## Task Condition

Are the vectors $a, b$ and $c$ coplanar?

$a=\{6 ; 3 ; 4\}$

$b=\{-1 ;-2 ;-1\}$

$c=\{2 ; 1 ; 2\}$
## Solution

For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.

$(a, b, c)=\left|\begin{array}{ccc}6 & 3 & 4 \\ -1 & -2 & -1 \\ 2 & 1 & 2\end{array}\right|=$
$=6 \cdot\left|\begin{array}{cc}-2 & -1 \\ 1 & 2\end{array}\right|-3 \cdot\left|\begin{array}{cc}-1 & -1 \\ 2 & 2\end{array}\right|+4 \cdot\left|\begin{array}{cc}-1 & -2 \\ 2 & 1\end{array}\right|=$

$=6 \cdot(-3)-3 \cdot 0+4 \cdot 3=-18-0+12=-6$

Since $(a, b, c)=-6 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.

## Problem Kuznetsov Analytic Geometry 6-27