Originally Posted by

**Harry1W** I had been under the impression that two points define a line, and three a plane.

So assuming we have three (non-collinear) points: A, B and C, with position vectors $\displaystyle \boldsymbol{a} , \boldsymbol{b} , $ and $\displaystyle \boldsymbol{c} $, respectively. Then, by considering a general point on the plane, P, with position vector $\displaystyle \boldsymbol{r} $, and making a parallelogram, it is possible to show that

$\displaystyle \begin{array}{ccc}

\vec{AP} &=& \lambda \vec{AC} + \mu \vec{AB} \\

\boldsymbol{r} - \boldsymbol{a} & = & \lambda \left( \boldsymbol{c} - \boldsymbol{a} \right) + \mu \left( \boldsymbol{b} - \boldsymbol{a} \right)

\end{array} $

So, in my eyes, this method seems to uniquely define the plane.

However, if we say that $\displaystyle \boldsymbol{a} = \left( \begin{array}{c}

a_1 \\

a_2 \\

a_3

\end{array} \right) $, and $\displaystyle \boldsymbol{b} = \left( \begin{array}{c}

b_1 \\

b_2 \\

b_3

\end{array} \right) $ and $\displaystyle \boldsymbol{c} = \left( \begin{array}{c}

c_1 \\

c_2 \\

c_3

\end{array} \right) $, then considering that the general form for a plane is $\displaystyle px + qy + rz = d $, we have but three equations:

$\displaystyle a_1 p + a_2 q + a_3 r = d $,

$\displaystyle b_1 p + b_2 q + b_3 r = d $ and

$\displaystyle c_1 p + c_3 q + c_3 r = d $,

and four variables. Therefore, there must exist multiple solutions. So, it seems to me, that this method does not define a plane uniquely - which leaves me a little confused, as I cannot imagine how three points could not define a plane uniquely.

Any suggestions much appreciated.