Given two linearly independent vectors $A$ and $B$ in $\mathbb R^3$. Let $C=(B\times A)-B$
(a) Prove that $A$ is orthogonal to $B+C$.
(b) Prove that the angle $\theta$ between $B$ and $C$ satisfies $\frac{1}{2}\pi \lt \theta \lt \pi$.
(c) If $||B||=1$ and $||B\times A||=2$, compute the length of $C$.
(a) is easy since $A$ is orthogonal to $B\times A$. However, I can't solve (b) and (c). I tried using the law of cosines and the identity $||A\times B||=||A||||B||\operatorname{sin}\theta$, however I don't know how to show that the angle must satisfy the above condition.
I'd appreciate some help on solving (b) and (c).
Hint: the dot product $B\cdot C$ is $$ B\cdot[B\times A-B]=B\cdot(B\times A)-B\cdot B=-B\cdot B <0 $$