Discussion
Let \(v\) be a unit speed smooth curve in \(\mathbf{R}^{3}\) with tangent \(\bar{t}\), normal \(\bar{n}\), binormal \(\bar{b}\). Then :
Answer: A
\(\bar{t}'\) is orthogonal to \(\bar{t}\) and \(\bar{b}\).
The first Frenet-Serret formula for a unit speed curve is \(\bar{t}' = \kappa\bar{n}\). So \(\bar{t}'\) is a multiple of the normal, and the normal is perpendicular to both the tangent and the binormal.
The orthogonality to \(\bar{t}\) can also be seen directly: \(\bar{t} \cdot \bar{t} = 1\), and differentiating gives \(2\bar{t} \cdot \bar{t}' = 0\).
\(\bar{t}'\) is parallel to \(\bar{n}\), so it is not orthogonal to \(\bar{n}\) (unless \(\kappa = 0\)), which rules out B and C, and D is wrong because it is orthogonal to \(\bar{t}\) and \(\bar{b}\).
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