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Prove that if $mathbfx,mathbfy,mathbfz in mathbbR^3$ are coplanar vectors and $mathbfN$ is a unit normal vector to the plane then $$(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbf0.$$

This is an elementary identity involving cross products which is used in the proof of the Gauss-Bonnet Theorem and whose proof was left as an exercise. I've tried it unsuccessfully. Initially I tried writing $mathbfN=fracmathbfxtimesmathbfy=fracmathbfytimesmathbfz=fracmathbfztimesmathbfx$ and substituting into the equation to get $| mathbfxtimesmathbfy|z +| mathbfytimesmathbfz|mathbfx+| mathbfztimesmathbfx|mathbfy=mathbf0$ but then I realised these terms are only correct up to $pm$ signs. You could write the norms in terms of sines of angles and divide by norms to get unit vectors with coefficients $sintheta,sinpsi,sin(theta+psi)$ (or $2pi -(theta+psi)$ I suppose) but I don't know what to do from there, especially when the terms are only correct up to sign. Any hints how to prove this identity? Perhaps there is a clever trick to it but I can't see it. Edit: Maybe writing $mathbfz=lambdamathbfx+mumathbfy$ will help.

Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$

You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$

Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$

Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.

Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.

In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.

If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.

Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.

Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.

Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.

If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.

Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.

NB this argument doesn't use any properties of $bf N$.

By the properties of the triple product ( circluar shift) we can rearrange formula:

$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $

All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$

lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.

So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.

Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.

Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$