$(\vec Ae^{kt} + \vec Be^{-kt})\times (\vec Ae^{kt} - \vec Be^{-kt})$ How do I evaluate this cross product? $$(\vec Ae^{kt} + \vec Be^{-kt})\times (\vec Ae^{kt} - \vec Be^{-kt})$$ where $\vec A, \vec B$ are constant vectors and $k, s$ are constant scalars.
 A: By expansion, knowing that $\vec{A} \times \vec{A}=\vec{B} \times \vec{B}=0$, it is 
$$-\vec{A} \times \vec{B} + \vec{B} \times \vec{A}=-2\vec{A} \times \vec{B}.$$
Remark: you mention in your question a variable $s$ that is not present.
A: Call $\vec{A}e^{kt} = \vec{C}$ say, and  $\vec{B}e^{-kt} = \vec{D}$. Then call the sum and the difference of these two vectors say $\vec{E} = \vec{C} + \vec{D}$ and $\vec{F} = \vec{C}-\vec{D}$. By doing the cross product and substituting you get $-2c_2d_3\vec{i} + 2c_1d_3\vec{j}-2c_1d_2\vec{k}$ , where $\vec{C} = (c_1,c_2,c_3)$ and $\vec{D} = (d_1,d_2,d_3)$.
At this point, substitute, giving a labelling such as $\vec{A} = (a_1,a_2,a_3)$ and $\vec{B} = (b_1,b_2,b_3)$ and you'll find the answer. Remember that you'll have to multiply by the exponentials.
For instance $c_1 = a_1e^{kt} + b_1e^{-kt}$ and $d_3 = a_3e^{kt} - b_3e^{-kt}$
So that the $i$-th component will be $-2(a_1e^{kt} + b_1e^{-kt})(a_3e^{kt} - b_3e^{-kt})= -2 ( a_1a_3e^{2kt} - a_1b_3+b_1a_3-b_1b_3e^{-2kt})$
A: $$\begin{align}
(\vec Ae^{kt} + \vec Be^{-kt})\times (\vec Ae^{kt} - \vec Be^{-kt})&=\color{blue}{(\vec Ae^{kt} + \vec Be^{-kt})\times (\vec Ae^{kt} + \vec Be^{-kt})}\\\\
&-\color{red}{2(\vec Ae^{kt} + \vec Be^{-kt})\times (\vec Be^{-kt})}\\\\
&=\color{blue}{0}-\color{red}{2(\vec A\times\vec B)}
\end{align}$$
