Given a nonzero vector $B$ and a vector-valued function $F$ such that $F(t)\cdot B=t$ for all $t$, and such that the angle between $F'(t)$ and $B$ is constant (independent of $t$). Prove that $F''(t)$ is orthogonal to $F'(t)$.
What I have so far is that $F'(t)\cdot B+F(t)\cdot B'=1$, so $F'\cdot B=1$. I need to show $F''\cdot F'=0$. But how do I get here from what I have? I would appreciate any solutions or hints.