A plane in 3D can be defined using various combinations of

  • points lying in the plane,
  • direction vectors in (or parallel to) the plane, and
  • normal vectors which are perpendicular to the plane.

For each of the following combinations of data,

  • does it define a unique plane?
  • does it define a family of possible planes? If so, can you describe the family?

The points \((-1,4,6)\), \((0,7,4)\) and \((3,-1,10)\)

The points \((-1,4,6)\), \((0,7,4)\) and \((-3,-2,10)\)

Generally, three points in space will uniquely define a plane whereas two points will not. However, the three points must all be distinct and they must not lie on a straight line or else they behave like two points.

The easiest way to check for collinearity is to compare the vectors joining pairs of the points.

For card (A), the vector from the first to the second point is \[\begin{pmatrix}0-(-1)\\7-4\\4-6\end{pmatrix} = \begin{pmatrix}1\\3\\-2\end{pmatrix}\] and from the first to the third is \[\begin{pmatrix}3-(-1)\\-1-4\\10-6\end{pmatrix} = \begin{pmatrix}4\\-5\\4\end{pmatrix}.\]

These are clearly not parallel vectors, so the three points are not in a line and they do define a unique plane.

For card (B), two of the points are the same as on card (A). The vectors between the points this time are \[\begin{pmatrix}1\\3\\-2\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}-2\\-6\\4\end{pmatrix}.\]

illustration of an book with pages spread open These vectors are parallel (the second is \(-2\) times the first) so the three points are collinear. There are an infinite number of possible planes that all contain these three points.

Imagine a thin book such as an exercise book, with the three fixed points lying along its spine. Allow the pages to spread out – each page represents part of a plane that passes through the three points. We could say that the three points on card (B) define a book of planes. This is sometimes called a fan, a sheaf or a pencil of planes.

The direction vectors \(\mathbf{i}+3\mathbf{j}+\mathbf{k}\) and \(-2\mathbf{i}-6\mathbf{j}-2\mathbf{k}\)

Since a vector has magnitude and direction but no fixed position, vectors alone can never uniquely define a plane.

In general, two vectors will define a family of planes that are all parallel to one another; we could call it a stack of planes.

But these two vectors are parallel. So as we found with card (B), they could define a book of planes, but in this case the spine of the book has no fixed position – it could be anywhere in space as long as it points in the direction of our two vectors. This is not just a book of planes, but a library!

The direction vectors \(\mathbf{i}+3\mathbf{j}\),
\(\mathbf{i}+\mathbf{j}+2\mathbf{k}\) and

The direction vectors \(\mathbf{i}+2\mathbf{j}+\mathbf{k}\), \(-2\mathbf{i}-4\mathbf{j}+3\mathbf{k}\) and \(3\mathbf{i}+6\mathbf{j}+5\mathbf{k}\)

As noted above, a set of vectors cannot uniquely define a plane. Neither of these cards contain parallel vectors but as there are three on each card, we should check whether they are coplanar.

On card (D), the first and third vectors have no \(\mathbf{k}\)-component so they lie in the \(x\text-y\) plane (and are not parallel). The second, however, does not lie in this plane. So these three vectors do not define a plane at all.

Why is it important that the first and third vectors on card (D) are not parallel? What would happen if they were?

illustration of a stack of planes The vectors on card (E) all have their \(\mathbf{i}\)- and \(\mathbf{j}\)-components in the ratio \(1:2\). So all the vectors do lie in (or at least, parallel to) a vertical plane. (If you looked down the \(z\)-axis you would see straight lines in the direction \(\mathbf{i}+2\mathbf{j}\).) Together they define a stack of planes.

The point \((1,5,3)\) and direction vectors \(\mathbf{i}-3\mathbf{j}+2\mathbf{k}\) and \(5\mathbf{i}+2\mathbf{j}+2\mathbf{k}\)

These two vectors are not parallel to one another, so together with a point, they uniquely define a plane.

The points \((1,5,8)\) and \((-2,17,3)\) and direction vector \(-3\mathbf{i}+2\mathbf{j}-\mathbf{k}\)

The points \((1,-6,2)\) and \((3,12,-4)\) and direction vector \(\mathbf{i}+9\mathbf{j}-3\mathbf{k}\)

Two points and a vector will usually be enough to uniquely define a plane. An exception is if the given vector is parallel to the vector between the two points, in which case we have a book of planes.

The vector joining the points on card (G) is \(-3\mathbf{i}+12\mathbf{j}-5\mathbf{k}\) which is not parallel to the given vector, so card (G) uniquely defines a plane.

The vector joining the points on card (H) is \(2\mathbf{i}+18\mathbf{j}-6\mathbf{k}\) which is parallel to \(\mathbf{i}+9\mathbf{j}-3\mathbf{k}\), so card (H) defines a book of planes.

The point \((7,-3,2)\) and normal vector \(-2\mathbf{i}+8\mathbf{j}-12\mathbf{k}\)

The normal vector defines the orientation of a plane and a single fixed point determines its location. This combination always uniquely defines a plane.

The normal vector \(-2\mathbf{i}+\mathbf{j}+4\mathbf{k}\) and points \((-3,3,4)\) and \((1,-1,7)\)

The normal vector \(-2\mathbf{i}+\mathbf{j}+4\mathbf{k}\) and points \((6,-3,5)\) and \((1,-1,7)\)

A normal vector plus one point uniquely defines a plane. A second point will either lie in the same plane or it will define a different plane. One way to check is to use the scalar product to find out whether the vector joining the two points is perpendicular to the normal vector.

On card (J) the vector joining the points is \(4\mathbf{i}-4\mathbf{j}+3\mathbf{k}\) which gives us a scalar product of \(4\times(-2)-4\times1+3\times4 = 0\). The vectors are therefore perpendicular and the two points lie in the same plane. Therefore (J) uniquely defines a plane.

On card (K) we have \(-5\mathbf{i}+2\mathbf{j}+2\mathbf{k}\) and a scalar product of \(-5\times(-2)+2\times1+2\times4 = 20\). The vectors are not perpendicular so card (K) does not define a plane at all.

The direction vector \(4\mathbf{i}+8\mathbf{j}-10\mathbf{k}\) and normal vector \(3\mathbf{i}+\mathbf{j}+2\mathbf{k}\)

This is similar to (J) and (K). The vector must be perpendicular to the normal vector if it is to define anything. The scalar product is \(3\times4+8+2\times(-10)=0\) so the directions are consistent. Since we have no fixed point, this defines a stack of planes.


Uniquely defines a plane Defines a family of planes Does not define a plane
A, F, G, I, J B (a book),
C (a library),
E (a stack),
H (a book),
L (a stack)
D, K