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There is a regular parametric curve $r(t)$ in $\Bbb{R}^3$,and $r'(t)\bullet a=0$,where $a$ is a fixed vector in $\Bbb{R}^3$, show that $r(t)$ is a plane curve. Thanks very much

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You assumed $r(t)=x(t)i+y(t)j+z(t)k$ an smooth function? –  B. S. Aug 28 '13 at 8:15
@BabakS. Yes, $r(t)$ is at least $\in C^3(\Bbb{R}^3)$ –  Laura Aug 28 '13 at 8:21

2 Answers 2

up vote 2 down vote accepted

Let $(a,b,c)$ be an orthonormal basis of $\mathbb{R}^3$ and write $r(t)=x(t)a+y(t)b+z(t)c$. Because $r'(t) \cdot a=0$, we deduce $x'(t)=0$ that is $x(t)$ is constant. Therefore, the curve lies on an affine plane parallel to the plane generated by $(b,c)$.

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Another way to view it: let $L(u) = a\bullet u$, and put $f(t) = L(r(t))$. Then $f'(t) = a \bullet r'(t) = 0$, which shows that $L$ is constant along the curve. In other words, the curve is contained in a level set of $L$, i.e. in a plane with normal vector $a$.

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