# Vanishing dot product

Why is the following implication true?

$${d\over dt}\vec v(t)=\vec n\times \vec v(t) \implies \vec n\cdot\vec v(t)=\vec n\cdot \vec v(0)$$

I think the result can be obtained by expressing $\vec v(t)$ in a Taylor series expansion in $t$. And from $\displaystyle {d\over dt}\vec v(t)=\vec n\times \vec v(t)$, we see that $\vec n\cdot \vec v^{(n)}(0)=0$ due to the $\times \vec n$ bit.

Is there a more direct way?

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## 1 Answer

$$\vec n\cdot\frac{d\vec v(t)}{dt}=\vec n\cdot(\vec n\times \vec v(t))=0$$ then $$\frac{d}{dt} \vec n\cdot\vec v(t)=0$$ that means $\vec n\cdot\vec v(t)=constant$. The constant can be fixed through $\vec v(0)$ and you are done.

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Thank you, Jon! –  Matthew 1629 Dec 9 '12 at 16:09
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