Bad proof? Derivative of a vector is always normal to vector

Presumably your class was using T to mean a vector-valued function, not just a vector, so it is not the case that T ⋅ T is necessarily a constant. If T ⋅ T does happen to be constant, for example if the value of T is always a unit vector, then the argument works and T' will be normal to T.
T is a vector valued function of time t. T•T' = 0 if T•T is constant with respect to t. But I don't know what you mean by "However, wouldn't this apply to any vector?" and "Likewise, if T was simply the position".

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