Equation of a plane

Just like a line, a plane can be described in several ways. We start with the simplest one, the parametric equations of a plane.

Parametric equations of a plane

First let’s recall how we found the parametric equations of a line p. We chose a point A that the line p passed through and then a direction vector $\overrightarrow{\mathbf{u}}$. If we added t times the direction vector (a scalar multiple) to the point A, we got a point on the line p. For every point X on the line we therefore had the equation

$$X=A+t\cdot \overrightarrow{\mathbf{u}},t\in \mathbb{R}.$$

A similar argument leads to the equation of a plane. To describe a plane we need two different non-zero vectors that are not parallel (not collinear), and of course a point A that the plane passes through. The vectors must not be parallel so that they do not lie along a single line. One line does not determine a plane, because a single line is contained in infinitely many planes.

We can also look at it another way – a line can be determined by two different points A, B. Infinitely many planes pass through these two points, but if we add a third point C that does not lie on the line AB, the plane is then determined uniquely.

So if we have three different points A, B, C, we can form two different vectors $\overrightarrow{\mathbf{u}}=\overrightarrow{AB}$ and $\overrightarrow{\mathbf{v}}=\overrightarrow{AC}$. These two vectors and, for example, the point A define a plane. Let’s recall how vectors are added:

We can see that adding two vectors gives a vector that points “between” the two. By adding different multiples of the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ we can fill the whole plane. So we can write the parametric equation of the plane that is determined by the vectors $\overrightarrow{\mathbf{u}},\overrightarrow{\mathbf{v}}$ and passes through the point A:

$$X=A+t\cdot \overrightarrow{\mathbf{u}}+s\cdot \overrightarrow{\mathbf{v}},t,s\in \mathbb{R}$$

where X is any point in the plane. For every point in the plane we can find t and s that make the equation true. We can also write this equation out as a system of equations. We assume that X(x, y, z), A(a₁, a₂, a₃), $\overrightarrow{\mathbf{u}}=(u_1,u_2,u_3)$ and $\overrightarrow{\mathbf{v}}=(v_1,v_2,v_3)$.

$$\begin{array}{rcl}x& =& a_1+t\cdot u_1+s\cdot v_1\\ y& =& a_2+t\cdot u_2+s\cdot v_2\\ z& =& a_3+t\cdot u_3+s\cdot v_3\end{array}$$

Example

Find out whether the point Q(0, 3, 5) lies in the plane determined by the points A(2, 3, 4), B(3, 7, −1), C(−5, 4, 4).

First we set up the equation of the plane. We find the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$, where $\overrightarrow{\mathbf{u}}=\overrightarrow{AB}$ and $\overrightarrow{\mathbf{v}}=\overrightarrow{AC}$. So:

$$\begin{array}{rcl}\overrightarrow{\mathbf{u}}& =& B-A=(3,7,-1)-(2,3,4)=(1,4,-5)\\ \overrightarrow{\mathbf{v}}& =& C-A=(-5,4,4)-(2,3,4)=(-7,1,0)\end{array}$$

The plane passes through A, so we write the system of equations:

$$\begin{array}{rcl}x& =& 2+t\cdot 1-s\cdot 7\\ y& =& 3+t\cdot 4+s\cdot 1\\ z& =& 4-t\cdot 5+s\cdot 0\end{array}$$

Now we find out whether the point Q(0, 3, 5) lies in this plane. We do this by substituting the numbers 0, 3, 5 for x,y,z and solving the system of equations. So we get the system:

$$\begin{array}{rcl}0& =& 2+t\cdot 1-s\cdot 7\\ 3& =& 3+t\cdot 4+s\cdot 1\\ 5& =& 4-t\cdot 5+s\cdot 0\end{array}$$

From the last equation we can make t the subject:

$$\begin{array}{rcl}5& =& 4-t\cdot 5+s\cdot 0\\ 5& =& 4-t\cdot 5\\ 1& =& -t\cdot 5\\ t\cdot 5& =& -1\\ t& =& -\frac{1}{5}\end{array}$$

We substitute t into the second equation:

$$\begin{array}{rcl}3& =& 3+t\cdot 4+s\cdot 1\\ 3& =& 3-\frac{1}{5}\cdot 4+s\cdot 1\\ 0& =& -\frac{4}{5}+s\cdot 1\\ s& =& \frac{4}{5}\end{array}$$

Finally we substitute the values of t and s into the first equation to see whether it holds:

$$\begin{array}{rcl}0& =& 2+t\cdot 1-s\cdot 7\\ 0& =& 2-\frac{1}{5}\cdot 1-\frac{4}{5}\cdot 7\\ 0& =& \frac{10}{5}-\frac{1}{5}-\frac{28}{5}\\ 0& =& -\frac{19}{5}\end{array}$$

The equation does not hold, so the whole system has no solution and the point Q(0, 3, 5) does not lie in the plane.