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Prove that if $mathbf{x},mathbf{y},mathbf{z} in mathbb{R}^3$ are coplanar vectors and $mathbf{N}$ is a unit normal vector to the plane then $$(mathbf{x}timesmathbf{y} cdot mathbf{N}) mathbf{z} + (mathbf{y}timesmathbf{z} cdot mathbf{N}) mathbf{x} + (mathbf{z}timesmathbf{x} cdot mathbf{N}) mathbf{y}=mathbf{0}.$$

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 $mathbf{N}=frac{mathbf{x}timesmathbf{y}}{| mathbf{x}timesmathbf{y}|}=frac{mathbf{y}timesmathbf{z}}{| mathbf{y}timesmathbf{z}|}=frac{mathbf{z}timesmathbf{x}}{| mathbf{z}timesmathbf{x}|}$ and substituting into the equation to get $| mathbf{x}timesmathbf{y}|z +| mathbf{y}timesmathbf{z}|mathbf{x}+| mathbf{z}timesmathbf{x}|mathbf{y}=mathbf{0}$ 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 $mathbf{z}=lambdamathbf{x}+mumathbf{y}$ 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
$$
(mathbf{x}timesmathbf{y} cdot mathbf{N}) mathbf{z} + (mathbf{y}timesmathbf{z} cdot mathbf{N}) mathbf{x} + (mathbf{z}timesmathbf{x} cdot mathbf{N}) mathbf{y}=mathbf{v}.
$$

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 $x=ahat{i}+bhat{j},,y=chat{i}+dhat{j},,z=ehat{i}+fhat{j},,N=Nhat{k}$ reduces the sum to $$N((ad-bc)(ehat{i}+fhat{j})+(cf-de)(ahat{i}+bhat{j})+(be-af)(chat{i}+dhat{j})).$$The $hat{i}$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hat{j}$ 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, $$begin{align}
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
end{align}$$
and similarly $$begin{align}
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
end{align}$$
So $$begin{align}
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
end{align} $$
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:

$ (mathbf{x}timesmathbf{y}) cdot mathbf{N}) mathbf{z} + (mathbf{y}timesmathbf{z}) cdot mathbf{N}) mathbf{x} + (mathbf{z}timesmathbf{x}) cdot mathbf{N}) mathbf{y} \ =(mathbf{N}timesmathbf{x}) cdot mathbf{y}) mathbf{z} + (mathbf{N}timesmathbf{y}) cdot mathbf{z}) mathbf{x} + (mathbf{N}timesmathbf{z}) cdot mathbf{x}) mathbf{y} $

All cross product vectors $$v_1=(mathbf{N}timesmathbf{x}),v_2=(mathbf{N}timesmathbf{y}), v_3=(mathbf{N}timesmathbf{z})$$

lie in the plane of coplanar vectors $mathbf{x},mathbf{y},mathbf{z}$ and they are vectors $mathbf{x},mathbf{y},mathbf{z}$ rotated by $pi/2$ in this plane.

So we can limit themselves to this plane and take any vectors with components $mathbf{x}=[ x_1 x_2]^T,mathbf{y}=[ y_1 y_2]^T,mathbf{z} =[ z_1 z_2]^T$.

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