## Task Condition

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

$a=\{3 ; 1 ; 0\}$

$b=\{-5 ;-4 ;-5\}$

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

$(a, b, c)=\left|\begin{array}{ccc}3 & 1 & 0 \\ -5 & -4 & -5 \\ 4 & 2 & 4\end{array}\right|=$

$=3 \cdot\left|\begin{array}{cc}-4 & -5 \\ 2 & 4\end{array}\right|-1 \cdot\left|\begin{array}{cc}-5 & -5 \\ 4 & 4\end{array}\right|+0 \cdot\left|\begin{array}{cc}-5 & -4 \\ 4 & 2\end{array}\right|=$

$=3 \cdot(-6)-1 \cdot 0+0 \cdot 6=-18-0+0=-18$

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

## Problem Kuznetsov Analytic Geometry 6-24