Composition of reflections equal iff $\bf a \perp b$ Given unit vectors $\bf a$ and $\bf b$, let $R_a$ be the reflection across the plane $\bf x\cdot a$ and similarly for $R_b$. Show that $R_aR_b=R_bR_a$ iff $\bf a \perp b$. (where $R_aR_b$ denotes a composition of respective reflections)
Here's the proof that I sketched and was wondering if it suffices?
Say $\bf a\perp b$, then both planes ($\bf x\cdot a$ and $\bf x\cdot b$) are orthogonal and hence they intersect. Since composition of reflections across two distinct intersecting planes is a rotation, so are $R_aR_b$ and $R_bR_a$.The angle of the rotation equals twice the angle between the intersecting planes - $\bf x\cdot a$ and $\bf x\cdot b$ - these two rotations are of equal angles and the result follows.
Now let $R_aR_b=R_bR_a$. Also, let R be an isometry such that $R({\bf 0})=\bf 0$. Then it preserves scalar products and thus 
$$||R_aR_b||^2=||R_bR_a||^2=R_aR_b\cdot R_bR_a=R_a\cdot R_b$$
But then the angle between $R_a$ and $R_b$ is $0^\circ$ and this is possible iff $\bf a \perp b$
 A: Since reflections wrt $a$ and $b$ do not change $(span\{a,b\})^{\perp}$, you may consider the question on the 2D plane, then this is very easy. 
A: The two planes are given as $a\cdot x=0$ and $b\cdot x=0$ (and not as "$x\cdot a$").
Concerning the first part: You have not really used the assumption $a\perp b$. It is true that the composition of reflection across intersecting planes is a rotation by twice the angle between the planes. But saying this one is not really taking care about the sense of this rotation, and this sense is reversed when the two reflections are interchanged. In order to understand what's going on look at the corresponding two-dimensional problem: One line of reflection is the $x$-axis, and the other line is an ascending line through the origin making an angle $\alpha\in\ \bigl]0,{\pi\over2}\bigr[\ $ with the $x$-axis. (This amounts to $a=(0,1,0)$ and $b=(-\sin\alpha,\cos\alpha,0)$ in the setting of your problem.)
Concerning the second part: Here you are way off. Note that $R_a$ and $R_b$ are not vectors, so there is no scalar product between them. I'd say that when you really have understood the first part it will be obvious to you that $R_aR_b\ne R_bR_a$ if $a\not\perp b$.
