Equations of Planes

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Parametric equations

So far we have learned the following:

• In \(\mathbb{R}^2\), given two non-parallel vectors \(\mathbf{u}\) and \(\mathbf{v}\), any vector can be written as \(r\mathbf{u} + s\mathbf{v}\) for some scalars \(r\) and \(s\).

• Any point on a line passing through a point \(P\) with position vector \(\mathbf{r}_0\) in the direction given by a vector \(\mathbf{v}\) in \(\mathbb{R}^3\) can be written as \(\mathbf{r}_0 + t\mathbf{v}\) for some scalar \(t\).

• A line is \(\mathbb{R}^3\) can be given by two points \(P\) and \(Q\). An equation of such line can be written in vector form as \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\) where \(\mathbf{r} = \left\langle x,y,z\right\rangle\), \(\mathbf{r}_0 = \left\langle x_0, y_0, z_0\right\rangle\) is the position vector of \(P\), and \(\mathbf{v} = \left\langle v_1,v_2,v_3\right\rangle = \overrightarrow{PQ}\).

• We can split the vector equation into components to get parametric equations of the line: \[\left\{\begin{aligned}x &= x_0 + tv_1\\y &= y_0 + tv_2\\z &= z_0 + tv_3\end{aligned}\right.\]

Suppose we have to find an equation of the plane containing three points \(P\), \(Q\) and \(R\). We can start by finding the vectors \(\mathbf{u} = \left\langle u_1,u_2,u_3\right\rangle = \overrightarrow{PQ}\) and \(\mathbf{v} = \left\langle v_1,v_2,v_3\right\rangle = \overrightarrow{PR}\). Then any point on the plane can be written as \(\mathbf{r}_0 + r\mathbf{u} + s\mathbf{v}\) for some scalars \(r\) and \(s\), where \(\mathbf{r}_0 = \left\langle x,y,z\right\rangle\) is the position vector of \(P\):

This will give us a parametric vector equation of the plane: \[\mathbf{r} = \mathbf{r}_0 + r\mathbf{u} + s\mathbf{v}.\]

This we can split into components to get the parametric equations:

\[\left\{\begin{aligned}x &= x_0 + r u_1 + s v_1\\y &= y_0 + r u_2 + s v_2\\z &= z_0 + r u_3 + s v_3\end{aligned}\right.\]

Each pair of values for the scalars \(r\) and \(s\) corresponds to exactly one point in the plane. We can think about it as mapping the two-dimensional \(rs\)-plane onto our plane in \(\mathbb{R}^3\) such that the origin maps to \(P\), the point \((1,0)\) maps to \(Q\), the point \((0,1)\) maps to \(R\), the \(r\)-axis maps to the line \(PQ\), and the \(s\)-axis maps to the line \(PR\):

An example

Implicit equations

There is another way to describe a plane: instead of giving a point and two “direction” vectors, we can instead give a point and one vector that is perpendicular to the plane. We call this vector a normal vector of the plane.

Suppose we have a point \(P\) with position vector \(\mathbf{r}_0 = \left\langle x_0, y_0, z_0\right\rangle\) and a normal vector \(\mathbf{n} = \left\langle n_1,n_2,n_3\right\rangle\). Then for any point \(Q(x,y,z)\) with the position vector \(\mathbf{r} = \left\langle x,y,z\right\rangle\), the point belongs to the plane if and only if the vector \(\overrightarrow{PQ}\) is orthogonal to \(\mathbf{n}\). Two vectors are orthogonal if and only if their dot product is zero, which gives us the implicit vector equation of the plane:

\[\mathbf{n}\cdot (\mathbf{r} - \mathbf{r}_0) = 0.\]

Writing this in components and reorganizing will give us the implicit equation of the plane:

\[n_1 x + n_2 y + n_3 z - (n_1 x_0 + n_2 y_0 + n_3 z_0) = 0.\]

An example

Try it yourself

Angle between planes, parallel planes

Two planes are parallel when their normal vectors are scalar multiples of each other.

If two planes are not parallel, they intersect, and the angle between them is the same as the angle between their normal vectors.

Try it yourself