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Math Help - Determining whether 4 points are coplanar. Answer check

  1. #1
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    Red face Determining whether 4 points are coplanar. Answer check

    The question asks, whether the points are (1,0,-1) , (0,2,3) and (-2,1,1) and (4,2,3) are coplanar.

    I let (1,0,-1) be point A
    let (0,2,3) point B
    let (-2,1,1) point C
    let (4,2,3) point D

    So vector AB, AC, AD are
    <br />
\begin{array}{l}<br />
 AB =  <  - 1,2,4 >  \\<br />
 AC =  <  - 3,1,2 >  \\<br />
 AD =  < 3,2,4 >  \\<br />
 \end{array}<br />

    than using the triple scalar product

    <br />
\begin{array}{l}<br />
 AB \bullet (AC \times AD) = \left| \begin{array}{l}<br />
  - 1,2,4 \\ <br />
  - 3,1,2 \\ <br />
 3,2,4 \\ <br />
 \end{array} \right| \\ <br />
  =  - 1\left| \begin{array}{l}<br />
 1,2 \\ <br />
 2,4 \\ <br />
 \end{array} \right| - 2\left| \begin{array}{l}<br />
  - 3,2 \\ <br />
 3,4 \\ <br />
 \end{array} \right| + 4\left| \begin{array}{l}<br />
  - 3,1 \\ <br />
 3,2 \\ <br />
 \end{array} \right| \\ <br />
  =  - 1[4 - 4] - 2[ - 12 - 6] + 4[ - 6 - 3] \\ <br />
  = 0 + 36 - 36 \\ <br />
  = 0 \\ <br />
 \end{array}<br />

    Considering that the triple scalar product is zero, would mean that the points are coplanar, correct?

    But the answer is given as though that are not coplanar
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  2. #2
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    Quote Originally Posted by Craka View Post
    The question asks, whether the points are (1,0,-1) , (0,2,3) and (-2,1,1) and (4,2,3) are coplanar.

    I let (1,0,-1) be point A
    let (0,2,3) point B
    let (-2,1,1) point C
    let (4,2,3) point D

    So vector AB, AC, AD are
    <br />
\begin{array}{l}<br />
AB = < - 1,2,4 > \\<br />
AC = < - 3,1,2 > \\<br />
AD = < 3,2,4 > \\<br />
\end{array}<br />

    than using the triple scalar product

    <br />
\begin{array}{l}<br />
AB \bullet (AC \times AD) = \left| \begin{array}{l}<br />
- 1,2,4 \\ <br />
- 3,1,2 \\ <br />
3,2,4 \\ <br />
\end{array} \right| \\ <br />
= - 1\left| \begin{array}{l}<br />
1,2 \\ <br />
2,4 \\ <br />
\end{array} \right| - 2\left| \begin{array}{l}<br />
- 3,2 \\ <br />
3,4 \\ <br />
\end{array} \right| + 4\left| \begin{array}{l}<br />
- 3,1 \\ <br />
3,2 \\ <br />
\end{array} \right| \\ <br />
= - 1[4 - 4] - 2[ - 12 - 6] + 4[ - 6 - 3] \\ <br />
= 0 + 36 - 36 \\ <br />
= 0 \\ <br />
\end{array}<br />

    Considering that the triple scalar product is zero, would mean that the points are coplanar, correct?

    But the answer is given as though that are not coplanar
    I'd just find the equation of the plane that three of the points lie in. Then test whether the fourth point satisfies the equation.
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  3. #3
    MHF Contributor

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    Paris, France
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    Quote Originally Posted by Craka View Post
    The question asks, whether the points are (1,0,-1) , (0,2,3) and (-2,1,1) and (4,2,3) are coplanar.

    I let (1,0,-1) be point A
    let (0,2,3) point B
    let (-2,1,1) point C
    let (4,2,3) point D

    So vector AB, AC, AD are
    <br />
\begin{array}{l}<br />
 AB =  <  - 1,2,4 >  \\<br />
 AC =  <  - 3,1,2 >  \\<br />
 AD =  < 3,2,4 >  \\<br />
 \end{array}<br />

    than using the triple scalar product

    <br />
\begin{array}{l}<br />
 AB \bullet (AC \times AD) = \left| \begin{array}{ccc}<br />
  - 1&2&4\\ <br />
  - 3&1&2\\ <br />
 3&2&4\\ <br />
 \end{array} \right| \\ <br />
  =  - 1\left| \begin{array}{cc}<br />
 1 &2 \\ <br />
 2 &4 \\ <br />
 \end{array} \right| - 2\left| \begin{array}{cc}<br />
  - 3& 2 \\ <br />
 3 &4 \\ <br />
 \end{array} \right| + 4\left| \begin{array}{cc}<br />
  - 3& 1 \\ <br />
 3 &2 \\ <br />
 \end{array} \right| \\ <br />
  =  - 1[4 - 4] - 2[ - 12 - 6] + 4[ - 6 - 3] \\ <br />
  = 0 + 36 - 36 \\ <br />
  = 0 \\ <br />
 \end{array}<br />

    Considering that the triple scalar product is zero, would mean that the points are coplanar, correct?

    But the answer is given as though that are not coplanar
    I did the same computation as you (i.e. check if \overrightarrow{AB},\overrightarrow{AC},\overright  arrow{AD} are linearly dependent by computing the determinant), and I confirm the points are coplanar.
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