Collinear points and coplanar points in 3D

Content updated: October 22, 2026

Collinearity of three points can be tested using two parallel vectors. For four points, determine whether a plane contains them all; this is different from determining a unique plane.

On this page: Idea · Method · Examples · Practice · Further study

1. Vectors with a common initial point

Let $A,B,C$ be three points. Set $\vec u=\overrightarrow{AB}$ and $\vec v=\overrightarrow{AC}$. If $A\ne B$, they are collinear if and only if some $t\in\mathbb R$ satisfies $\vec v=t\vec u$. Then $C=A+t\overrightarrow{AB}$ lies on the line $AB$.

Treat coincident points separately. When $A=B$, the set $A,B,C$ has at most two distinct points and therefore lies on a line, but $AB$ does not name a line because those two points coincide. If all three points coincide, infinitely many lines pass through the common point.

2. Collinearity and coplanarity criteria

A compact condition, avoiding division by a coordinate that may equal $0$, is

Here collinear means that the points lie on a line, including when some points coincide. A cross product different from $\vec0$ ensures that the three points are distinct and noncollinear, so they determine a unique plane.

For four points $A,B,C,D$, also set $\vec w=\overrightarrow{AD}$. They are coplanar if and only if the vectors $\vec u,\vec v,\vec w$ span a space of dimension at most $2$. Equivalently,

If $\vec u,\vec v$ are independent, their cross product is normal to $(ABC)$. The dot product condition $0$ says that $D$ lies in that plane as well. In this case one can also find $s,t$ such that $\vec w=s\vec u+t\vec v$.

If $\vec u,\vec v$ are parallel, not every coplanar vector can necessarily be expressed in terms of just those two vectors. However, adding one vector $\vec w$ raises the number of independent directions to at most $2$, so the four points still lie in a plane. To determine a unique plane, find two independent vectors among the three under consideration.

3. Worked examples

Example 1. Three collinear points

Let $A(1;3;1)$, $B(0;1;2)$, and $C(3;7;-1)$. Then

Since $A\ne B$, the point $C$ lies on the line $AB$. The three points are therefore collinear. The negative multiplier indicates opposite orientations and does not prevent parallelism.

Example 2. Four noncoplanar points

Let $A(1;-1;1)$, $B(1;3;1)$, $C(4;3;2)$, and $D(4;1;3)$. Then $\vec u=(0;4;0)$, $\vec v=(3;4;1)$, and $\vec w=(3;2;2)$. Calculation yields

Thus the four points are not coplanar. We can also use a system of equations: $\vec v\times\vec w=(6;-3;-6)\ne\vec0$, so coplanarity of the three vectors would imply $\vec u=x\vec v+y\vec w$ for real numbers $x,y$. The coordinate system is

The first two equations imply $x=2$ and $y=-2$; the left-hand side of the last equation is then $-2$, so it is not satisfied. The system has no solution, agreeing with the scalar triple product test.

Example 3. Two parallel vectors do not form a basis

Let $A(0;0;0)$, $B(1;0;0)$, $C(2;0;0)$, and $D(0;1;0)$. All four points lie in $(Oxy)$. But $\overrightarrow{AD}=(0;1;0)$ cannot be expressed in terms of $\overrightarrow{AB}=(1;0;0)$ and $\overrightarrow{AC}=(2;0;0)$. A basis of directions for this plane is $\overrightarrow{AB},\overrightarrow{AD}$.

4. Practice

1. Let $A(0;1;2)$, $B(1;3;1)$, and $C(2;5;0)$. Check that the points are collinear.
2. Let $A(0;0;0)$, $B(1;0;0)$, $C(0;1;0)$, and $D(1;1;m)$. Find $m$ such that the points are coplanar.
3. Let $A=B=(1;2;3)$, $C(2;2;3)$, and $D(1;2;4)$. Are the four points coplanar? Do they determine a unique plane?

5. Common mistakes

• Dividing coordinate components to test proportionality when a denominator may be $0$.
• Naming a line through two coincident points.
• Using two parallel vectors as a basis for all vectors in a plane.
• Confusing the existence of a plane containing the points with the existence of a unique plane determined by them.

6. Related lessons

• Vector coordinates from two endpoints (English lesson)
• The cross product and its applications (English lesson)
• Normal vectors to a plane (English lesson)
• Equations of planes (English lesson)

Original practice resource on Scribd.