<s>[INST] Vector Analysis Identity One Question given in class was to prove that:
$$\mathbf{(A \cdot B \times C)(a \cdot b \times c)} $$
is equal to
$$\begin{vmatrix} \mathbf A \cdot a & \mathbf A \cdot b & \mathbf A \cdot c \\
\mathbf B \cdot a & \mathbf B \cdot b & \mathbf B \cdot c \\ \mathbf C \cdot a & \mathbf C \cdot b & \mathbf C \cdot c \end{vmatrix}$$ 
I tried to write everything out in components, but that went no where fast. Anywhere to start? I am only allowed to use vector analysis and properties of the dot product and cross product. However, I do know that 
$$\mathbf{(A \cdot B \times C)} $$
is equal to
$$\begin{vmatrix} \mathbf A_x & \mathbf A_y & \mathbf A_z\\
\mathbf B_x & \mathbf B_y & \mathbf B_z\\ \mathbf C_x & \mathbf C_y & \mathbf C_z \ \end{vmatrix}$$ 
any help? It looks like it has some cool geometric properties. 
 [/INST] Just for reference in case someone want to see the actual algebraic manipulation.
In addition to some symmetry relations among dot and cross products
$$p \times q = -q \times p \quad\text{ and }\quad p\cdot(q \times r) = q \cdot( r \times p) = r \cdot( p \times q)$$
The key identities we need are:
$$\begin{align}
p \times (q \times r) &= (p\cdot r) q - (p \cdot q)r &\tag{*1}\\
(p \times q) \cdot ( r \times s) &= (p\cdot r)(q \cdot s) - (p\cdot s)(q \cdot r)& \tag{*2}
\end{align}$$
Using $(*2)$, we can rewrite the determinant at hand as
$$\left|\begin{matrix}
A\cdot a & A\cdot b & A\cdot c\\
B\cdot a & B\cdot b & B\cdot c\\
C\cdot a & C\cdot b & C\cdot c\\
\end{matrix}\right|
= \begin{array}{rl}
  & A\cdot a((B\cdot b)(C\cdot c)- (B\cdot c)(C\cdot b))\\
- & B\cdot a((A\cdot b)(C\cdot c)-(A\cdot c)(C\cdot b))\\
+ & C\cdot a((A\cdot b)(B\cdot c)-(A\cdot c)(B\cdot b))\end{array}
= \begin{array}{rl}
 & A\cdot a((B\times C)\cdot(b \times c))\\
-& B\cdot a((A\times C)\cdot(b \times c))\\
+& C\cdot a((A\times B)\cdot(b \times c))
\end{array}
$$
The determinant now has the form
$$u \cdot (b \times c)
\quad\text{ where }\quad
u = A\cdot a(B\times C)
- B\cdot a (A \times C)
+ C\cdot a (A \times B)$$
Using $(*1)$, we have
$$\begin{align}A\cdot a(B\times C) - B\cdot a(A\times C)
&= ((A\cdot a)B - (B\cdot a)A)\times C\\
&\stackrel{(*1)}{=} -(a \times (A \times B))\times C\\
&= C \times (a \times (A \times B))\\
&\stackrel{(*1)}{=} (C\cdot(A \times B)) a - (C\cdot a)(A\times B)
\end{align}
$$
Adding $(C\cdot a)(A \times B)$ to both sides, we obtain
$$u = (C\cdot (A \times B)) a = (A\cdot(B \times C)) a$$
and results follows.
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