}\), then \(B\) divides \(AC\) in the ratio \(AB:BC=1:2\).
What is the difference between parallel and collinear?
Parallel vectors point along the same line direction but can sit anywhere. Points are collinear when the vectors are parallel and also share a common point, forcing all the points onto one line.
Can the dot product test for collinearity?
No. A zero dot product means the vectors are perpendicular. For collinearity use a scalar multiple, \(\overrightarrow{AC}=\lambda\,\overrightarrow{AB}\), or a zero cross product in 3D.