1. ## 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
$\displaystyle \begin{array}{l} AB = < - 1,2,4 > \\ AC = < - 3,1,2 > \\ AD = < 3,2,4 > \\ \end{array}$

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

2. 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
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.

3. 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

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

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
I did the same computation as you (i.e. check if $\displaystyle \overrightarrow{AB},\overrightarrow{AC},\overright arrow{AD}$ are linearly dependent by computing the determinant), and I confirm the points are coplanar.

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# how to show that 4 points are coplanar

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