TITLE: Dot product of two vectors obtained by cross product
QUESTION [7 upvotes]: How can I prove the following identity:
$$(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d}) = (\vec{c}\cdot\vec{a})(\vec{b}\cdot\vec{d})-(\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{d})$$

REPLY [3 votes]: We have the following identities for vectors $\vec{x}$,$\vec{y}$,$\vec{z}$:
$$\vec{x}\cdot(\vec{y}\times\vec{z})=\vec{y}\cdot(\vec{z}\times\vec{x})=\vec{z}\cdot(\vec{x}\times\vec{y})$$
and
$$\vec{x}\times (\vec{y}\times\vec{z})=\vec{y}\;(\vec{x}\cdot \vec{z})-\vec{z}\;(\vec{x}\cdot \vec{y})$$
Let $\vec{a}\times\vec{b} = \vec{x}$:
$$(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=\vec{x}\cdot(\vec{c}\times\vec{d})$$
By using the first identity,
$$\vec{x}\cdot(\vec{c}\times\vec{d}) = \vec{c}\cdot(\vec{d}\times\vec{x})$$
Which can be expanded by using the second identity:
$$\vec{c}\cdot(\vec{d}\times(\vec{a}\times\vec{b}))=\vec{c}\cdot(\vec{a}\;(\vec{d}\cdot\vec{b})-\vec{b}\;(\vec{d}\cdot\vec{a}))$$
Giving us the desired identity
$$(\vec{c}\cdot\vec{a})(\vec{d}\cdot\vec{b})-(\vec{c}\cdot\vec{b})(\vec{d}\cdot\vec{a})$$