How can I prove the following identity:
$$(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d}) = (\vec{c}\cdot\vec{a})(\vec{b}\cdot\vec{d})-(\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{d})$$
$\endgroup$ 32 Answers
$\begingroup$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
$$(\vec{c}\cdot\vec{a})(\vec{d}\cdot\vec{b})-(\vec{c}\cdot\vec{b})(\vec{d}\cdot\vec{a})$$
$\endgroup$ $\begingroup$Here's another way of proving the identity using index notation, Einstein summation convention, and the Levi-Civita symbol's magic.
The $i$-th components of the cross products can be written as,
$$ (\vec{a} \times \vec{b})_i = \epsilon_{ijk}a_jb_k $$
$$ (\vec{c} \times \vec{d})_i = \epsilon_{ipq}c_pd_q $$
The dot product between these cross products is,$$ \begin{align*} (\vec{a} \times \vec{b})_i (\vec{c} \times \vec{d})_i &= \epsilon_{ijk}a_jb_k \epsilon_{ipq}c_pd_q \\ &= \epsilon_{ijk}\epsilon_{ipq} a_jb_k c_pd_q \end{align*} $$
By the properties of the Levi-Civita, we have that,
$$ \epsilon_{ijk}\epsilon_{lpq} = \delta^{ijk}_{lpq} $$
Contracting by setting $i = l$, we obtain the contracted epsilon identity,
$$ \epsilon_{ijk}\epsilon_{ipq} = \delta^{j}_{p}\delta^{k}_{q} - \delta^{j}_{q}\delta^{k}_{p} $$
Making use of this in our dot product,
$$ \begin{align*} (\vec{a} \times \vec{b})_i (\vec{c} \times \vec{d})_i &= (\delta^{j}_{p}\delta^{k}_{q} - \delta^{j}_{q}\delta^{k}_{p}) a_jb_k c_pd_q \\ &= \delta^{j}_{p}\delta^{k}_{q} a_jb_k c_pd_q - \delta^{j}_{q}\delta^{k}_{p} a_jb_k c_pd_q \end{align*} $$
The Kronecker deltas force indices together and eliminate off-diagonals so,
$$ \delta^{j}_{p}\delta^{k}_{q} a_jb_k c_pd_q = \delta^{k}_{q} a_jb_k c_jd_q = a_jb_k c_jd_k $$
$$ \delta^{j}_{q}\delta^{k}_{p} a_jb_k c_pd_q = \delta^{k}_{p} a_jb_k c_pd_j = a_jb_k c_kd_j $$
Therefore our result becomes,
$$ \begin{align*} (\vec{a} \times \vec{b})_i (\vec{c} \times \vec{d})_i &= a_jb_k c_jd_k - a_jb_k c_kd_j \\ &= a_jc_jb_kd_k - a_jd_jb_kc_k \end{align*} $$
By definition of the dot product, we see that the R.H.S contains products of dot products so writing everything in vector notation, we finally have,
$$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) $$
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