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Thread: Defining a plane

  1. #1
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    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 $\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.
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  2. #2
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    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)


    note: I could be wrong about this it has been a while.
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  3. #3
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    Quote Originally Posted by Harry1W View Post
    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.
    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.
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  4. #4
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    Ah that makes a lot of sense. Thank you!
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