## Task Condition

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

$a=\{4 ; 3 ; 1\}$

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

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

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
4 & 3 & 1 \\
1 & -2 & 1 \\
2 & 2 & 2
\end{array}\right|= \\
& =4 \cdot\left|\begin{array}{cc}
-2 & 1 \\
2 & 2
\end{array}\right|-3 \cdot\left|\begin{array}{cc}
1 & 1 \\
2 & 2
\end{array}\right|+1 \cdot\left|\begin{array}{cc}
1 & -2 \\
2 & 2
\end{array}\right|= \\
& =4 \cdot(-6)-3 \cdot 0+1 \cdot 6=-24-0+6=-18
\end{aligned}
$$

Since $(a, b, c)=-18 \neq 0$, the vectors $a, b$, and $c$ are not coplanar.

## Problem Kuznetsov Analytic Geometry 6-7