Is the equation presented (that the time-derivative of a given vector in such a scenario is equal to its angular frequency vector cross the vector itself) true in the case of a vector whose origin is(adsbygoogle = window.adsbygoogle || []).push({}); noton the axis of rotation?

The way I'm visualizing this, if we take such a displaced origin but keep it at rest so that it may still act as an inertial reference frame, we would find that the speed measured from this new reference frame would be the same but that the direction may be in the minus frequency cross vector direction in the case of us displacing our origin far enough so that if we were to construct a cylinder using that given circle of rotation, our origin would beexteriorto this cylinder.

Please tell me where my thinking is flawed.

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# I Quick Question: Rate of Change of a Rotating Vector

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