# Finding the affine equation of a plane given three points

First of all, I am not sure the correct word is "affine" in english however it is "affin" in Swedish and I can't find the english counterpart. Affine form is basically the non-parametric equation of a plane: a

$$ax + by + cz + d = 0$$

I am given three points:

$$P:(1,2,0)$$ $$Q:(-1,3,1)$$ $$R:(-1,1,2)$$

and asked to find the "non-parametric" equation for the plane.

My attempt is to simply solve the system of equations: $$ax + by + cz + d = 0$$

$$\begin{cases} a + 2b + d = 0 \\ -a + 3b + c + d = 0 \\ -a+b+2c+d = 0 \end{cases}$$

The process is quite tedious so I won't bore you with it, but I know it's correct and I end up with:

$$\begin{cases} a = -2b-d \\ b = ? \\c=2b \\ d=-\frac{7}{2}b \end{cases}$$

There is too many unknown variables too solve the system of equations, but in my book the author simply assigns $b = 2$ and solves it. Why can this be done?

• It's called Cartesian equation of a plane. – Silvia Ghinassi Dec 22 '15 at 16:27
• The degree of freedom is from scaling the equation $ax+but+cz+d=0$. So, you can choose any value you wish. – Michael Burr Dec 22 '15 at 16:29

The Cartesian equation of a plane in $\Bbb R^3$ can neatly be expressed as $\vec n\cdot\vec x=\vec n\cdot\vec x_0$ where $\vec n$ is a vector normal to the plane, $\vec x=\langle x,y,z\rangle$, and $\vec x_0$ is a point on the plane. In our case, the two directions \begin{align*} \vec{PQ} &= \langle -2,1,1\rangle & \vec{PR} &= \langle-2,-1,2\rangle \end{align*} are both in the plane. Therefore we may take $$\vec n=\vec{PQ}\times\vec{PR}=\langle3,2,4\rangle$$ We may also take $\vec x_0=P$. This gives $$3\,x+2\,y+4\,z=7$$

$$P-Q = (2,-1,-1) \\ Q-R=(0,2,-1) \\ \hat n = (P-Q)\times (Q-R) =(3,2,4)$$ now $$(\hat r - Q)\cdot\hat n = 0$$ i.e. $$3x+2y+4z=7$$

It is because the parameters that describe the plane are not unique. If $ax+by+cz+d=0$ then also $2ax+2by+2cz+2d=0$. Of course the $2$ there could be any non-zero real number.

The work you did shows us that if $b=0$ then all of $a$, $c$ and $d$ would also be zero. This shows that in case of this particular plane $b$ cannot be zero. And by the earlier argument we can choose it arbitrarily.

Better to choose an equation of a plane with three constants like:

$$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$$

( eliminating $d$ ), where (a,b,c) can be solved in the same way.

Geometrically it means we are not leaving out minimum distance related to $d$ to the origin arbitrarily.