## Condition of the problem

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

$$
\begin{aligned}
& a=\{3 ; 3 ; 1\} \\
& b=\{1 ;-2 ; 1\} \\
& c=\{1 ; 1 ; 1\}
\end{aligned}
$$
## 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.

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

$=3 \cdot\left|\begin{array}{cc}-2 & 1 \\ 1 & 1\end{array}\right|-3 \cdot\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|+1 \cdot\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=$

$=3 \cdot(-3)-3 \cdot 0+1 \cdot 3=-9-0+3=-6$

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

## Problem Kuznetsov Analytic Geometry 6-5