## Task Condition

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

$a=\{3 ; 7 ; 2\}$

$b=\{-2 ; 0 ;-1\}$

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

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 7 & 2 \\
-2 & 0 & -1 \\
2 & 2 & 1
\end{array}\right|= \\
& =3 \cdot\left|\begin{array}{cc}
0 & -1 \\
2 & 1
\end{array}\right|-7 \cdot\left|\begin{array}{cc}
-2 & -1 \\
2 & 1
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
-2 & 0 \\
2 & 2
\end{array}\right|= \\
& =3 \cdot 2-7 \cdot 0+2 \cdot(-4)=6-0-8=-2
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 6-10