# Proving $(\bf x\times y\cdot N)\ z+(y\times z\cdot N)\ x+(z\times x \cdot N)\ y= 0$ when $\bf x,y,z$ are coplanar and $\bf N$ is a unit normal vector

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 $$(\mathbf{x}\times\mathbf{y} \cdot \mathbf{N})\ \mathbf{z} + (\mathbf{y}\times\mathbf{z} \cdot \mathbf{N})\ \mathbf{x} + (\mathbf{z}\times\mathbf{x} \cdot \mathbf{N})\ \mathbf{y}=\mathbf{0}.$$

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 $$\mathbf{N}=\frac{\mathbf{x}\times\mathbf{y}}{\| \mathbf{x}\times\mathbf{y}\|}=\frac{\mathbf{y}\times\mathbf{z}}{\| \mathbf{y}\times\mathbf{z}\|}=\frac{\mathbf{z}\times\mathbf{x}}{\| \mathbf{z}\times\mathbf{x}\|}$$ and substituting into the equation to get $$\| \mathbf{x}\times\mathbf{y}\|z +\| \mathbf{y}\times\mathbf{z}\|\mathbf{x}+\| \mathbf{z}\times\mathbf{x}\|\mathbf{y}=\mathbf{0}$$ 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 $$\sin\theta,\sin\psi,\sin(\theta+\psi)$$ (or $$2\pi -(\theta+\psi)$$ 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 $$\mathbf{z}=\lambda\mathbf{x}+\mu\mathbf{y}$$ will help.

• What does $x \times y \cdot N$ mean? Dot product $(x \times y) \cdot N$ ? – Widawensen Mar 20 '19 at 12:51
• @Widawensen Yes, what else could it mean? – Marc van Leeuwen Mar 20 '19 at 12:53
• @MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes. – Taladris Mar 21 '19 at 2:54

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 $$(\mathbf{x}\times\mathbf{y} \cdot \mathbf{N})\ \mathbf{z} + (\mathbf{y}\times\mathbf{z} \cdot \mathbf{N})\ \mathbf{x} + (\mathbf{z}\times\mathbf{x} \cdot \mathbf{N})\ \mathbf{y}=\mathbf{v}.$$

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.

• Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(\alpha x)=\alpha (Qx)$. – AlephNull Mar 20 '19 at 11:55
• Oh I see, you're talking about the elements, not the terms. I understand the solution now. – AlephNull Mar 20 '19 at 13:42
• By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...") – John Hughes Mar 20 '19 at 16:51

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 $$w\times v =-(v\times w)$$ 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.

• very nice solution! – John Hughes Mar 20 '19 at 16:54
• Indeed, this is very elegant. So my last remark had some significance! – AlephNull Mar 20 '19 at 17:06
• Thank you both :-) – Song Mar 21 '19 at 7:07

Writing $$x=a\hat{i}+b\hat{j},\,y=c\hat{i}+d\hat{j},\,z=e\hat{i}+f\hat{j},\,N=N\hat{k}$$ reduces the sum to $$N((ad-bc)(e\hat{i}+f\hat{j})+(cf-de)(a\hat{i}+b\hat{j})+(be-af)(c\hat{i}+d\hat{j})).$$The $$\hat{i}$$ coefficient is $$N(ade-bce+acf-ade+bce-acf)=0$$. The $$\hat{j}$$ coefficient can be handled similarly.

• I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer. – AlephNull Mar 20 '19 at 13:45

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:

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

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

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

Another approach to the problem uses a formula for triple product.

$$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c}) = \det \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{bmatrix}$$

Then consider determinant

$$\begin{vmatrix} n_1 & x_1 & y_1 & z_1 \\ n_2 & x_2 & y_2 & z_2 \\ n_3 & x_3 & y_3 & z_3 \\ n_1 & x_1 & y_1 & z_1 \end{vmatrix}$$

where columns consist of vectors $$\mathbf{N} ,\mathbf{x},\mathbf{y},\mathbf{z}$$ components (the fourth row repeats the first one).

Of course such determinant equals to $$0$$.
Developing the determinant along the fourth row we obtain:

$$-n_1\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \\ \end{vmatrix} +x_1\begin{vmatrix} n_1 & y_1 & z_1 \\ n_2 & y_2 & z_2 \\ n_3 & y_3 & z_3 \\ \end{vmatrix} -y_1\begin{vmatrix} n_1 & x_1 & z_1 \\ n_2 & x_2 & z_2 \\ n_3 & x_3 & z_3 \\ \end{vmatrix} +z_1\begin{vmatrix} n_1 & x_1 & y_1 \\ n_2 & x_2 & y_2 \\ n_3 & x_3 & y_3 \\ \end{vmatrix}=0$$

from which the formula for the first component of the vector given in the question follows

(the first summand is equal to $$0$$ as the vectors $$\mathbf{x},\mathbf{y},\mathbf{z}$$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)

Similarly the determinants

$$\begin{vmatrix} n_1 & x_1 & y_1 & z_1 \\ n_2 & x_2 & y_2 & z_2 \\ n_3 & x_3 & y_3 & z_3 \\ n_2 & x_2 & y_2 & z_2 \end{vmatrix}$$ and $$\begin{vmatrix} n_1 & x_1 & y_1 & z_1 \\ n_2 & x_2 & y_2 & z_2 \\ n_3 & x_3 & y_3 & z_3 \\ n_3 & x_3 & y_3 & z_3 \end{vmatrix}$$

give the second and the third component of the question vector, equal to $$0$$.