This page of vector identities lists the following (among many other identities):
$$
(\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}))\,\mathbf{D}= (\mathbf{A}\cdot\mathbf{D} )\left(\mathbf{B}\times\mathbf{C}\right)+\left(\mathbf{B}\cdot\mathbf{D}\right)\left(\mathbf{C}\times\mathbf{A}\right)+\left(\mathbf{C}\cdot\mathbf{D}\right)\left(\mathbf{A}\times\mathbf{B}\right)
$$
which is presumably supposed to hold for vectors $\mathbf{A,B,C,D} \in \Bbb R^3$. Unlike the other identities, this one is given without justification or citation. With this in mind, my questions are:
Is the identity true?(proven in answers below)- Is the identity well-known? Is there a citation that can be used here?
- How can we prove it?
Some answers have been given, but alternate approaches would be interesting to see.
Thank you for your consideration.
Quick thoughts on the problem:
- $\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})$ is a scalar-triple product and can be rewritten as
$$
\det \pmatrix{\mathbf{A}& \mathbf{B} & \mathbf{C}}
$$ - I have a hunch Cauchy-Binet can be applied here somehow
- This amounts to a statement about the map
$$
D \mapsto [(A \times B)(C\cdot D) + (B \times C)(A\cdot D) + (C \times A)(B\cdot D)]
$$ - A proof in Levi-Cevita notation might be quick.
Best Answer
By formula number 8 in the above link, we may derive from $$A\times((B\times C)\times D)=-A\times (D\times(B\times C))$$ $$\Leftrightarrow (A\cdot D)(B\times C)-(A\cdot(B\times C))D=-A\times((D\cdot C)B-(D\cdot B)C),$$ from which the result follows.