## Task Condition

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

$a=\{1 ;-1 ; 4\}$

$b=\{1 ; 0 ; 3\}$

$c=\{1 ;-3 ; 8\}$
## 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.

$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -3 & 8\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}1 & 3 \\ 1 & 8\end{array}\right|+4 \cdot\left|\begin{array}{cc}1 & 0 \\ 1 & -3\end{array}\right|=$

$=1 \cdot 9+1 \cdot 5+4 \cdot(-3)=9+5-12=2$

Since $(a, b, c)=2 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.

## Problem Kuznetsov Analytic Geometry 6-26