## Task Condition

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

$a=\{3 ; 10 ; 5\}$

$b=\{-2 ;-2 ;-3\}$

$c=\{2 ; 4 ; 3\}$
## 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.

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 10 & 5 \\
-2 & -2 & -3 \\
2 & 4 & 3
\end{array}\right|= \\
& =3 \cdot\left|\begin{array}{cc}
-2 & -3 \\
4 & 3
\end{array}\right|-10 \cdot\left|\begin{array}{cc}
-2 & -3 \\
2 & 3
\end{array}\right|+5 \cdot\left|\begin{array}{cc}
-2 & -2 \\
2 & 4
\end{array}\right|= \\
& =3 \cdot 6-10 \cdot 0+5 \cdot(-4)=18-20=-2
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 6-16