Equations of planes
Master the equations of planes for Year 12 Specialist Mathematics in Queensland (QCAA). A plane is fixed by one point and a normal vector perpendicular to it, and is written either as the vector equation \(\mathbf{r}\cdot\mathbf{n}=d\) or the equivalent Cartesian equation \(ax+by+cz=d\).
You will learn to read a normal straight off a Cartesian equation, build a plane from a point and a normal, find the normal to a plane through three points using the cross product \(\overrightarrow{AB}\times\overrightarrow{AC}\), convert between the two forms, and test whether a point lies in a plane — core skills for the 3D vectors topic and everything that follows.
Theory
An equation of a plane is fixed by one point on the plane and a normal vector perpendicular to it. In Year 12 Specialist Mathematics (QCAA, Queensland) a plane is written as the vector (normal) equation \(\mathbf{r}\cdot\mathbf{n}=d\) or the equivalent Cartesian equation \(ax+by+cz=d\), and the normal to a plane through three points comes from a cross product.
A plane is fixed by one point on it together with a normal vector \(\mathbf{n}\) — a vector perpendicular to every direction lying in the plane. If \(\mathbf{a}\) is the position vector of a known point and \(\mathbf{r}\) is the position vector of a general point, then \(\mathbf{r}-\mathbf{a}\) lies in the plane, so it is perpendicular to \(\mathbf{n}\):
\(\mathbf{n}\cdot(\mathbf{r}-\mathbf{a})=0\).
Expanding gives the vector (normal) equation \(\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}\), usually written \(\mathbf{r}\cdot\mathbf{n}=d\) where the scalar \(d=\mathbf{a}\cdot\mathbf{n}\).
Writing \(\mathbf{r}=(x,y,z)\) and \(\mathbf{n}=(a,b,c)\) turns the dot product into the Cartesian equation \(ax+by+cz=d\). The coefficients of \(x,y,z\) are exactly the components of the normal, so a normal can be read straight off any Cartesian equation.
Given three points \(A,B,C\), the edge vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) both lie in the plane, so a normal is their cross product \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}\). Two planes are parallel when their normals are scalar multiples, and a plane is perpendicular to a line when the line's direction is the plane's normal.
The vector (normal) equation of the plane through the point with position vector \(\mathbf{a}\) with normal \(\mathbf{n}\) is:
With \(\mathbf{r}=(x,y,z)\) and \(\mathbf{n}=(a,b,c)\), the Cartesian equation is:
For a plane through three points \(A,B,C\), a normal is the cross product of two edge vectors:
How to find the equation of a plane
- Find a point and a normal. Use a given point for \(\mathbf{a}\). For three points, take \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}\); for a plane perpendicular to a line, use the line's direction as \(\mathbf{n}\); for a plane parallel to a given plane, reuse that plane's normal.
- Compute \(d=\mathbf{a}\cdot\mathbf{n}\) by substituting the point into the dot product, which gives the vector equation \(\mathbf{r}\cdot\mathbf{n}=d\).
- Write the Cartesian equation \(ax+by+cz=d\), reading \((a,b,c)\) straight off the components of \(\mathbf{n}\).
- Test a point if asked: substitute its coordinates into \(ax+by+cz\) and the point lies in the plane only if the result equals \(d\).
Use \(\mathbf{n}\cdot(\mathbf{r}-\mathbf{a})=0\); the coefficients are the components of \(\mathbf{n}\):
| \(2(x-3)+1(y-1)+2(z+2)\) | \(=\) | \(0\) |
| \(2x-6+y-1+2z+4\) | \(=\) | \(0\) |
| \(2x+y+2z-3\) | \(=\) | \(0\) |
| \(2x+y+2z\) | \(=\) | \(3\) |
\(2x+y+2z=3\).
Form two edge vectors from \(A\); a normal is their cross product:
| \(\overrightarrow{AB}\) | \(=\) | \(\mathbf{b}-\mathbf{a}=(2,1,0)\) |
| \(\overrightarrow{AC}\) | \(=\) | \(\mathbf{c}-\mathbf{a}=(0,2,2)\) |
| \(\mathbf{n}\) | \(=\) | \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 0 \\ 0 & 2 & 2 \end{vmatrix}\) |
| \(=\) | \(\mathbf{i}(2-0)-\mathbf{j}(4-0)+\mathbf{k}(4-0)\) | |
| \(=\) | \((2,-4,4)\) |
Divide by \(2\) to use \(\mathbf{n}=(1,-2,2)\); then \(d=\mathbf{n}\cdot\mathbf{a}\) with \(A(1,0,2)\):
| \(d\) | \(=\) | \((1)(1)+(-2)(0)+(2)(2)\) |
| \(=\) | \(1+4=5\) | |
| \(x-2y+2z\) | \(=\) | \(5\) |
\(x-2y+2z=5\).
With \(\mathbf{r}=(x,y,z)\), the dot product gives the Cartesian form:
| \(\mathbf{r}\cdot(3,-1,2)\) | \(=\) | \(7\) |
| \(3x-y+2z\) | \(=\) | \(7\) |
Substitute \(P(1,-2,1)\) into the left-hand side and compare with \(7\):
| \(3x-y+2z\) | \(=\) | \(3(1)-(-2)+2(1)\) |
| \(=\) | \(3+2+2\) | |
| \(=\) | \(7=\text{RHS}\) |
\(3x-y+2z=7\); since the left side equals \(7\), \(P\) lies in the plane.
A plane perpendicular to a line takes the line's direction as its normal, \(\mathbf{n}=(1,-2,2)\):
| \(1(x-3)-2(y-1)+2(z-0)\) | \(=\) | \(0\) |
| \(x-3-2y+2+2z\) | \(=\) | \(0\) |
| \(x-2y+2z-1\) | \(=\) | \(0\) |
| \(x-2y+2z\) | \(=\) | \(1\) |
\(x-2y+2z=1\).
Common pitfalls
Frequently asked questions
What is the equation of a plane?
A plane through the point \(A\) with normal \(\mathbf{n}\) has vector equation \(\mathbf{r}\cdot\mathbf{n}=d\), where \(d=\mathbf{a}\cdot\mathbf{n}\), and the equivalent Cartesian equation \(ax+by+cz=d\) with \(\mathbf{n}=(a,b,c)\).
How do you find the normal to a plane through three points?
Form two edge vectors \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) and \(\overrightarrow{AC}=\mathbf{c}-\mathbf{a}\); a normal is their cross product \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}\), which is perpendicular to both edges.
How do you read a normal off a Cartesian equation?
In \(ax+by+cz=d\), the coefficients of \(x,y,z\) are the components of a normal, so \(\mathbf{n}=(a,b,c)\). For example \(3x-y+2z=7\) has normal \((3,-1,2)\).
How do you convert between the vector and Cartesian forms of a plane?
Write \(\mathbf{r}=(x,y,z)\) and expand the dot product: \(\mathbf{r}\cdot(a,b,c)=d\) becomes \(ax+by+cz=d\), and reversing this reads the normal and constant back out.
How do you tell whether a point lies in a plane?
Substitute the point into \(ax+by+cz\); the point lies in the plane only if the result equals \(d\). If it gives any other value, the point is off the plane.
When are two planes parallel, and when is a plane perpendicular to a line?
Two planes are parallel when their normals are scalar multiples of each other. A plane is perpendicular to a line exactly when the line's direction vector is a normal to the plane.