## Task Condition

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

$a=\{1 ; 5 ; 2\}$

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

$c=\{1 ; 1 ; 1\}$
## 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.

$$
(a, b, c)=\left|\begin{array}{ccc}
1 & 5 & 2 \\
-1 & 1 & -1 \\
1 & 1 & 1
\end{array}\right|=
$$

$=1 \cdot\left|\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right|-5 \cdot\left|\begin{array}{cc}-1 & -1 \\ 1 & 1\end{array}\right|+2 \cdot\left|\begin{array}{cc}-1 & 1 \\ 1 & 1\end{array}\right|=$

$=1 \cdot 2-5 \cdot 0+2 \cdot(-2)=2-0-4=-2$

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

## Problem Kuznetsov Analytic Geometry 6-3