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Let $\alpha$ be a planar curve in $\mathbb R^3$, contained in plane $P$. Show that its normal vector $N$ at every point is in the plane P also.

The only thing I know is that the torsion is 0, but I don't how to relate it to N. Please give me some idea.

JSCB
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    In my view, it's a bit odd to speak of a vector lying "in a plane". Vectors don't have locations. I'd be inclined to say that the normal vector is "parallel" to the plane $P$, not "in" it. – bubba Feb 02 '14 at 03:33
  • Yes, @bubba, of course, from a pedantic perspective you're correct. We are sloppy with "free vectors." Of course $N$ lies in the subspace parallel to $P$. – Ted Shifrin Feb 02 '14 at 03:58

2 Answers2

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Hint: Write down what it means for an arclength-parametrized curve $\alpha$ to be in a plane, and differentiate, using the Frenet equations.

Ted Shifrin
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  • What if it's not an arc-length-parametrized curve ? – JSCB Feb 02 '14 at 08:20
  • Jason, for all theoretical problems like this, we always assume $\alpha$ is arclength parametrized. Any regular parametrized curve (that is, one with everywhere nonzero speed) can always be so reparametrized. You could also, if you insist, use the chain rule to correct the Frenet equations, but why do all that work? :) – Ted Shifrin Feb 02 '14 at 12:20
  • @TedShifrin Maybe for practical reasons? We theorists know that is practically impossible to determine an arc-length parametrizations even for simple curves such as $(a\cos(t),b\sin(t))$. All we need is the derivation in respect of arc-length $\partial_c$ of a curve $c$ defined by $\partial_c=(\dots)’/|c'|$. – Michael Hoppe Feb 02 '14 at 14:30
  • @MichaelHoppe: Of course, and that's what my differential geometry students have been doing for homework. But, given a theoretical problem, not a specific parametric curve, one should work with an arclength parametrization. – Ted Shifrin Feb 02 '14 at 14:43
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Here's how I would do it:

Let the equation of the plane $P$ be given as

$(\mathbf r - \mathbf r_0) \cdot \nu = 0, \tag{1}$

where $\mathbf r_0 \in P$ is a fixed point, $\nu$ is a unit vector normal to $P$, and $\mathbf r \in P$ is an arbitrary point. Let $\alpha(s)$ be an arc-length parametrized curve lying in $P$; then by (1),

$(\alpha(s) - \mathbf r_0) \cdot \nu = 0, \tag{2}$

and if we differentiate (2) with respect to $s$ we obtain

$\alpha'(s) \cdot \nu = 0 \tag{3}$

since $\mathbf r_0$ and $\nu$ are constants. Since $\alpha(s)$ is parametrized by arc-length, we have $\alpha'(s) = T(s)$, where $T(s)$ is the unit tangent vector to $\alpha$ and a member of the Frenet frame. (3) thus becomes

$T(s) \cdot \nu = 0, \tag{4}$

and if we differentiate again and use the Frenet equation $T'(s) = \kappa N(s)$, where $\kappa$ is the curvature of, and $N$ is the unit normal to, the curve $\alpha(s)$, we see that

$\kappa N(s) \cdot \nu = 0, \tag{5}$

assuming of course $T'(s) \ne 0$, which assumption is vital to the definition of $\kappa$ and the Frenet normal $N$. Under these conditions, $\kappa \ne 0$ and so (5) becomes

$N(s) \cdot \nu = 0, \tag{6}$

showing that $N(s)$ is tangent to, i.e., lies in, the plane $P$. And that's as far as I'm gonna take it.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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