Finding the affine equation of a plane given three points

Question

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?

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
Does this answer your question? Forming equation of a plane by solving linear equation set — Anne Bauval, Mar 27 '23 at 15:57

Brian Fitzpatrick · Accepted Answer · 2015-12-22T19:16:45.470

2

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 $$ 3\,x+2\,y+4\,z=7 $$

edited Dec 22 '15 at 19:16

answered Dec 22 '15 at 16:48

Brian Fitzpatrick

26,559

This is another method. It does not answer the OP's question. – Anne Bauval Mar 27 '23 at 15:24
Apparently the OP disagrees with you (eight years ago)... – Brian Fitzpatrick Mar 27 '23 at 19:15
Yes I saw that the OP contradicted him or herself, which I find wrong towards the other answerers and the at-that-time-future readers. – Anne Bauval Mar 27 '23 at 20:55

Jorik · Answer 2 · 2015-12-22T17:31:00.750

1

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.

edited Dec 22 '15 at 17:31

answered Dec 22 '15 at 16:29

Jorik

1,411

score 1 · Answer 3 · answered Dec 22 '15 at 16:44

1

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

answered Dec 22 '15 at 16:44

David Holden

18,040

This is another method. It does not answer the OP's question. – Anne Bauval Mar 27 '23 at 15:29

Finding the affine equation of a plane given three points

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