Question about unit normal of a sphere

I have question, I want some one explain how to compute the unit normal of sphere notice, gives $N=\displaystyle \pm \frac{1}{r}(h-a)$ where $a$ is center point of sphere $h(t)$ is unit speed of curve on sphere thank you

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I edited your question in $\LaTeX$, please check that I didn't alter the meaning of the question. – Git Gud Mar 27 '13 at 7:28
@GitGud: It looks right now. – Babak S. Mar 27 '13 at 7:33
you did right this i mean – leena adam Mar 27 '13 at 7:58

I assume you are talking in terms of the Frenet Frame. Supposing you do not what the explain it with equations, this would be my answer:

To understand the Frenet Frame let us make an analogy. Suppose we are flying a plane. We can go forward and backwards, left and right and up and down.

The tangent unit vector by definition describes how much we go up and how much we go down. the Normal N describes how much we go left or right. The Binormal describes how much we go up and down.

Now image flying on a sphere. Let us start of with a curve that is the intersection of the sphere and a plane trought the origin. Zoom in on one instant. At that time you are on the sphere. At the next instant you also have to be on the sphere. The only way to do that is to remain equally far from the center. If you have to stay equally far from a tower with your plane to only way to do that is to constantly turn towards the tower. If not you are flying away from it. So the turning left and right amounts to turning towards the center. Therefore N directs to the center.

The plus minus signs depends on what you call left and what you call right. The actual size of the vector is something you have to compute.

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but how we compute N – leena adam Mar 27 '13 at 11:38
is there any formula to compute the unit normal – leena adam Mar 27 '13 at 11:39
Oops, I forgot to post the link. Frenet-Serret. If you have a curve parametrized such that the speed is always 1, the formula is quite simple. You need formula 1 and 2. – MrOperator Mar 27 '13 at 15:27

The unit normal at the point $p$ on a surface $M$ by definition is perpendicular to the tangent plane $T_p(M)=\{\alpha^{'}(0)|\alpha:(-\epsilon,\epsilon) \to M,\,\alpha(0)=p \}$.

Let $h$ be a unit speed curve $p$, $\langle h(s)-a,h(s)-a\rangle=r^2$. Differentiate both sides, we get $\langle h^{'}(s),h(s)-a\rangle=0$. So $h(s)-a$ is perpendicular to any $h^{'}(s)$, in particular it is normal to the tangent plane. So the normal vector should be parallel to $h(s)-a$. Then we normalize it and get the desired result.

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