# Cross sectional curve with helix $\beta$ satisfying $\beta'(t) \cdot u = \cos \theta$. Is $u$ unit magnitude?

Apologies if the title is sort of confusing. This is a HW problem, and I am stuck at exactly the magnitude of a vector.

$$\begin{array}{l}{\text { Problem 1. Let } \beta \text { be a unit speed cylindrical helix with } \beta^{\prime}(t) \cdot u=\cos \theta \text { for fixed } \theta} \\ {\text { and } u . \text { Consider the cross sectional curve }}\end{array}$$

$$\gamma(t)=\beta(t)-t \cos (\theta) u$$

$$\text { (1) Prove that } \gamma \text { lies in a plane orthogonal to } u$$

I claim that $$\gamma'(t) \cdot u = 0$$ (to prove that it is orthogonal), and that claim simplifies to $$\cos \theta - \cos \theta \cdot |u|^2 = 0$$. I obviously want $$u$$ to be a unit vector. $$\beta^{\prime}(t) \cdot u=\cos \theta$$ looks suspiciously like a property inherent to helixes, so how can I show $$u$$ to be of unit magnitude?

• What is your definition of cylindrical helix? Typically, it's that $\beta'\cdot u=\cos\theta$ for a fixed $\theta$ and $u$ unit. – cmk Sep 11 '19 at 20:27
• @cmk I think that is the definition, but I am unsure how I can just conclude that these two $u$ are the same (because the problem does not say that, but am I reading into the question too much?) – runeblaze Sep 11 '19 at 20:48
• I think so. Since they already introduced $u$ earlier in the problem as the one corresponding to the cylindrical helix, I think it's safe to assume that they're talking about the same $u$. – cmk Sep 11 '19 at 20:49
• ok thanks! I read the statement again and now it does seem to be that case. – runeblaze Sep 11 '19 at 22:19

So in the end this seems like I did not read the problem properly. Yes, I can assume $$u$$ to be a unit vector by the phrasing and the definition of cylindrical helixes: see cmk's comments for more details.