## Task Condition

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

$a=\{4 ; 2 ; 2\}$

$b=\{-3 ;-3 ;-3\}$

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

$(a, b, c)=\left|\begin{array}{ccc}4 & 2 & 2 \\ -3 & -3 & -3 \\ 2 & 1 & 2\end{array}\right|=$

$$
\begin{aligned}
& =4 \cdot\left|\begin{array}{cc}
-3 & -3 \\
1 & 2
\end{array}\right|-2 \cdot\left|\begin{array}{cc}
-3 & -3 \\
2 & 2
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
-3 & -3 \\
2 & 1
\end{array}\right|= \\
& =4 \cdot(-3)-2 \cdot 0+2 \cdot 3=-12-0+6=-6
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 6-19