Explanation of a cross product result In my book the result $$(u\times v)\cdot(x\times y)=\begin{vmatrix} u\cdot x & v\cdot x \\u \cdot y & v \cdot y\end{vmatrix},$$ where u, v, x and y are arbitrary vectors, is stated (here '$\cdot$' means the dot product and '$\times$' is the cross product). The book very briefly says that this can be easily done by observing that both sides are linear in u, v, x and y.
I know that if I expand and simply the LHS using the components of a vector the result will be true. However, I don't really understand what it means when the book says ' both sides are linear in u, v, x and y ' and how by noticing this fact, makes this relation easier to prove.
Any help will be greatly appreciated.
 A: It means that you can verify the relation just using the standard basis $\{e_1, e_2, e_3 \}$ of three dimensional space. For example you should check 
$(e_1 \times e_2)\cdot (e_2 \times e_3) =0$ which is the same as the right hand side.
A: Suppose we have two linear functions, $f$ and $g$, which agree on all the basis vectors of some space. Then they must agree for every vector on that space, because they are both linear, and a linear function is completely determined by its values on the basis.
In gory detail, suppose that we know that $f(\vec{e_i}) = g(\vec{e_i})$ for each basis vector $\vec{e_i}$. 
Consider some vector $\vec v$.  We can express $\vec v$ a a linear combination of basis vectors, say as $$\vec v = c_1\vec{e_1} + \cdots + c_n\vec{e_n}.$$ Then we know that 
$$\begin{align}
f(\vec v) & = f(c_1\vec{e_1} + \cdots + c_n\vec{e_n})  \\
& = c_1f(\vec{e_1}) + \cdots + c_nf(\vec{e_n}) & \text{(linearity of $f$)} \\
& = c_1g(\vec{e_1}) + \cdots + c_ng(\vec{e_n}) & \text{($f=g$ for basis vectors)} \\
& = g(c_1\vec{e_1} + \cdots + c_n\vec{e_n}) & \text{(linearity of $g$)}\\
& = g(\vec v)
\end{align}
$$
One can similarly show an analogous fact for functions of several variables. For example, if $f(u,v,x,y)$ and $g(u,v,x,y)$ are linear functions of $u, v, x,$ and $y$, and if they agree on all combinations of the basis vectors for some space, then they agree  on every vector in that space.
Now take $f(u,v,x,y) = (u\times v)\cdot(x\times y)$ and $g(u,v,x,y) =\begin{vmatrix} u\cdot x & v\cdot x \\u \cdot y & v \cdot y\end{vmatrix}$. These are easily seen to be linear, or easy to show to be linear if you don't see it, using properties of cross and dot products (for $f$) and of determinants and dot products (for $g$).  So if you can show that they are equal when $u,v,x,$ and $y$ are basis vectors, you are done. And for most choices of basis vectors as arguments, both sides are equal to zero, so this is quick to verify.
