Suppose
\[\omega\]
is a real valued function of vectors \[ \mathbf{a}, \: \mathbf{b} \in \mathbb{R}^3\]
such that \[\omega( \mathbf{a}, \mathbf{b})= \omega( \mathbf{b}, \mathbf{a})\]
and that \[\omega\]
is linear in both arguments.There is a vector
\[\mathbf{c}\]
such that \[\omega( \mathbf{a}, \mathbf{b})= det (\mathbf{a} , \mathbf{b}, \mathbf{c})\]
for all vectors \[\mathbf{a}, \: \mathbf{b}\]
.Proof
Let
\[\mathbf{e}_1=(1,0,0)^T, \: \mathbf{e}_2=(0,1,0)^T, \: \mathbf{e}_3=(0,0,1)^T\]
be the basis for \[\mathbb{R}^3\]
.Then we can write
\[\mathbf{a}=a_1 \mathbf{e}_1 +a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3 \]
\[\mathbf{b}=b_1 \mathbf{e}_1 +b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3 \]
\[\omega\]
is linear and symmetric in \[\mathbf{a} \: \mathbf{b}\]
so\[\begin{equation} \begin{aligned} \omega(\mathbf{a}, \mathbf{b}) &= \omega(a_1 \mathbf{e}_1 +a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3 ,b_1 \mathbf{e}_1 +b_2 \mathbf{e}_2 + b_3 \mathbf{e}_3 ) \\ &= a_1 b_2 \omega(\mathbf{e}_1 , \mathbf{e}_2) + a_1 b_3 \omega(\mathbf{e}_1 , \mathbf{e}_3)+a_2 b_1 \omega(\mathbf{e}_2 , \mathbf{e}_1) \\ &+ a_2 b_3 \omega(\mathbf{e}_2 , \mathbf{e}_3) + a_3 b_1 \omega(\mathbf{e}_3 , \mathbf{e}_1) + a_3 b_2 \omega(\mathbf{e}_3 , \mathbf{e}_2) \\ &=
(a_2 b_3-a_3 b_2) \omega(\mathbf{e}_2, \mathbf{e}_3), + (a_3 b_1-a_1b_3) \omega(\mathbf{e}_3, \mathbf{e}_1) \\ &+ (a_1 b_2-a_2b_1) \omega(\mathbf{e}_1, \mathbf{e}_2), \end{aligned} \end{equation}\]
(1)Let
\[\omega(\mathbf{e}_2 , \mathbf{e}_3) =c_1, \: \omega(\mathbf{e}_3 , \mathbf{e}_1) =c_2 , \: \omega(\mathbf{e}_1 , \mathbf{e}_2) =c_3 \]
With these substitutions we can write (1) as
\[\begin{equation} \begin{aligned} & (a_2 b_3-a_3b_2) c_1 + (a_3 b_1-a_1b_3) c_2 + (a_1 b_2-a_2b_1) c_3 \\ &= det
\left( \begin{array}{ccc} a_1 &b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right) \\ &=det (\mathbf{a} , \mathbf{b}, \mathbf{c})
\end{aligned} \end{equation}\]