Can I find $b$ from $a = b \times c$?

Question

So I just started studying the TNB (or Frenet-Serret) frame, where B = T × N. Then my book also goes on to say that T = N × B and N = B × T. Basically, we can find a new valid cross-product equation by "shifting" the terms over the equality sign. My question is, is this property general for all cross products a = b × c, or just something unique to the TNB frame (or any other set of 3 vectors that meet particular criteria - if so, I'd like to know those criteria)?

P.S. New to this site, so might be disregarding a lot of expected niceties. Thanks in advance for your consideration and time!

Answer 1

It is very special indeed. This is the situation when you have a (right-handed) orthonormal set of vectors, i.e., three unit vectors that are perpendicular to one another (taken in the "right-handed" order). [Can you prove that this is a necessary condition for it to work?]

With regard to the title of your question (which seems totally different), the answer is no. You can add any multiple of $\mathbf c$ to $\mathbf b$ without changing the cross product.

Answer 2

Here's the answer: $$ v \times w = sp(v) \cdot w $$ with $$ sp(v) = \begin{bmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{bmatrix} $$ so we have: $$ a = b \times c = -c \times b = -sp(c) \cdot b $$ Now we would look for the inverse, but $$ det(sp(c)) = 0 $$ so no inverse exists