## Task Condition

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

$a=\{5 ; 3 ; 4\}$

$b=\{4 ; 3 ; 3\}$

$c=\{9 ; 5 ; 8\}$
## 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.

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
5 & 3 & 4 \\
4 & 3 & 3 \\
9 & 5 & 8
\end{array}\right|= \\
& =5 \cdot\left|\begin{array}{ll}
3 & 3 \\
5 & 8
\end{array}\right|-3 \cdot\left|\begin{array}{ll}
4 & 3 \\
9 & 8
\end{array}\right|+4 \cdot\left|\begin{array}{ll}
4 & 3 \\
9 & 5
\end{array}\right|= \\
& =5 \cdot 9-3 \cdot 5+4 \cdot(-7)=45-15-28=2
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 6-21