1. ## Defining a plane

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 $\boldsymbol{a} , \boldsymbol{b} ,$ and $\boldsymbol{c}$, respectively. Then, by considering a general point on the plane, P, with position vector $\boldsymbol{r}$, and making a parallelogram, it is possible to show that

$\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 $\boldsymbol{a} = \left( \begin{array}{c}

a_1 \\
a_2 \\
a_3

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

b_1 \\
b_2 \\
b_3

\end{array} \right)$
and $\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 $px + qy + rz = d$, we have but three equations:

$a_1 p + a_2 q + a_3 r = d$,
$b_1 p + b_2 q + b_3 r = d$ and
$c_1 p + c_3 q + c_3 r = d$,

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.

2. Hmm from what I remember in multivariable calculus, planes are defined by their normal vector, and the 2 vectors that sweep out the plane, the only problem is that you can have a lot of vectors that have the same span (like if you take i and j that sweeps out the xy plane but you can take i,j and rotate them clockwise 45 degrees and they still sweep out the same plane)

3. 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 $\boldsymbol{a} , \boldsymbol{b} ,$ and $\boldsymbol{c}$, respectively. Then, by considering a general point on the plane, P, with position vector $\boldsymbol{r}$, and making a parallelogram, it is possible to show that

$\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 $\boldsymbol{a} = \left( \begin{array}{c}

a_1 \\
a_2 \\
a_3

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

b_1 \\
b_2 \\
b_3

\end{array} \right)$
and $\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 $px + qy + rz = d$, we have but three equations:

$a_1 p + a_2 q + a_3 r = d$,
$b_1 p + b_2 q + b_3 r = d$ and
$c_1 p + c_3 q + c_3 r = d$,

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.
There exist multiple solutions and therefore multiple equations but only one plane. If px+ qy+ rz= d, then multiplying each number by the same thing, say a, gives (ap)x+ (aq)y+ (ar)z= ad, a different equation for the same plane.

4. Ah that makes a lot of sense. Thank you!