Originally Posted by

**Craka** 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

$\displaystyle

\begin{array}{l}

AB = < - 1,2,4 > \\

AC = < - 3,1,2 > \\

AD = < 3,2,4 > \\

\end{array}

$

than using the triple scalar product

$\displaystyle

\begin{array}{l}

AB \bullet (AC \times AD) = \left| \begin{array}{l}

- 1,2,4 \\

- 3,1,2 \\

3,2,4 \\

\end{array} \right| \\

= - 1\left| \begin{array}{l}

1,2 \\

2,4 \\

\end{array} \right| - 2\left| \begin{array}{l}

- 3,2 \\

3,4 \\

\end{array} \right| + 4\left| \begin{array}{l}

- 3,1 \\

3,2 \\

\end{array} \right| \\

= - 1[4 - 4] - 2[ - 12 - 6] + 4[ - 6 - 3] \\

= 0 + 36 - 36 \\

= 0 \\

\end{array}

$

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