# How do I find the Cartesian equation of a plane containing three given points?

Jun 5, 2015

The Cartesian equation of a plane $P$ is $a x + b y + c z + d = 0$, where $a , b , c$ are the coordinates of the normal vector $\vec{n} = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$

Let $A , B \mathmr{and} C$ be three noncolinear points, $A , B , C \in P$
Note that $A , B \mathmr{and} C$ define two vectors $\vec{A B}$ and $\vec{A C}$ contained in the plane $P$. We know that the cross product of two vectors contained in a plane defines the normal vector of the plane.

Now, we can use an example to illustrate the solution. Assume that the coordinates of the three points are the following:
$A \left(1 , 2 , 3\right) , B \left(- 2 , 1 , 0\right)$ and $C \left(0 , 3 , 2\right)$

From the coordinates of the points $A , B$ and $C$ we can find the vectors $\vec{A B}$ and $\vec{A C}$:

$\vec{A B} = \left({x}_{b} - {x}_{a}\right) \cdot \hat{i} + \left({y}_{b} - {y}_{a}\right) \cdot \hat{j} + \left({z}_{b} - {z}_{a}\right) \cdot \hat{k}$
$\vec{A C} = \left({x}_{c} - {x}_{a}\right) \cdot \hat{i} + \left({y}_{c} - {y}_{a}\right) \cdot \hat{j} + \left({z}_{c} - {z}_{a}\right) \cdot \hat{k}$
where $\hat{i} , \hat{j}$ and $\hat{k}$ are the unit vectors on the cartesian axes of coordinates $O x , O y$ and $O z$.

After plugging in the values of the coordinates, we have:
$\vec{A B} = \left(- 2 - 1\right) \cdot \hat{i} + \left(1 - 2\right) \cdot \hat{j} + \left(0 - 3\right) \cdot \hat{k}$
So, $\vec{A B} = - 3 \hat{i} - \hat{j} - 3 \hat{k}$

$\vec{A C} = \left(0 - 1\right) \cdot \hat{i} + \left(3 - 2\right) \cdot \hat{j} + \left(2 - 3\right) \cdot \hat{k}$.
So, $\vec{A C} = - \hat{i} + \hat{j} - \hat{k}$

Next, we'll find the normal vector from the cross product of $\vec{A B}$ and $\vec{A C}$

$\vec{n} = \vec{A B}$ x $\vec{A C} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 3 , - 1 , - 3\right) , \left(- 1 , 1 , - 1\right) |$ $=$ $4 \hat{i} - 4 \hat{k}$

Therefore, the equation of the plane is $4 x - 4 z + d = 0$. In order to find $d$, we can plug in the coordinates of point $A$:
$d = 4 z - 4 x \implies d = 4 \cdot 3 - 4 \cdot 1 = 8$

So, the cartesian equation of the plane is $4 x - 4 z + 8 = 0$.