Vector Analysis Identity

Question

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.

Answer 1

Hints: Use your known identity to express $\mathbf{(A \cdot B \times C)}$ and $\mathbf{(a \cdot b \times c)}$ as determinants. What is the relationship between the product of two matrices and their determinants? And what two matrices would you need to multiply in order to get $\mathbf A\cdot\mathbf a$ in the upper-right corner?

Answer 2

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.