Vector product of two vector products

Question

For all vectors u,v,w,z in $ \mathbb R^3$ , how to show that theoretically,

$(u\times v)\times (w\times z)=(z\cdot(u \times v))*w-(w\cdot(u\times v))*z\tag{1}$ and that

$(u\times v)\times (w\times z)=(u\cdot(w\times z))*v-(v\cdot(w\times z))*u\tag{2}$

Why do both equations make sense geometrically?

Answer:- Suppose $\vec{u}=[4,3,2],\vec{v}=[7,8,3],\vec{w}=[5,10,1], \vec{z}=[6,9,11]$

After plugging in these vectors in equation(1) and (2) i got L.H.S.= R.H.S. that is $\vec{r}=[509,1006,141]$ on both sides of equations.

But i don't understand how to prove these two equations theoretically.

I also don't understand how these two equations make sense geometrically?

If any member knows correct answers to both questions, may answer these questions.

Answer 1

In (1), we seek a vector perpendicular to $u\times v$, and hence in the plane $u,\,v$ span, i.e. a linear combination of $u,\,v$, as in (2). Yet it's required by the same logic to be a linear combination of $w,\,z$, as in (1). In each case, each of the vectors required in the linear combination needs a scalar coefficient linear in three vectors, so it must be a scalar triple product to within a multiplicative constant. Antisymmetry under exchanging $u$ with $v$ or $w$ with $z$ determines the right-hand sides of (1), (2) to within a multiplicative constant. It would be surprising if, for example, we also needed a factor of $3$ (although a $-1$ might still take us by surprise). For a full proof to deal with such an overall factor, however, you'd want to start with$$[(u\times v)\times(w\times z)]_i=\underbrace{\epsilon_{ijk}\epsilon_{jlm}}_{\delta_{kl}\delta_{im}-\delta_{km}\delta_{il}}\epsilon_{kno}u_lv_mw_nz_o.$$Edit: $(i\times j)\times j\times k=k\times i=j=1j-0i=(i\cdot j\times k)j-(j\cdot j\times k)i$ validates (2) without tensor notation. Then$$(u\times v)\times(w\times z)=-(w\times z)\times(v\times u)=(z\cdot(u\times v))w-(w\cdot(u\times v))z$$ proves (1).

Answer 2

Disclaimer: this is not intended for the OP, who had satisfactory answers to their question. I'm just posting this as a reference if someone who could be interested stumbles upon this question. It does answer the question though, so I don't think it's off-topic.

My point is that this formula is a little subtly more general than it seems, even for the trained eye. Usually, if $V$ is a $K$-vector space, we would say that the correct generalization of the vector product is just the wedge product $u\wedge v\in \bigwedge^2 V$, where $u,v\in V$. And then we would add that if $\dim(V)=3$ and we fix both an isomorphism $\phi: V\to V^*$ and a nonzero element $\delta\in \bigwedge^3 V$, then using $\delta$ we have $\bigwedge^2 V\simeq V^*$, and using $\phi$ gives $\bigwedge^2 V\simeq V$, which is how we can define $u\times v\in V$. This works in particular when $V$ is a real vector space with a scalar product (which yields $\phi$) and an orientation (which yields $\delta$), which both come for free when $V=\mathbb{R}^3$.

But actually we can make sense of the formula in the question using just $\delta$ (we don't need $\phi$). Indeed, using some nonzero $\delta\in \bigwedge^3 V$, we get two (related, of course) duality isomorphisms $$\Phi: \bigwedge^2 V\to V^*$$ and $$\Psi: V\to \bigwedge^2(V^*)\simeq \left(\bigwedge^2 V\right)^*.$$ To make the final formula closer to the orginal one, I will also write $x\wedge y = \langle x,y\rangle\delta$ if $x\in V$ and $y\in \bigwedge^2V$ (in other words $\langle x,y\rangle=\Psi(x)(y)=\Phi(y)(x)$). Then we have the formulas $$ \Phi(u\wedge v)\wedge \Phi(w\wedge z) = \Psi\left(\langle z,u\wedge v\rangle w - \langle w,u\wedge v\rangle z \right)$$ and $$ \Phi(u\wedge v)\wedge \Phi(w\wedge z) = \Psi\left(\langle u,w\wedge z\rangle v - \langle v,w\wedge z\rangle u \right).$$

For a quick-and-dirty proof, we can write $\delta=e_1\wedge e_2\wedge e_3$ and reduce to the case where $u,v,w,z$ are all some $e_i$.