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Let $\mathbf{u}$ and $\mathbf{v}$ be two 3D unit vectors. The transform that performs reflection in the plane normal to $\mathbf{u}$ is given by

$$

T_{\mathbf{u}}(\mathbf{x}) = \mathbf{x} – 2(\mathbf{x} \cdot \mathbf{u})\mathbf{u}

$$

and similarly, reflection in the plane normal to $\mathbf{v}$ is performed by

$$

T_{\mathbf{v}}(\mathbf{x}) = \mathbf{x} – 2(\mathbf{x} \cdot \mathbf{v})\mathbf{v}

$$

Let $\theta$ be the angle between $\mathbf{u}$ and $\mathbf{v}$, and let $\mathbf{n}$ be the unit vector in the direction of $\mathbf{u} \times \mathbf{v}$. So, then we know that $\cos\theta = \mathbf{u} \cdot \mathbf{v}$, and $\mathbf{u} \times \mathbf{v} = (\sin\theta)\mathbf{n}$.

The composition of these two reflections is a rotation around $\mathbf{n}$ by an angle of $2\theta$ (I believe), and that rotation is given by Rodrigues’ formula:

$$

R(\mathbf{x}) = (\cos2\theta)\mathbf{x} +

(1 – \cos2\theta)(\mathbf{x} \cdot \mathbf{n})\mathbf{n} +

(\sin 2\theta)(\mathbf{x} \times \mathbf{n})

$$

It seems to me that we ought to be able to prove from first principles that

$$

T_{\mathbf{u}}\big( T_{\mathbf{v}}(\mathbf{x}) \big) = R(\mathbf{x})

$$

I’ve slogged through pages of vector algebra for a few hours, but to no avail. It’s depressing — I used to be good at this stuff, but apparently not any more. I’d like a proof that uses nothing but elementary vector arithmetic, and I’d like it to be coordinate-free, please.

**Edit**

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As a couple of people have mentioned, it seems sensible to work in the $\mathbf{u}\text{-}\mathbf{v}\text{-}\mathbf{n}$ coordinate system. This doesn’t violate my “coordinate free” requirement as long as we don’t start writing out explicit coordinates for $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{n}$. The vector $R(\mathbf{x}) – \mathbf{x}$ should be entirely in the

$\mathbf{u}\text{-}\mathbf{v}$ plane, so all of its $\mathbf{n}$ terms must vanish, and we should be left with an expression that involves only $\mathbf{u}$ and $\mathbf{v}$, which (I hope) will give us the link to the reflections. The algebraic grunt-work involved is what’s giving me trouble.

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Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have …

$$\begin{align}

T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\

R(\mathbf{x}) &= (2c^2-1)\mathbf{x} + 2 s^2 (\mathbf{x}\cdot\mathbf{n}) \mathbf{n} + 2 s c (\mathbf{x}\times\mathbf{n}) \\

&= (2c^2-1)\mathbf{x} + 2 (\mathbf{x}\cdot\mathbf{w})\mathbf{w} + 2 c (\mathbf{x}\times\mathbf{w}) \\

\end{align}$$

Decomposing $\mathbf{x}$ as $p\mathbf{u} + q\mathbf{v} + r \mathbf{w}$, we can get fairly directly …

$$\mathbf{x}\cdot\mathbf{u} = p + q c \qquad \mathbf{x}\cdot\mathbf{v}=pc+q \qquad \mathbf{x}\cdot\mathbf{w}=rs^2 \qquad (\star)$$

$$\mathbf{x}\times\mathbf{w} = \mathbf{x}\times \left(\mathbf{u}\times\mathbf{v}\right) = (\mathbf{x}\cdot\mathbf{v})\mathbf{u}-(\mathbf{x}\cdot\mathbf{u})\mathbf{v} \qquad (\star\star)$$

Then it’s straightforward to show that the difference of the transformations vanishes:

$$\begin{align}

T(\mathbf{x}) – R(\mathbf{x}) &=\mathbf{x}-2(p+qc)\mathbf{u}-2(pc+q)\mathbf{v} + 4c(pc+q)\mathbf{u}\\

&-\left( (2c^2-1)\mathbf{x} + 2 r s^2 \mathbf{w} + 2 c \left( (pc+q)\mathbf{u} – (p+qc)\mathbf{v} \right) \right) \\[6pt]

&= (2-2c^2)\;\mathbf{x} + 2 \left(-p-qc+2pc^2+2qc-pc^2-qc\right)\;\mathbf{u} \\

&+ 2\left(-pc-q+pc+qc^2\right)\mathbf{v} – 2 r s^2 \mathbf{w} \\[6pt]

&= 2 s^2 \left( \mathbf{x} – p\mathbf{u} – q \mathbf{v} – r\mathbf{w} \right) \\[6pt]

&= 0

\end{align}$$

and we conclude that the transformations are equivalent. $\square$

**Edit.** Without jumping immediately to the decomposition of $\mathbf{x}$, we can use the expansion in $(\star\star)$ to write

$$\begin{align}

\frac{T(\mathbf{x})-R(\mathbf{x})}{2s^2} \;\;&=\;\; \mathbf{x}

\;-\; \left( \; \frac{\mathbf{x}.( \mathbf{u} – c \mathbf{v} )}{s^2}\;\mathbf{u}

\;+\; \frac{\mathbf{x}.( \mathbf{v} – c \mathbf{u} )}{s^2}\;\mathbf{v}

\;+\; \frac{\mathbf{x}.\mathbf{w}}{s^2}\;\mathbf{w} \;\right)

\end{align}$$

If you can “see” that the coefficients of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are the components of $\mathbf{x}$ —which would be clear for orthogonal $\mathbf{u}$ and $\mathbf{v}$, for which $c=0$ and $s=1$— then you’re done. If not, note that you can arrive at this insight by solving the dot-product equations $(\star)$ for $p$, $q$, $r$.

As long as you are not obliged to use your formulae for $T_{\bf u}({\bf x})$ and so on, there is a very easy approach to this.

Any reflection in ${\Bbb R}^3$ is represented by an orthogonal matrix of determinant $-1$. Any rotation in ${\Bbb R}^3$ is represented by an orthogonal matrix of determinant $+1$. Multiply two of the former and you get one of the latter. . .

Not sure if that satisfies the requirements in your last sentence though.

Restricted to any plane perpendicular to the line of intersection of the reflection planes, the reflections are a pair of orientation reversing isometries fixing the same point. Their composition is thus an orientation preserving isometry fixing one point: a rotation. To see what the rotation angle must be, track the effect of each rotation on angles around the fixed point.

This argument works for pairs of orthogonal reflections across $(n-1)$ dimensional hyperplanes in $n$ dimensional Euclidean space, since the $(n-2)$ dimensions in the intersection of the planes are frozen out leaving a $2$-d problem.

Reflection operators $P$ and $Q$ on $\mathbb{R}^n$ satisfy

$\qquad P = P^T$

$\qquad Q = Q^T$

$\qquad P^2 = Q^2 = I$

$\Rightarrow$

$\qquad (PQ)^T PQ = Q^T P^T PQ = Q^T(P^T P)Q = Q^TQ = I$

i.e. $PQ$ is orthogonal. It also has determinant 1, so is a

rotation .

(By the way, “*Every* rotation is the result of … an even number of reflections”

— reflection .)

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