Cross/dot product identity

Question

In order to prove

$[A\times(B\times C)]\cdot \{[B\times (C\times A)] \times [C\times(A\times B)]\}=0$

With the triple product I got

$[(A\cdot C)B -(A\cdot B)C]\cdot \{[(A\cdot B)C-(B\cdot C)A)] \times [(B\cdot C)A-(A\cdot C)B)]\}=0$

Which can be written as

$[J-K] \cdot \{[K-I]\times[I-J]\}=0$, here $J=[(A\cdot C)B]$ and so on

$[I-J] \cdot \{[J-K]\times[K-I]\}=0$

Vector J is a combination of A and C, vector K of A and B, but that minus confused me. I feel that I'm missing a property to continue the proof. What is it?

Answer 1

We can use the two identities (which aren't difficult to prove):

$$\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=\mathbf{B}(\mathbf{A}\cdot\mathbf{C})-\mathbf{C}(\mathbf{A}\cdot\mathbf{B})\tag{1}$$

$$\mathbf{A}\cdot(\mathbf{\mathbf{B}}\times{\mathbf{C}})=\mathbf{\mathbf{C}}\cdot(\mathbf{A}\times{\mathbf{B}})=\mathbf{B}\cdot(\mathbf{C}\times{\mathbf{A}})\tag{2}$$

So we have from $\text{(1)}$:

\begin{equation*} \left(\mathbf{A}\times(\mathbf{B}\times\mathbf{C})\right)\cdot\left[(\mathbf{B}\times(\mathbf{C}\times\mathbf{A}))\times(\mathbf{C}\times(\mathbf{A}\times\mathbf{B}))\right] = \\ \underbrace{\left(\mathbf{B}(\mathbf{A}\cdot\mathbf{C})-\mathbf{C}(\mathbf{A}\cdot\mathbf{B})\right)}_{\mathbf{U}}\cdot\underbrace{\left[\left(\mathbf{C}(\mathbf{B}\cdot\mathbf{A})-\mathbf{A}(\mathbf{B}\cdot\mathbf{C})\right)\times\left(\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-\mathbf{B}(\mathbf{C}\cdot\mathbf{A})\right)\right]}_{\mathbf{V}} \end{equation*}

Expanding the factor denoted $\mathbf{V}$ gives us:

\begin{align*} \require{cancel} \mathbf{V}=(\mathbf{A}\cdot\mathbf{B})(\mathbf{B}\cdot\mathbf{C})\mathbf{C}\times\mathbf{A}-(\mathbf{A}\cdot\mathbf{B})(\mathbf{A}\cdot\mathbf{C})\mathbf{C}\times\mathbf{B}&-\\\cancelto{0}{(\mathbf{B}\cdot\mathbf{C})(\mathbf{B}\cdot\mathbf{C})\mathbf{A}\times\mathbf{A}}+ (\mathbf{B}\cdot\mathbf{C})(\mathbf{A}\cdot\mathbf{C})\mathbf{A}\times\mathbf{B} \end{align*}

When now expanding the dot product $\mathbf{U}\cdot\mathbf{V}$, note that terms like $\mathbf{B}\cdot(\mathbf{C}\times\mathbf{B})$ trivially vanish since they are dot products of a vector with some vector that's perpendicular to it. So avoiding writing such terms in advance, we are left with:

\begin{align*} \mathbf{U}\cdot\mathbf{V} &= (\mathbf{A}\cdot\mathbf{C})(\mathbf{A}\cdot\mathbf{B})(\mathbf{B}\cdot\mathbf{C})\mathbf{B}\cdot(\mathbf{C}\times\mathbf{A})-(\mathbf{A}\cdot\mathbf{B})(\mathbf{B}\cdot\mathbf{C})(\mathbf{A}\cdot\mathbf{C})\mathbf{C}\cdot(\mathbf{A}\times\mathbf{B}) \end{align*}

Observing that the coefficients of both expressions above are equal to $(\mathbf{A}\cdot\mathbf{B})(\mathbf{B}\cdot\mathbf{C})(\mathbf{A}\cdot\mathbf{C})$ and using identity $\text{(2)}$ we obtain that indeed $\mathbf{U}\cdot\mathbf{V}=0 \ \ \ \ \ \Box$.

Answer 2

If you can prove that $I \cdot (J \times K)= J \cdot (K \times I)$ and that adding one of the vectors to another in the triple box product preserves the total, then you have that:

$I \cdot ((J-K) \times (K-I))=I \cdot ((J-K) \times K)=I \cdot (J \times K)$

and

$J \cdot ((J-K) \times (K-I))=J \cdot ((-K) \times K-I)=J\cdot (-K \times -I)=I\cdot (J \times K)$.

Subtracting these gives the desired result.